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Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 6th Lesson Current Electricity Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 6th Lesson Current Electricity

Very Short Answer Questions

Question 1.
Define mean free path of electron in a conductor.
Answer:
The average distance transversed by an electron during successive collisions in a conductor is called mean free path of electron in a conductor.

Question 2.
State Ohm’s law and write its mathematical form.
Answer:
At constant temperature, the strength of the current (I) in a conductor is directly proportional to the potential difference (V) between its ends.
∴ I ∝ V ⇒ I = \(\frac{\mathrm{V}}{\mathrm{R}}\) ⇒ V = IR (Mathematical form)
where R is constant, it is called the resistance of the conductor.

Question 3.
Define resistivity or specific resistance.
Answer:
Resistivity or specific resistance (ρ) : The resistance of a conductor of unit length and unit area of cross-section is called resistivity.
If l = 1, A = 1 ⇒ ρ = \(\frac{\mathrm{R} \times 1}{1}\) = ρ ⇒ R

Question 4.
Define temperature coefficient of resistance.
Answer:
Temperature coefficient of resistance (α) : The ratio of the change in resistance per 1°C rise in temperature to the resistance at 0°C is called the temperature coefficient of resistance.
α = \(\frac{R_t-R_0}{R_0 t}\)

Question 5.
Under what conditions is the current through the mixed grouping of cells maximum ?
Answer:
The current through the mixed grouping of cells maximum, when

1. Effective emf of all the cells is high.
2. The value of external resistance is equal to the total internal resistance of all the cells.

Question 6.
If a wire is stretched to double its original length without loss of mass, how will the resistivity of the wire be influenced ?
Answer:
Resistivity of the wire remains unchanged as it does not change with change in dimensions of a material without change in its temperature.

Question 7.
Why is manganin used for making standard resistors ?
Answer:
Due to high resistivity and low temperature coefficient of resistance, manganin wire (Cu – 84% + Mn – 12% + Ni – 4%) is used in the preparation of standard resistances.

Question 8.
The sequence of bands marked on a carbon resistor are: Red, Red, Red, Silver. What is its resistance and tolerance ?
Answer:
The resistance of a carbon resistor marked with Red, Red, Red = 22 × 10²Ω = 2.2kΩ = 2200Ω

[∵ Sequence number for Red = 2 and multiplication factor = 10²]
The tolerance of carbon resistor = ± 10%

Question 9.
Write the color code of a carbon resistor of resistance 23 kilo ohms.
Answer:
Color code of a carbon resistor of 23 Kilo Ohms (= 23 × 10³Ω) are Red, Orange, Orange
[∵ Sequence number 2 for Red, 3 for orange, multiplication factor 10³ for orange]

Question 10.
If the voltage V applied across a conductor is increased to 2V, how will the drift velocity of the electrons change ?
Answer:
The drift velocity, V_d = \(\frac{\mathrm{eE}}{\mathrm{m}}\) \(\tau^{\prime}\) = \(\frac{\mathrm{eV}}{\mathrm{mL}} \tau\)
\(\frac{\mathrm{V}_{\mathrm{d}_1}}{\mathrm{~V}_{\mathrm{d}_2}}\) = \(\frac{v_1}{v_2}\)
Here V₁ = V, V₂ = 2V
\(\frac{\mathbf{V}_{\mathrm{d}_1}}{\mathrm{~V}_{\mathrm{d}_2}}\) = \(\frac{V}{2 V}\)
∴ \(\mathrm{V}_{\mathrm{d}_2}\) = \(2 \mathrm{~V}_{\mathrm{d}_1}\)
∴ Drift velocity is increased by twice.

Question 11.
Two wires of equal length, of copper and, manganin, have the same resistance. Which wire is thicker ?
Answer:
R = \(\frac{\rho \mathrm{A}}{l}\) ⇒ A = \(\frac{\mathrm{R} l}{\rho}\)
Since ρ_cu < p_manganin, copper wire is thicker than manganin wire.

Question 12.
Why are household appliances connected in parallel ?
Answer:
In parallel, the voltage (V) across each appliance is same. The current (I) through them depends upon the power (P) of the appliance. The higher power appliance draws more current and lower power appliance draws less current.
(∵ P = VI or I ∝ P)

Question 13.
The electron drift speed in metals is small (~ms^-1) and the charge of the electron is also very small (~10^-19C), but we can still obtain a large amount of current in a metal. Why ?
Answer:
Current through a metal, I = n A eV_d.
A is the area of cross-section of the metal. The electron drift speed, V_d (~10^-5, ms^-1) is small. The charge of electron, e (~1.6 × 10^-19C) is also very small. But we can still obtain a large amount of current in a metal due to presence of large number of free electrons (n) is a conductor (~ 10²⁹ m^-3).

Short Answer Questions

Question 1.
A battery of emf 10V and internal resistance 3Ω is connected to a resistor R.

1. If the current in the circuit is 0.5 A. Calculate the value of R.
2. What is the terminal voltage of the battery when the circuit is closed.

Answer:
Given, E = 10V, r = 3Ω, I = 0.5A, R = ?, V = ?

1. E = I(R + r) or R + r = \(\frac{E}{I}\) = \(\frac{10}{0.5}\) =20Ω ⇒ R = 20 – 3 = 17Ω
2. Terminal voltage, V = IR = 0.5 × 17 = 8.5Ω

Question 2.
Draw a circuit diagram showing how a potentiometer may be used to find Internal resistance of a cell and establish a formula for It.
Answer:
Measurement of internal resistance (r) with potentiometer:

1. Potentiometer to measure internal resistance (r) of a cell (ε) is shown in diagram.
   
2. The cell (emf ε) whose internal resistance (r) is to be determined is connected across a resistance box (RB) through a key K₂.
3. With key K₂ open, balance is obtained at length
   l₁ (AN₁). Then ε = ϕl₁ —– (1)
4. When key K₂ is closed, the cell sends a current (I) through the resistance box (R.B).
5. If V is the terminal potential difference of the cell and balance is obtained at length l₂ (AN₂). Then V = ϕl₂ —– (2)
6. \(\frac{(1)}{(2)}\) ⇒ \(\frac{\varepsilon}{\mathrm{V}}\) = \(\frac{l_1}{l_2}\) —- (3)
7. But ε = I(r + R) and V = IR. This gives
   \(\frac{\varepsilon}{\mathrm{V}}\) = \(\frac{(\mathrm{r}+\mathrm{R})}{\mathrm{R}}\)
   \(\frac{I_1}{I_2}\) = \(\left(\frac{r}{R}+1\right)\) [∵ from (3)]
   ∴ r = R\(\left(\frac{l_1}{l_2}-1\right)\)

Question 3.
Derive an expression for the effective resistance when three resistors are connected in
(i) series
(ii) parallel. (T.S. Mar. ’19)
Answer:
Effective resistance when three resistors are connected:
(i) In series:

1. Three resistors R₁, R₂ and R₃ are connected in series as shown in fig. V₁, V₂, V₃ are the potential differences across R₁, R₂ and R₃. I is the current flowing through them.
2. Applying Ohm’s law to R₁, R₂ and R₃, Then V₁ = IR₁, V₂ = IR₂, V₃ = IR₃
3. In series, V = V₁ + V₂ + V₃
   IR_S = IR₁ + IR₂ + IR₃ [∵ V = IR_S]
   ∴ R_S = R₁ + R₂ + R₃

ii) In parallel:

1. Three resistors, R₁, R₂ and R₃ are connected in parallel as shown in fig. Potential differences across each resistor is V. I₁, I₂, I₃ are the currents flowing through them.
2. Applying Ohmes law to R₁, R₂ and R₃, then
   V = I₁R₁ = I₂R₂ = I₃R₃
   ⇒ I₁ = \(\frac{\mathrm{V}}{\mathrm{R}_1}\), I₂ = \(\frac{\mathrm{V}}{\mathrm{R}_2}\), I₃ = \(\frac{\mathrm{V}}{\mathrm{R}_3}\)
3. In parallel, I = I₁ + I₂ + I₃
   ⇒ \(\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{p}}}\) = \(\frac{\mathrm{V}}{\mathrm{R}_1}\) + \(\frac{\mathrm{V}}{\mathrm{R}_2}\) + \(\frac{\mathrm{V}}{\mathrm{R}_3}\) [∵ I = \(\frac{V}{R_p}\)]
   ∴ \(\frac{1}{\mathrm{R}_{\mathrm{p}}}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\)

Question 4.
‘m’ cells each of emf E and internal resistance ‘r’ are connected in parallel. What is the total emf and internal resistance ? Under what conditions is the current drawn from mixed grouping of cells a maximum ?
Answer:
Cells in parallel:

1. When ‘m’ identical cells each of emf ‘ε’ and internal resistance r are connected to the external resistor of resistance R as shown in fig., then the cells are connected in parallel.
2. As the cells are connected in parallel, their equivalent internal resistance r_p is given by
   \(\frac{1}{\mathrm{r}_{\mathrm{p}}}\) = \(\frac{1}{\mathrm{r}}\) + \(\frac{1}{\mathrm{r}}\) + ….. upto m terms = \(\frac{\mathrm{m}}{\mathrm{r}}\)
   ∴ r_p = \(\frac{\mathrm{r}}{\mathrm{m}}\)
   
3. As R and r_p are in series, so total resistance in the circuit = R + \(\frac{\mathrm{r}}{\mathrm{m}}\).
4. In parallel combination of identical cells, the effective emf in the circuit is equal to the emf due to a single cell, because in this combination, only the size of the electrodes increases but not emf.
5. Therefore, current in the resistance R is given by I = \(\frac{\varepsilon}{\mathrm{R}+\frac{\mathrm{r}}{\mathrm{m}}}\) = \(\frac{\mathrm{m} \varepsilon}{\mathrm{m} R+\mathrm{r}}\)
6. When the external resistance is negligible is comparison to the internal resistance (R<<r), the current drawn from mixed grouping of cells a maximum.

Question 5.
Define electric resistance and write it’s SI unit. How does the resistance of a conductor vary if
(a) Conductor is stretched to 4 times of it’s length
(b) Temperature of a conductor is increased.
Answer:
Electric resistance (R) : The resistance offered by a flow of electrons in a conductor is called electric resistance.
S.l unit of resistance is ohm (Ω).
The resistance of a conductor
R = \(\frac{\rho l}{\mathrm{~A}}\) = \(\frac{\rho l^2}{\mathrm{~V}}\) ⇒ R ∝ l²
a) in first case, R₁ = R, l₁ = l
b) In second case, l₂ = 4l, R₂ =?
\(\frac{R_2}{R_1}\) = \(\frac{l_2^2}{l_1^2}\) ⇒ \(\frac{R_2}{R}\) = \(\left(\frac{4l}{l_4}\right)^2\) ∴ R₂ = 16R

b) Variation of Resistance with temperature is given by R_t = R₀ (1 + α t)
If temperature increases, resistance also increases.

Question 6.
When the resistance connected In series with a cell is halved, the current is equal to or slightly less or slightly greater than double. Why?
Answer:
‘When he resistance R is connected to cell of emf, ε in series, the current is given by
I = \(\frac{\varepsilon}{\mathrm{R}+\mathrm{r}}\)

where r is internal resistance of cell.
When the resistance is halved \(\left(\frac{\mathrm{R}}{2}\right)\), the current flows through the circuit is I’ = \(\frac{\varepsilon}{\frac{R}{2}+r}\)

1. If r is negligible comparison with \(\frac{\mathrm{R}}{2}\), I¹ = \(\frac{2 \varepsilon}{R}\)
   ∴ I¹ = 2 I [∵ \(\frac{\varepsilon}{R}\) also equal to 1]
2. If r < < \(\frac{\mathrm{R}}{2}\), the current I¹ is slightly greater than 2.
3. If r is just slightly greater than R, the current (I¹) is slightly less than 2.

Question 7.
Two cells of emfs 4.5V and 6.0V and infernal resistance 6Ω and 3Ω respectively have their negative terminals joined by a wire of 18Ω and positive terminals by a wire of 12Ω resistance. A third resistance wire of 24Ω connects middle points of these wires. Using Kirchhoffs laws, find the potential difference at the ends of this third wire.
Answer:

1. Let the currents through the various arms, of the network be as shown in fig:
2. Applying KVL to closed mesh ABCDA, we have
   4.5 – 6I₁ – 18I₁ – 24 (I₁ + I₂) = 0
   ⇒ 48I₁ + 24I₂ = 4.5 —–> (i)
3. For a closed mesh CDEFC, we have
   – 24(I₁ + I₂) – 12I₂ + 6 – 3I₂ = 0
   24I₁ + 39I₂ = 6 —–> (ii)
   
4. (ii) × 2 – (i) ⇒
   
   ∴ I₂ = \(\frac{7.5}{54}\) = 0.139 A —–> (iii)
   Substituting (iii) in (i), we get
   48I₁ + 78 × 0.139 = 12
   48I₁ = 12 – 10.84 = 1.158
   I₁ = \(\frac{1.158}{48}\) = 0.024 A
5. Potential difference across third wire = (I₁ + I₂) × 24 = 0.163 × 24 = 3.912 Volt.

Question 8.
Three resistors each of resistance 10 ohm are connected, in turn, to obtain
(i) minimum resistance
(ii) maximum resistance. Compute
(a) The effective resistance in each case
(b) The ratio of minimum to maximum resistance so obtained.
Answer:
Given, Resistance of each resistor R = 10Ω, no. of resistors, n = 3
i) If three resistors are connected in parallel, we get minimum resistance.
∴ Minimum resistance R_min = R_p = \(\frac{\mathrm{R}}{\mathrm{n}}\) = \(\frac{10}{3} \Omega\) = 3.33Ω

ii) If three resistors are connected in series, we get maximum resistance.
∴ Maximum resistance R_max = R_s = n R = 3 × 10 = 30Ω

a) The effective resistance to get minimum resistance,
R_eff = \(\frac{\mathrm{R}}{\mathrm{n}}\) = \(\frac{10}{3}\) = 3.33Ω (In parallel)
The effective resistance to get maximum resistance
R_eff = n R = 3 × 10 = 30Ω (In series)

b) \(\frac{R_{\min }}{R_{\max }}\) = \(\frac{\left(\frac{10}{3}\right)}{(3 \times 10)}\) = \(\frac{10}{90}\) ∴ \(\frac{\mathrm{R}_{\min }}{\mathrm{R}_{\max }}\) = \(\frac{1}{9}\)

Question 9.
State Kirchhoffs law for an elêctrical network. Using these laws deduce the condition for balance in a Wheatstone bridge. (Ã.P. Mar. 19, ‘16 & Mar. 14)
Answer:
1) Kirchhoff s first law (Junction rule or KCL) : The algebraic sum of the currents at any junction is zero. ∴ ΣI = 0
(or)
The sum of the currents flowing towards a junction is equal to the sum of currents away from the junction.

2) Kirchhoffs second law (Loop rule or KVL): The algebraic sum of potential around any closed loop is zero.
∴ Σ(IR) + ΣE = 0
Wheatstone bridge : Wheatstone’s bridge circuit consists of four resistances R₁, R₂, R₃ and R₄ are connected to form a closed path. A cell of emf e is connected between the point A and C and a galvanometer is connected between the points B and D as shown in fig. The current through the various branches are indicated in the figure. The current through the galvanometer is I_g and the resistance of the galvanometer is G.
Applying Kirchhoffs first law at the junction D, I₁ – I₃ – I_g = 0 —— (1)
at the junction B, I₂ + I_g – I₄ = 0 —— (2)
Applying Kirchhoffs second law to the closed path ADBA,
-I₁R₁ + I₂R₂ = 0
or
⇒ I₁R₁ + I_gG = I₂R₂ —– (3)
applying kirchhoffs second law to the closed path DCBD,
-I₃R₃ + I₄R₄ + I_gG = 0
⇒ I₃R₃ – I_gG = I₄R₄ —– (4)
When the galvanometer shows zero deflection the points D and B are at the same potential. So I_g = 0.

Substituting this value in (1), (2), (3) and (4).
I₁ = I₃ ——- (5)
I₂ = I₄ —— (6)
I₁R₁ = I₂R₂ —– (7)
I₃R₃ = I₄R₄ —– (8)
Dividing (7) by (8)
\(\frac{\mathrm{I}_1 \mathrm{R}_1}{\mathrm{I}_3 \mathrm{R}_3}\) = \(\frac{I_2 R_2}{I_4 R_4}\) ⇒ \(\frac{R_1}{R_3}\) = \(\frac{R_2}{R_1}\) [∵ I₁ = I₃ & I₂ = I₄]
∴ Wheatstone’s Bridge principle : R₄ = R₃ × \(\frac{\mathrm{R}_2}{\mathrm{R}_1}\)

Question 10.
State the working principle of potentiometer. Explain with the help of circuit diagram how the emf of two primary cells are compared by using the potentiometer. (T.S. Mar. 19 & A.P. Mar. 16)
Answer:
Working principle of potentiometer : The potential difference across a length of the potentiometer wire is directly proportional to its length (or) when a steady current is passed through a uniform wire, potential drop per unit length or potential gradient is constant,
i.e. ε ∝ l ⇒ ε = ϕl where ϕ is potential gradient.
Comparing the emf of two cells ε₁ and ε₂ :

1. To compare the emf of two cells of emf E₁ and E₂ with potentiometer is shown in diagram.
   
2. The points marked 1, 2, 3 form a two way key.
3. Consider first a position of the key where 1 and 3 are connected so that the galvanometer is connected to ε₁.
4. The Jockey is moved along the wire till at a point N₁ at a distance l₁ from A, there is no deflection in the galvanometer. Then ε₁ ∝ l₁ ⇒ ε₁ = ϕl₁ —– (1)
5. Similarly, if another emf ε₂ is balanced against
   l₂ (AN₂) then ε₂ ∝ l₂ ⇒ ε₂ = ϕl₂ —— (2)
6. \(\frac{(1)}{(2)}\) ⇒ \(\frac{\varepsilon_1}{\varepsilon_2}\) = \(\frac{l_1}{l_2}\)

Question 11.
State the working principle of potentiometer explain with the help of circuit diagram how the potentiometer is used to determine the internal resistance of the given primary cell. (A.P. & T.S. Mar. ’15)
Answer:
Working principle of potentiometer : The potential difference across a length of the potentiometer wire is directly proportional to its length (or) when a steady current is passed through a uniform wire, potential drop per unit length or potential gradient is constant.
i.e. E ∝ l ⇒ e = ϕl
where ϕ is potential gradient.

Measurement of internal resistance (r) with potentiometer:

1. Potentiometer to measure internal resistance (r) of a cell (ε) is shown in diagram.
2. The cell (emf ε) whose internal resistance (r) is to be determined is connected across a resistance box (R.B) through a key K₂.
   
3. With key K₂ open, balance is obtained at length l₁(AN₁). Then ε = ϕl₁ ——-> (1)
4. When key K₂ is closed, the cell sends a current (T) through the resistance box (R.B).
5. If V is the terminal potential difference of the cell and balance is obtained at length l₂ (AN₂).
   Then V = ϕl₂ ——> (2)
6. \(\frac{(1)}{(2)}\) ⇒ \(\frac{\varepsilon}{\mathrm{V}}\) = \(\frac{l_1}{l_2}\) —– (3)
7. But ε = I (r + R) and V = IR. This gives
   \(\frac{\varepsilon}{V}\) = \(\frac{(\mathrm{r}+\mathrm{R})}{\mathrm{R}}\)
   \(\frac{l_1}{l_2}\) = \(\left(\frac{r}{R}+1\right)\) [∵ from (3)]
   ∴ r = \(\mathrm{R}\left(\frac{l_1}{l_2}-1\right)\)

Question 12.
Show the variation of current versus voltage graph for GaAs and mark the
(i) Non-linear region
(ii) Negative resistance region.
Answer:
The relation between V and I is not unique. That is, there is more than one value of V for the same current I. material exhibiting such behaviour is GaAs (i.e., a light emitting diode).

Question 13.
A student has two wires of iron and copper of equal length and diameter. He first joins two wires in series and passes an electric current through the combination which increases gradually. After that he joins two wires in parallel and repeats the process of passing current. Which wire will glow first in each case ?
Answer:
1) In series combination, there will be same current through Iron and as well as copper wire. Since the rate of heat production, P = I² R or P ∝ R (for the given value of I). The resistance of Iron wire is more than that of copper for the given length and diameter. Hence in Iron wire, the rate of heat production increases gradually. In series combination Iron will glow first.

2) In parallel combination of Iron and copper wire, there will be same P.D (V) across them.
Since the rate of heat production, P = \(\frac{\mathrm{V}^2}{\mathrm{R}}\) or P ∝ \(\frac{1}{R}\) (for the given value of V). The resistance of Iron is more than that of copper for the given length and diameter. Hence in copper wire, the rate of heat production is more.
In parallel combination copper will glow first.

Question 14.
Three identical resistors are connected in parallel and total resistance of the circuit is R/3. Find the value of each resistance.
Answer:
Given three resistances are identical.
Hence R₁ = R₂ = R₃ = x (say)
Total resistance in parallel, R_p = \(\frac{\mathrm{R}}{3}\)
If three identical resistances are connected in parallel, then
\(\frac{1}{R_p}\) = \(\frac{1}{\mathrm{R}_1}\) + \(\frac{1}{\mathrm{R}_2}\) + \(\frac{1}{\mathrm{R}_3}\)
\(\frac{1}{\left(\frac{\mathrm{R}}{3}\right)}\) = \(\frac{1}{x}\) + \(\frac{1}{x}\) + \(\frac{1}{x}\) ⇒ \(\frac{3}{\mathrm{R}}\) = \(\frac{1+1+1}{\mathbf{x}}\)
∴ x = R.

Long Answer Questions

Question 1.
Under what condition is the heat produced in an electric circuit
a) directly proportional
b) inversely proportional to the resistance of the circuit ?
Compute the ratio of the total quantity of heat produced in the two cases.
Answer:
Expression of heat produced by electric current:
Consider a conductor AB of resistance R.

Let V = P.D applied across the ends of AB.
I = current flowing through AB.
t = time for which the current is flowing.
∴ Total charge flowing from A to B in time t is q = It. By definition of P.D, work done is carrying unit charge from A to B = V
Total work done in carrying a charge q from A to B is
W = V × q = V It = I² Rt
(∵ V = IR)
This work done is called electric work done. If this electric work done appears as heat, then amount of heat produced (H) is given by H = W = I² Rt Joule.
This is a statement of Joule’s law of heating.
a) If same current flows through an electric circuit, heat is developed.
i.e., H ∝ R.

b) If same P.D applied across the the electric circuit heat is developed.
i.e., H₂ ∝ \(\frac{1}{\mathrm{R}}\).

c) The ratio of H₁ and H₂ is given by
\(\frac{\mathrm{H}_1}{\mathrm{H}_2}\) = \(\frac{\mathrm{R}}{\frac{1}{\mathrm{R}}}\)
∴ \(\frac{\mathrm{H}_1}{\mathrm{H}_2}\) = R²

Question 2.
Two metallic wires A and B are connected in parallel. Wire A has length L and radius r, wire B has a length 2L and radius 2r. Compute the ratio of the total resistance of the parallel combination and resistance of wire A.
Answer:
1) For metal Wire ‘A’
Length = L
Radius = r
Area = πr²
Resistance, R_A = \(\frac{\rho_{\mathrm{A}} \mathrm{L}}{\pi \mathrm{r}^2}\) —– (i)
Where ρ_A is specific resistance.
For metal Wire ‘B’
Length = 2L
Radius = 2r
Area = π(2r)² = 4πr²
Resistance, R_B = \(\frac{\rho_{\mathrm{B}}(2 \mathrm{~L})}{4 \pi \mathrm{r}^2}\) = \(\frac{\rho_{\mathrm{B}} \mathrm{L}}{2 \pi \mathrm{r}^2}\) —– (ii)
Where ρ_B is specific resistance.

2) Total resistance of wire A and wire B in parallel combination is given by

3)

4) The ratio of the total resistance parallel combination to resistance of wire A, is given by

5)
∴ \(\frac{R_p}{R_A}\) = \(\frac{\rho_{\mathrm{B}} \pi \mathrm{r}^2}{\mathrm{~L}\left(2 \rho_{\mathrm{A}}+\rho_{\mathrm{B}}\right)}\)

Question 3.
In a house three bulbs of 100W each are lighted for 4 hours daily and six tube lights of 20W each are lighted for 5 hours daily and a refrigerator of 400W is worked for 10 hours daily for a month of 30 days. Calculate the electricity bill if the cost of one unit is Rs. 4.00.
Answer:
No. of bulbs in a house, N = 3
Rated power on each bulb, P = 100 W
Time of lighted t = 4H
Energy consumption of 3 bulbs per day = \(\frac{\mathrm{Npt}}{1000}\) KWH
Energy consumption of 3 bulbs for 30 days = \(\frac{30 \mathrm{~N} \mathrm{Pt}}{1000}\) KWH
E_B = \(\frac{30 \times 3 \times 100 \times 4}{1000}\) = 36KWH
Similarly, Energy consumption of 6 tube lights for 30 days, ET = \(\frac{30 \times 6 \times 20 \times 5}{1000}\) KWH = 18 KWH.
And similarly, Energy consumption of one Refrigerator for 30 days,
E_R = \(\frac{30 \times 400 \times 10}{1000}\)KWH = 120 KWH
∴ The total energy consumption, E = E_B + E_T + E_R = (36 + 18 + 120) KWH
∴ E = 174 KWH = 174 units [∵ 1KWH = 1 unit]
Cost of 1 unit = Rs. 4.00/-
Cost of 174 units = No. of units × cost of 1 unit
= 174 × 4 = Rs. 696/-
∴ Electricity bill for one month of that house = Rs. 696/-

Question 4.
Three resistors of 4 ohms, 6 ohms and 12 ohms are connected in parallel. The combination of above resistors is connected in series to a resistance of 2 ohms and then to a battery of 6 volts. Draw a circuit diagram and calculate.
a) Current in main circuit.
b) Current flowing through each of the resistors in parallel.
c) p.d and the power used by the 2 ohm resistor.
Answer:
Circuit diagram for the given data is

a) Effective resistance when R₁, R₂ and R₃ are connected in parallel is given by
\(\frac{1}{R_p}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\) ⇒ \(\frac{1}{R_p}\) = \(\frac{1}{4}\) + \(\frac{1}{6}\) + \(\frac{1}{12}\)
⇒ \(\frac{1}{\mathrm{R}_{\mathrm{p}}}\) = \(\frac{3+2+1}{12}\) = \(\frac{6}{12}\) = \(\frac{1}{12}\)
∴ R_p = 2Ω
Total resistance in the circuit R = R_p + R₄ = 2 + 2 = 4Ω
∴ Current in mam circuit I = \(\frac{\mathrm{V}}{\mathrm{R}}\) = \(\frac{6}{4}\) = 1.5A

b) Current flowing through R₁, I₁ = \(\frac{I R_P}{R_1}\) = \(\frac{1.5 \times 2}{4}\) = 0.75A
Current flowing through R₂, I₂ = \(\frac{\mathrm{IR}_{\mathrm{P}}}{\mathrm{R}_2}\) = \(\frac{1.5 \times 2}{6}\) = 0.5A
Current flowing through R₃, I₃ = \(\frac{\mathrm{IR}_{\mathrm{P}}}{\mathrm{R}_3}\) = \(\frac{1.5 \times 2}{12}\) = 0.25A

c) RD across 2Ω resistor (i.e., R₄), V₄ = IR₄ = 1.5 × 2 = 3 Volt.
Power used by 2Ω resistor, P = V₄I = 3 × 1.5 = 4.5 W.

Question 5.
Two lamps, one rated 100 Ω at 220 V and the other 60W at 220 V are connected in parallel to a 220 volt supply. What current is drawn from the supply line?
Answer:
Data for

Since R₁ and R₂ are connected in parallel effective resistance

Question 6.
Estimate the average drift speed of conduction electrons in a copper wire of cross – sectional area 3.0 × 10^-7 m² carrying a current of 5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 10³ kg/m³ and its atomic mass is 63.5 u.
Answer:
Given, Cross-sectional area of copper wire, A = 3 × 10^-7m²
carrying current of copper, I = 5A
Charge of electron, e = 1.6 × 10^-19C
Density of conduction electrons = No. of atoms per cubic meter,

∴ Average drift speed of conduction electrons.
V_d = \(\frac{\mathrm{I}}{\mathrm{neA}}\) = \(\frac{5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 3 \times 10^{-7}}\)
⇒ V_d = \(\frac{5}{8.5 \times 1.6 \times 3 \times 10^2}\) = 0.1225 × 10^-2m/s
∴ V_d = 1.225 mm/s

Question 7.
Compare the drift speed obtained above with
i) Thermal speed of copper atoms at ordinary temperatures.
ii) Speed of propagation of electric field along the conductor which causes the drift motion.
Answer:
i) At a temperature T, the thermal speed of a copper atom of mass M is obtained from

∴ drift speed of electron (V_d) = 1.047 × 10^-8
= 10^-8 times of thermal speed at ordinary temperature.

ii) The electric field travels along conductór with speed of EMW
C = 3 × 10⁸ m/s
V_d = 1.225 × 10^-3m/s
\(\frac{\mathrm{V}_{\mathrm{d}}}{\mathrm{C}}\) = \(\frac{1.225 \times 10^{-3}}{3 \times 10^8}\)
V_d = 0.408 × 10^-11 C
∴ Drift speed is, in compansion of C, extremely smaller by a factor of 10^-11.

Problems

Question 1.
A 10Ω thick wire is stretched so that its length becomes three times. Assuming that there Is no change in its density on stretching, calculate the resistance of the stretched wire.
Solution:
Given R₁ = 10Ω,
l₁ = 1
l₂ = 3l, R₂ ?

R₁ = \(\frac{\rho}{\mathrm{V}} l_1^2\)
R₂ = \(\frac{\rho}{\mathrm{V}} l_2^2\)
R₃ = \(\left(\frac{l_2}{l_1}\right)^2\) ⇒ \(\frac{\mathrm{R}_2}{10}\) = \(\left(\frac{3 l}{l}\right)^2\)
∴ R₂ = 10 × 9 = 90Ω.

Question 2.
A wire of resistance 4R is bent in the form of a circle. What is the effective resistance between the ends of the diameter? (A.P. Mar. ’19 & T.S. Mar. ’16, Mar. ’14)
Solution:
Resistance of long wire = 4R
Hence the resistance of half wire =
\(\frac{4 R}{2}\) = 2R

Now these two wire are connected in parallel. Hence the effective resistance between the ends of the diameter
R_P = \(\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}\) ⇒ R_p = \(\frac{2 \mathrm{R} \times 2 \mathrm{R}}{2 \mathrm{R}+2 \mathrm{R}}\)
∴ R_p = R.

Question 3.
Find the resistivity of a conductor which carries a current of density of 2.5 × 10⁶A m^-2 when an electric field of 15 Vm^-1 is applied across it.
Solution:
Given current density
J = \(\frac{1}{A}\) = 2.5 × 1o⁶ Am^-2
Applied electric field E = 15Vm^-1
Resistivity of conductor,
ρ = \(\frac{E}{J}\) = \(\frac{15}{2.5 \times 10^6}\)
∴ ρ = 6 × 10^-6Ωm.

Question 4.
What is the color code for a resistor of resistance 350mΩ with 5% tolerance ?
Solution:
Resistance of resistor = 350mΩ with 5% tolerance
= 350 × 10^-3Ω
= 35 × 10^-2Ω
First significant figure (3) indicates 1^st band
Second significant figure (5) indicates 2^nd band
Third significant figure (10^-2) indicates 3^rd band
We know that
0 1 2 3 4 5 6 7 8
B B R O Y of Great Britian has Very Good Wife wearing Gold silver Necklace
9 10^-1 ← 10^-2
3 indicates orange
5 indicates green
10^-2 indicates Silver
5% tolerance substance is Gold.
∴ Color code of given resistor is orange, green, silver, gold.

Question 5.
You are given 8Ω resistor. What length of wire of resistivity 120 Ωm should be joined in parallel with it to get a value of 6Ω ?
Solution:
Given, Resistance of resistor R = 8Ω

Resisty of wire ρ = 120
Let l length of the resistance x is to be connected to get effective resistance,
R_p = 6Ω
Then \(\frac{1}{\mathrm{R}}\) + \(\frac{1}{x}\) = \(\frac{1}{\mathrm{R}_{\mathrm{p}}}\)
\(\frac{1}{8}\) + \(\frac{1}{x}\) = \(\frac{1}{6}\) ⇒ \(\frac{1}{x}\) = \(\frac{1}{6}\) – \(\frac{1}{8}\) = \(\frac{2}{48}\)
∴ x = 24Ω
And x l = ρ
24 l = 120
∴ l = 5m

Question 6.
Three resistors 3Ω, 6Ω and 9Ω are connected a battery. In which of them will the power dissipation be maximum if:
a) They all are connected in parallel
b) They all are connected in series ? Give reasons.
Solution:
Given R₁ = 3Ω, R₂ = 6Ω, R₃ = 9Ω
a) Effective resistance in parallel is given by
\(\frac{1}{R_p}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\) = \(\frac{1}{3}\) + \(\frac{1}{6}\) + \(\frac{1}{9}\)
\(\frac{1}{R_p}\) = \(\frac{6+3+2}{18}\)
∴ R_P = \(\frac{18}{11} \Omega\)

∴ Dissipated power in parallel,
P_P ∝ \(\frac{1}{\mathrm{R}_{\mathrm{P}}}\) ⇒ P_P ∝ \(\frac{1}{\left(\frac{18}{11}\right)}\) ∴ P_P ∝ \(\frac{11}{18}\) —– (1)

b) Effective resistance in series is given by R_s = R₁ + R₂ + R₃ = 3 + 6 + 9 = 18Ω
∴ Dissipated power in series,
P_S ∝ R_S ⇒ P_S ∝ 18 —- (2)
From equations (1) and (2) power dissipation is maximum in series and minimum in parallel.

Reasons:

1. In series connection, P ∝ R and V ∝ R. Hence dissipated power (P) and potential difference (V) is more because current is same across each resistor.
2. In parallel connection, P ∝ \(\frac{1}{R}\) and I ∝ \(\frac{1}{R}\). Hence dissipated power (P) and potential difference (V) is less because voltage is same across each resistor.

Question 7.
A silver wire has a resistance of 2.1Ω at 27.5°C and a resistance of 2.7Ω at 100°C. Determine the temperature coeff. of resistivity of silver.
Solution:
For silver wire, R₁ = 2.1Ω, t₁ = 27.5°C
R₂ = 2.7Ω, t₂ = 100°C, α = ?
α = \(\frac{R_2-R_1}{R_1 t_2-R_2 t_1}\) = \(\frac{2.7-2.1}{2.1 \times 100-2.7 \times 27.5}\)
= \(\frac{0.6}{210-74.25}\) = \(\frac{0.6}{135.75}\)
∴ Temperature coefficient of resistivity
∝ = 0.443 × 10^-2/°C

Question 8.
If the length of a wire conductor is doubled by stretching it while keeping the potential difference constant, by what factor will the drift speed of the electrons change ?
Solution:
Taking l₁ = l,
l₂ = 2l
Since V_d ∝ l, \(\frac{\mathrm{v}_{\mathrm{d}_2}}{\mathrm{~V}_{\mathrm{d}_1}}\) = \(\frac{l_2}{l_1}\)
\(\frac{\mathrm{V}_{\mathrm{d}_2}}{\mathrm{~V}_{\mathrm{d}_1}}\) = \(\frac{2 l}{l}\)
∴ \(\mathrm{V}_{\mathrm{d}_2}\) = \(2 \mathrm{~V}_{\mathrm{d}_1}\)
∴ Drift speed of electrons changes by a factor 2.

Question 9.
Two 120V light bulbs, one of 25W and’ another of 200W are connected in series. One bulb burnt out almost instantaneously. Which one was burnt and why ?
Solution:
Given, For first bulb,
P₁ = 25W,
V₁ = 120V

Resistance of first bulb R₁ = \(\frac{\mathrm{v}_1^2}{\mathrm{P}_1}\)
R₁ = \(\frac{(120)^2}{25}\) —– (1)
For second bulb, P₂ = 200W, V₂ = 120V
Resistance of second bulb,
R₂ = \(\frac{(120)^2}{200}\) —— (2)
\(\frac{(1)}{(2)}\) ⇒ \(\frac{\mathrm{R}_1}{\mathrm{R}_2}\) = 8 ⇒ R₁ = 8R₂
As R₁ > R₂, 25 W bulb burnt out almost instantaneously, since two bulbs have rated at same voltage.

Question 10.
A cylindrical metallic wire is stretched to increase its length by 5%. Calculate the percentage change in resistance.
Solution:
Given, % change in length, \(\frac{\mathrm{d} l}{l}\) = 5%
Resistance of wire R = \(\frac{\rho l^2}{\mathrm{~V}}\)
% Change in Resistance of wire,
\(\frac{d R}{R}\) = 2\(\frac{\mathrm{d} l}{l}\) = 2 × 5% = 10%

Question 11.
Two wires A and B of same length and same material, have their cross sectional areas in the ratio 1:4. What would be the ratio of heat produced in these wires when the Voltage across each is constant ?
Solution:
Given l_A = l_B, ρ_A = ρ_B, V_A = V_B,
A_A : A_B = 1 : 4
Rate of heat produced in a wire,
H = i²R = \(\frac{\mathrm{V}^2}{\mathrm{R}^2}\) = \(\frac{V^2 A}{\rho l}\)
Since V, ρ, l are same for both wires A and B, H ∝ A (area of crossection)
For two wires A and B,
\(\frac{\mathrm{H}_{\mathrm{A}}}{\mathrm{H}_{\mathrm{B}}}\) = \(\frac{A_A}{A_B}\) = \(\frac{1}{4}\)
∴ H_A : H_B = 1 : 4.

Question 12.
Two bulbs whose resistances are in the ratio of 1:2 are connected in parallel to a source of constant voltage. What will be the ratio of power dissipation in these ?
Solution:
Given, R₁ : R₂ = 1 : 2, In parallel series
Dissipated power P = \(\frac{V^2}{R}\)
⇒ P ∝ \(\frac{\mathrm{I}}{\mathrm{R}}\) [ ∵ V = constant]
The ratio of dissipated powers in two bulbs is given by,
\(\frac{P_1}{P_2}\) = \(\frac{\mathrm{R}_2}{\mathrm{R}_1}\) = \(\frac{2}{1}\)
∴ P₁ : P₂ = 2 : 1

Question 13.
A potentiometer wire is 5m long and a potential difference of 6 V is maintained between its ends. Find the emf of a cell which balances against a length of 180cm of the potentiometer wire. (A.P. Mar. ’16)
Solution:
Length of potentiometer wire L = 5m
Potential difference V = 6 Volt
Potential gradient ϕ = \(\frac{\mathrm{V}}{\mathrm{L}}\) = \(\frac{6}{5}\) = 1.2 V / m
Balancing length l = 180cm
= 1.80m
Emf of the cell E = ϕl
= 1.2 × 1.8 = 2.16V

Question 14.
A battery of emf 2.5 V and internal resistance r is connected in series with a resistor of 45 ohm through an ammeter of resistance 1 ohm. The ammeter reads a current of 50 mA. Draw the circuit diagram and calculate the value of r.
Solution:
Circuit diagram for the given data is shown below.

Given, E = 2.5 V; R = 4 5Ω;
r_A = 1A; I = 50mA;
r = ?
E = I (R + r_A + r)
2.5 = 50 × 10^-3 (45 + 1 + r)
46 + r = \(\frac{2.5}{50 \times 10^{-3}}\) = \(\frac{2.5 \times 10^3}{50}\) = 50
∴ r = 50 – 46 = 4Ω.

Question 15.
Amount of charge passing through the cross section of a wire is q(t) = at² + bt + c. Write the dimensional formula for a, b and c. If the values of a, b and c in SI unit are 6, 4, 2 respectively, find the value of current at t = 6 seconds.
Solution:
Charge passing through wire is given by q(t) = at² + bt + c
According to principle of homogenity, Dimensional formula of q(t) = dimensional formula of at²
IT = aT²
∴ Dimensional formula a = IT^-1
Dimensional formula of q(t) = Dimensional formula of bt
IT = bT
∴ Dimensional formula of b = I
Dimensional formula of q(t) = Dimensional formula of C.
IT = C
∴ Dimensonal formula of C = IT
Current, I = \(\frac{\mathrm{dq}(\mathrm{t})}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(at² + bt + c)
= 2at + b
Here a = 6 and b = 4
⇒ I = 12t + 4
∴ Current at t = 6 sec,
I = 12 × 6 + 4 = 76 A.

Textual Exercises

Question 1.
The storage battery of a car has an emf of 12V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery ?
Solution:
Here E = 12 V, r = 0.4Ω
Maximum Current, I_max = \(\frac{\mathrm{E}}{r}\) = \(\frac{12}{0.4}\) = 30A

Question 2.
A battery of emf 10V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?
Solution:
Here E = 10 V, r = 3Ω, I = 0.5 A, R = ?, V = ?
I = \(\frac{E}{(R+r)}\) or (R + r) = \(\frac{E}{I}\) = \(\frac{10}{0.5}\) = 20 or
R = 20 – r = 20 – 3 = 17Ω
Terminal voltage V = IR = 0.5 × 17 = 8.5 Ω.

Question 3.
a) Three resistors 1Ω, 2Ω, and 3Ω are combined in series. What is the total resistance of the combination ?
Solution:
Here R₁ = 1Ω, R = 2 Ω, R₃ = 3Ω, V = 12V
In series, total resistance R_S = R₁ + R₂ + R₃ = 1 + 2 + 3 = 6Ω.

b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Solution:
Current through the circuit I = V/Rs = 12/6 = 2A
∴ Potential drop across R₁ = IR₁ = 2 × 1 = 2V
Potential drop across R₂ = IR₂ = 2 × 2 = 4V
Potential drop across R₃ = IR₃ = 2 × 3 = 6V

Question 4.
a) Three resistors 2Ω, 4Ω and 5Ω are combined in parallel. What is the total resistance of the combination ?
Solution:
Here R₁ = 2Ω, R₂ = 4Ω, R₃ = 5Ω, V = 20V
In parallel combination total resistance R_P is given by
\(\frac{1}{\mathrm{R}_{\mathrm{P}}}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + \(\frac{1}{R_3}\) = \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\) = \(\frac{10+5+4}{20}\) = \(\frac{19}{20}\) or R_P = \(\frac{20}{19} \Omega\)

b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Solution:
Current through R₁ = \(\frac{\mathrm{V}}{\mathrm{R}_1}\) = \(\frac{20}{2}\) = 10A
Current through R₂ = \(\frac{20}{4}\) = 5 A
Current through R₃ = \(\frac{20}{5}\) = 4A
Total current = \(\frac{20}{\left(\frac{20}{9}\right)}\) = 19A

Question 5.
At room temperature (27.0°C) the resistance of a heating element is 100Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10^-4°C^-1.
Solution:
Here R₂₇ = 100Ω, R₁ = 117Ω, t = ? α = 1.70 × 10^-4°C^-1
We know that
α = \(\frac{R_t-R_{27}}{R_{27}(t-27)}\) or t – 27 = R_t – R₂₇
t = \(\frac{\mathrm{R}_1-\mathrm{R}_2}{\mathrm{R}_{27} \times \alpha}\) + 27 = \(\frac{117-100}{100 \times 1.7 \times 10^{-4}}\) + 27
= 1000 + 27 = 1027°C

Question 6.
A negligibly small current is passed through a wire of length 15m and uniform cross-section 6.0 × 10^-7 m², and its resistance is measured to be 5.0Ω. What is the resistivity of the material at the temperature of the experiment ?
Solution:
Here L = 15 m, A = 6.0 × 10^-7m², R = 5.0 Ω, ρ = ?

Question 7.
A silver wire has a resistance of 2.1 Ω at 27.5°C, and a resistance of 2.7 Ω at 100°C. Determine the temperature coefficient of resistivity of silver.
Solution:
Here R_27.5 = 2.1Ω, R₁₀₀ = 2.7Ω; α = ?
α = \(\frac{\mathrm{R}_{100}-\mathrm{R}_{27.5}}{\mathrm{R}_{27.5} \times(100-27.5)}\) = \(\frac{2.7-2.1}{2.1 \times(100-27.5)}\) = 0.0039°C^-1

Question 8.
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0°C ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^-4°C^-1.
Solution:

Question 9.
Determine the current in each branch of the network shown in Fig.

Solution:
The current’s through the various arms of the circuit have been shown in figure.
According to Kirchhoff’s second law;
-10 + 10 (i₁ + i₂) + 10i₁ + 5(i₁ – i₃) = 0
(or) 10 = 25i + 10i – 5i₃
(or) 2 = 5i₁ + 2i₂ – i₃
(or) 2 = 5i₁ + 2i₂ – i₃ ………. → (i)
In a closed circuit ABDA ~
10i₁ + 5i₃ – 5i₂ = 0
(or) 2i₁ + i₃ – i₂ = 0
(or) i₂ = 2i₁ + i₃ …….. → (ii)
In a closed circuit BCDB
5(i₁ – i₃) – 10 (i₂ + i₃) – 5i₃ = 0
(or) 5i₁ – 10i₂ = 20i₃ =0
i₁ = 2i₁ + 4i₃…….. → (iii)
From (ii) and (iii)
i₁ = 2 (2i₁ – i₃) + 4i₃ = 4i₁ + 6i₃
(or) 3i₁ = -6i₃
(or) i₁ = -2i₃
Putting this value in (ii) : i₂ = 2(-2i₃) i₃ = -3i₃
Putting values in (i)
2 = 5(-2i₃) + 2(-3i₃) – i₃ (or) 2 = -17i₃
(or) i₃ = -2/17A
From (iv) i₁ = -(-2/17) = -4/17A
from (v) i₂ = 3(-2/17) = 6/17A
i₁ + i₂ = (4/17) + (6/17) = (10/17)A
i₁ + i₃ = (4/17) + (-2/17) = (6/17) A
i₂ + i₃ = (6/17) + (-2/17) = 4/17A.

Question 10.
a) In a metre bridge the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips ?

A meter bridge. Wire Ac is 1m long R is a resistance to be measured and S is a standard resistant
Solution:
Here l = 39.5cm, R = X = ?, S = Y = 12.5 Ω
As S = \(\frac{100-l}{l} \times \mathrm{R}\)
∴ 12.5 = \(\frac{100-39.5}{39.5} \times \mathrm{X}\)
or X = \(\frac{12.5 \times 39.5}{60.5}\) = 8.16Ω
Thick copper strips are used to minimise resistance of the connections which are not accounted in the formula.

b) Determine the balance point of the bridge above if X and Y are interchanged.
Solution:
As X and Y are interchanged therefore, l₁ and l₂ (i.e.) lengths are also interchanged.
Hence L = 100 – 39.5 = 60.5 cm.

c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge ? Would the galvanometer show any current ?
Solution:
The galvanometer will show no current.

Question 11.
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5Ω. What is the terminal voltage of the battery during charging ? What is the purpose of having a series resistor in the charging circuit ?
Solution:
Here emf of the the battery = 8.0V; voltage of d.c. supply = 120V
Internal resistance of battery r = 0.5Ω; external resistance R = 15.5Ω
Since a storage battery of emf 8V is charged with a.d.c supply of 120 V the effective emf in the circuit is given by ε = 120 – 8 = 112 V
Total resistance of the circuit = R + r = 15.5 + 0.5 = 16.0Ω
∴ Current in the circuit during charging is given by
I = \(\frac{\varepsilon}{R+r}\) = \(\frac{112}{16}\) = 7.0A
∴ Voltage across R = IR = 7.0 × 15.5 = 108.5 V
During charging the voltage of the d.c supply in a circuit must be equal to the sum of the voltage drop across R and terminal voltage of the battery
∴ 120 = 108.5 V or V= 120 – 108.5 = 11.5V
The series resistor limits the current drawn from the external source of d.c supply. In its absence the current will be dangerously high.

Question 12.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell ?
Solution:
Here ε₁ = 1.25 V, l₁ = 35.0 cm, ε₂ = ?.l₂ = 63.0 cm.
As \(\frac{\varepsilon_2}{\varepsilon_1}\) = \(\frac{l_2}{l_1}\) or
ε₂ = \(\frac{\varepsilon_1 \times l_2}{l_1}\) = \(\frac{1.25 \times 63}{35}\) = 2.25V

Question 13.
The number density of free electrons in a copper conductor estimated in textual example 6.1 is 8.5 × 10²⁸ m^-3. How long does an electron take to drift from one end of a wire 3.0m long to its other end ? The area of cross-section of the wire is 2.0 × 10^-6 m² and it is carrying a current of 3.0 A.
Solution:
Here n = 8.5 × 10²⁸ m^-3; L = 3.0 m; A = 2.0 × 10^-6 m²; I = 3.0A, t = ?
As I = n A eVd
∴ V_d = \(\frac{1}{\mathrm{nAe}}\)
Now, t = \(\frac{1}{\mathrm{v}_{\mathrm{d}}}\)
=
= \(\frac{3.08 \times 8.5 \times 10^{28} \times 2.0 \times 10^{-6} \times 1.6 \times 10^{-19}}{3.0}\)
= 2.72 × 10⁴ S
= 7hour 33 minutes.

Additional Exercises

Question 1.
The earth’s surface has a negative surface charge density of 10^-9 C m^-2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface ? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10⁶m).
Solution:
Here r = 6.37 × 10⁶ m; Q = 10^-9 cm²; I = 1800 A
Area of the globe A = 4πr² = 4 × 3.14 × (6.37 × 10⁶)²
= 509.64 × 10³C
t = \(\frac{Q}{I}\) = \(\frac{509.64 \times 10^3}{1800}\) = 283.1 S

Question 2.
a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015Ω are joined in series to provide a supply to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?
Solution:
Here ε = 2.0V; n = 6; r = 0.015Ω; R = 8.5 Ω
Current I = \(\frac{n E}{R+n r}\) = \(\frac{6 \times 2.0}{8.5+6 \times 0.015}\) = 1.4A
Terminal voltage,V = IR = 1.4 × 8.5 = 11.9V.

b) A secondary cell after long use has an emf 011.9 V and a large internal resistance of 380Ω. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car?
Solution:
Here E = 1.9 V; r = 380Ω
I_max = \(\frac{\varepsilon}{\mathrm{r}}\) = \(\frac{1.9}{380}\) = 0.005A
This amount of current cannot start a car because to start the motor, the current required is 100 A for few seconds.

Question 3.
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρ_Al = 2.63 × 10^-8 Ωm, ρ_Cu = 1.72 × 10^-8 Ωm, Relative density of Al = 2.7, of Cu = 8.9.)
Solution:
Given, for aluminium wire; R₁ = R; l₁ = l
Relative density d₁ = 2.7.
For copper wire R₂ = R, t₂ = 1, d₂ = 8.9
Let A₁, A₂ be the area of cross section for aluminium wire and copper wire.
We know, R₁ = \(\rho_1 \frac{l_1}{\mathrm{~A}_1}\) = \(\frac{2.63 \times 10^{-8} \times l}{A_1}\)
and mass of the aluminium wire m₁ = A₁l₁ × d₁ = A₁l₁ × 2.7
R₂ = ρ₂ = \(\frac{l_2}{\mathrm{~A}_2}\) = \(\frac{1.72 \times 10^{-8} \times l}{\mathrm{~A}_2}\)
Mass of copper wire m₂ = A₂l₂ × d₂ = A₂l × 8.9
Since two wires are of equal resistance R₁ = R₂
\(\frac{2.63 \times 10^{-8} \times 1}{\mathrm{~A}_1}\) = \(\frac{1.72 \times 10^{-8} \times l}{\mathrm{~A}_2}\) or \(\frac{\mathrm{A}_2}{\mathrm{~A}_1}\) = \(\frac{1.72}{2.63}\)
from (ii) and (iv) we have
\(\frac{\mathrm{m}_2}{\mathrm{~m}_1}\) = \(\frac{\mathrm{A}_2 l \times 8.9}{\mathrm{~A}_1 l \times 2.7}\) = \(\frac{8.9}{2.7} \times \frac{\mathrm{A}_2}{\mathrm{~A}_1}\)
= \(\frac{8.9}{2.7} \times \frac{1.72}{2.63}\) = 2.16

It shows that copper wire is 2.16 times heavier than aluminium wire since for the same value of length and resistance aluminium wire has lesser mass than copper wire, therefore aluminium wire is preferred for overhead power cables. A heavy cable may sag down owing to its own weight.

Question 4.
What conclusion can you draw from the following observations on a resistor made of alloy manganin ?

Answer:
Since the ratio of voltage and current for different readings is same so ohm’s law is valid to high accuracy. The resistivity of the alloy manganin is nearly independent of temperature.

Question 5.
Answer the following Questions.
a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed ?
Solution:
Only current through the conductor of non-uniform area of cross section is constant as the remaining quantities vary inversly with the area of cross-section of the conductor.

b) Is Ohm’s law.universally applicable for all conducting elements ? If not, give examples of elements which do not obey Ohm’s law.
Solution:
Ohm’s law is not applicable for non-ohmic elements. For example, vaccum tubes, semi-conducting diode, liquid electrolyte etc.

c) A low voltage supply from which one needs high currents must have very low internal resistance. Why ?
Solution:
As, I_max = emf internal resistance so for maximum current internal resistance should be least.

d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why ?
Solution:
A high tension supply must have a large internal resistance otherwise, if accidently the circuit is shorted, the current drawn will exceed safety limit and will cause damage to circuit.

Question 6.
Choose the correct alternative :
a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10²²/10²³).
Solution:
a) Greater
b) Lower
c) Nearly independent
d) 10²²

Question 7.
a) Given n resistors each of resistance R, how will you combine them to get the
i) maximum,
ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance ?
Solution:
For maximum effective resistance the n resistors must be connected in series.
Maximum effective resistance R_S = nR
For minimum effective resistance the n resistors must be connected in parallel.
Maximum effective resistance R_P = R/n
∴ \(\frac{R_S}{n p}\) = \(\frac{\mathrm{nR}}{\mathrm{R} / \mathrm{n}}\) = n²

b) Given the resistances of 1Ω, 2Ω, 3Ω, how will be combine them to get an equivalent resistance of
(i) (11/3) Ω
(ii) (11/5) Ω,
(iii) 6Ω,
(iv) (6/11)Ω?
Solution:
It is to be noted that
a) the effective resistance of parallel combination of resistors is less than the individual resistance and
b) the effective resistance of series combination of resistors is more than individual resistance.

case (i) Parallel combination of 1Ω and 2Ω is connected in series with 3Ω.
Effective resistance of 1Ω and 2Ω in parallel will be given by
RP = \(\frac{1 \times 2}{1+2}\) = \(\frac{2}{3} \Omega\)
∴ Equivalent resistance of \(\frac{2}{3} \Omega\) and 3] and 3Ω in series
= \(\frac{2}{3}\) + 3 = \(\frac{11}{3}\)Ω

Case(ii) : Parallel combination of 2Ω and 3Ω is connected in series with 1Ω.
Equivalent resistance of 2Ω and 3Ω in parallel
= \(\frac{2 \times 3}{2+3}\) = \(\frac{6}{5} \Omega\)
Equivalent resistance of \(\frac{6}{5} \Omega\) and 1Ω in series = \(\frac{6}{5}\) + 1 = \(\frac{11}{5}\)Ω

Case (iii) : All the resistances are to be connected in series now
∴ Equivalent resistance = 1 + 2 + 3 = 6Ω

Case (iv) : All the resistances are to be connected in parallel
∴ Equivalent resistance (R) is given by
\(\frac{1}{\mathrm{R}}\) = \(\frac{1}{1}\) + \(\frac{1}{2}\) + \(\frac{1}{3}\)
= \(\frac{6+3+2}{6}\) = \(\frac{11}{6}\) (Or) r = \(\frac{6}{11} \Omega\)

c) Determine the equivalent resistance of networks shown in Fig.

Answer:
a) The given network is a series combination of 4 equal units. Each unit has 4 resistances in which 2 resistances (1Ω each in series) are in parallel with 2 other resistances (2Ω each in series).
∴ Effective resistances of two resistances (each of 1Ω) in series = 1 + 1 = 2Ω.
Effective Resistance of two resistances (each of 2Ω) in series = 2 + 2 = 4Ω
If R is the resistance of one unit of resistances then
\(\frac{1}{\mathrm{R}_{\mathrm{P}}}\) = \(\frac{1}{2}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\) or R_P = \(\frac{4}{3} \Omega\)
∴ Equivalent resistance in network = 4 R_P = 4 × \(\frac{4 \Omega}{3}\) = \(\frac{16 \Omega}{3}\)

b) Total resistances each of value R are connected in series. Their effective resistance = 5R.

Question 8.
Determine the current drawn from a 12 V supply with internal resistance 0.5Ω by the Infinite network shown in Fig. Each resistor has 1Ω resistance.

Solution:
Let x be the equivalent résistance of infinite network. Since the net work is infinite, therefore, the addition of one more unit of three resistances each of value of 1Ω across the terminals will not alter the total resistance of network i.e. it should remain x.

Therefore, the network would appear as shown in the figure and its total resistance should remain x.
There the parallel combination of x and 1Ω is in series with two resistors of 1Ω each.
The resistance of parallel combination is

\(\frac{1}{\mathbf{R}_{\mathbf{P}}}\) = \(\frac{1}{x}\) + \(\frac{1}{1}\) = \(\frac{1+x}{x}\)
R_p = \(\frac{x}{(1+x)}\)
∴ Total resistance of network will be given by
x = 1 + 1 + \(\frac{x}{x+1}\) = 2 + \(\frac{x}{x+1}\)
x(x + 1) = 2(x + 1) + x
or x² + x = 2x + 2 + x or x² – 2x – 2 = 0
or x = \(\frac{2 \pm \sqrt{4+8}}{2}\) = \(\frac{2 \pm \sqrt{12}}{2}\)
Total resistance of the circuit shows a full scale deflection for a current of 2.5 mA. How will you convert the meter into
= \(\frac{2 \pm 2 \sqrt{3}}{2}\) = 1 ± \(\sqrt{3}\)
The value of resistance cannot be negative, therefore the resistance of network
= 1 + \(\sqrt{3}\) = 1 + 1.73 Ω = 2.73 Ω
Total resistance of the cfrcuit = 2.73 + 0.5
= 3.23Ω
∴ Current draw I = \(\frac{12}{3.23}\) = 3.72 amp

Question 9.
Figure shows a potentiometer with a cell of 2.0’V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

a) What is the value ε ?
Solution:
Here ε₁ = 1.02 V, L₁ = 67.3cm, ε₂ = e = ?, L₂ = 82.3 cm
Since \(\frac{\varepsilon_2}{\varepsilon_1}\) = \(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\)
∴ ε = \(\frac{\mathrm{L}_2}{\mathrm{~L}_1} \times \varepsilon_1\) = \(\frac{82.3}{67.3} \times 1.02\) = 1.247V

b) What purpose does the high resistance of 600 kΩ have ?
Answer:
The purpose of using high resistance of 600k Ω is to allow very small current through the galvanometer when the movable contact is far from the balance point.

c) Is the balance point affected by this high resistance ?
Answer:
No, the balance point is not affected by the presence of this resistance.

d) Is the balance point affected by the internal resistance of the driver cell ?
Answer:
No, the balance point is not affected by the internal resistance of the driver cell.

e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V ?
Answer:
No, the method will not work as the balance point will not be obtained on the potentiometer wire if the e.m.f of the driver cell is less than the emf of the other cell.

f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple) ? If not, how will you modify the circuit ?
Answer:
The circuit will not work for measuring extremely small emf because in that case the balance point will be just close to the end A. To modify the circuit we have to use a suitable high resistance in series with the cell of 2.0V This would decrease the current in the potentiometer wire. Therefore potential difference 1cm df wire will decrease. Hence extremely small emf can be measured.

Question 10.
Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε ?

Answer:
Here L₁ = 58.3cm; L₂ = 68.5 cm; R = 10Ω; I X = ?. Let be the current in the potentiometer wire and ε₁ and ε₂ be the potential drops across R and X respectively when connected in circuit by closing respective keky. Then

If there is no balance point with given cell of emf it means potential drop across R or X greater than the potential drop across the potentiometer wire AB. In order to obtain the balance point, the potential drops across R and X are to be reduced which is possible by reducing the current in R and X for that either suitable resistance should be put in series with R and X or a cell of smaller emf E should be used. Another possible way is to increase the potential drop across the potentiometer wire by increasing the voltage of driver cell.

Question 11.
Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Solution:
Here l₁ = 76.3 cm, l₂ = 64.8 cm.
r = ?, R = 9.5 Ω
Now, r = \(\left(\frac{l_1-l_2}{l_2}\right) R\) = \(\left(\frac{76.3-64.8}{64.8}\right)\) 9.5 = 1.68 Ω

Textual Exercises

Question 1.
a) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 × 10^-7 m² carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 10³ kg/m³, and its atomic mass is 63.5 u.
b) Compare the drift speed obtained above with,
i) thermal speeds of copper atoms at ordinary temperatures,
ii) speed of propagation of electric field along the conductor which causes the drift motion.
Solution:
a) The direction of drift velocity of conduction electrons is opposite to the electric field direction, i.e., electrons drift in the direction of increasing potential. The drift speed ud is given by Eq.
IΔt = + neA /v_d/Δt
V_d = (I/neA)
Now, e = 1.6 × 10^-19 C, A = 1.0 × 10^-7 m², I = 1.5 A. The density of conduction electrons, n is equal to the number of atoms per cubic meter (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). A cubic metre of ‘copper has a mass of 9.0 × 10³ kg. Since 6.0 × 10²³ copper atoms have a mass of 63.5 g,
n = \(\frac{6.0 \times 10^{23}}{63.5}\) × 9.0 × 10⁶
= 8.5 × 10²⁸ m^-3 Which gives,
\(v_{\mathrm{d}}\) = \(\frac{1.5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10 \times 10^{-7}}\)
= 1.1 × 10^-3 m s^-1 = 1.1 mm s^-1

b) i) At a temperature T, the thermal speed of a copper atom of mass M is obtained from
[<(1/2) Mυ² > = (3/2) K_BT] and is thus typically of the order of \(\sqrt{\mathrm{k}_{\mathrm{B}} \mathrm{T} / \mathrm{M}}\), where K_B is the Boltzmann constant. For copper at 300 K, this is about 2 × 10² m/s. This figure indicates the random vibrational speeds of copper atoms in a conductor. Note that the drift speed of electrons is much smaller, about 10^-5 times the typical thermal speed at ordinary temperatures,

ii) An electric field travelling along the conductor has a speed of an electromagnetic wave, namely equal to 3.0 × 10⁸ m s^-1. The drift speed is, in comparison, extremely small, smaller by a factor of 10^-11.

Question 2.
a) In Textual Example 1, the electron drift speed is estimated to be only a few mm s^-1 for currents in the range of a few amperes ? How then is current established almost the instant a circuit is closed ?
b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed ?
c) If the electron drift speed is so small, and the electron’s charge is small, how can we still obtain large amounts of current in a conductor ?
d) When electrons drift in a metal from lower to higher potential, does it mean that all the ‘free’ electrons of the metal are moving in the same direction ?
e) Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the i) absence of electric field,
ii) presence of electric field ?
Solution:
a) Electric field is established throughout the circuit, almost instantly (with the speed of light) causing at every point a local electron drift. Establishment of a current does not have to wait for electrons from one end of the conductor travelling to the other end. However it does take a little while for the current to reach its steady value.
b) Each ‘free’ electron does accelerate, increasing its drift speed after collision but starts to accelerate and increases its drift speed again only to suffer a collision again and so on. On the average, therefore, electrons acquire only a drift speed.
c) Simple, because the electron number density is enormous, ~10²⁹ m^-3.
d) By no means. The drift velocity is superposed over the large random velocities of electrons.
e) In the absence of electric field, the paths are straight lines, in the presence of electric field, the paths are, in general curved.

Question 3.
An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (27.0 °C) is found to be 75.3 Ω. When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element ? The temperature coefficient of resistance of nichrome averaged over the temperature range involved, is 1.70 × 10^-4 °C^-1.
Solution:
When the current through the element is very small, heating effects can be ignored and the temperature T₁ of the element is the same as room temperature. When the toaster is connected to the supply, its initial current will be slightly higher than its steady value of 2.68 A. But due to heating effect of the current, the temperature will rise. This will cause an increase in resistance and a slight decrease in current. In a few seconds, a steady state will be reached when temperature will rise no further, and both the resistance of the element and the current drawn will achieve steady values. The resistance R₂ at the steady temperature T₂ is
R₂ = \(\frac{230 \mathrm{~V}}{2.68 \mathrm{~A}}\) = 85.8Ω
Using the relation
R₂ = R₁ [1 + α(T₂ – T₁)]
with α = 1.70 × 10^-4°C^-1, we get
T₂ – T₁ = \(\frac{(85.8-75.3)}{(75.3) \times 1.70 \times 10^{-4}}\) = 820°C
that is, T₂ = (820 + 27.0)°C = 847 °C

Question 4.
The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 Ω and at steam point is 5.39 Ω. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is 5.795 Ω. Calculate the temperature of the bath.
Solution:
R₀ = 5Ω, R₁₀₀ = 5.23 Ω and R_t = 5.795Ω
Now, t = \(\frac{R_t-R_0}{R_{100}-R_0} \times 100\), R_t = R₀ (1 + αt)
= \(\frac{5.795-5}{5.23-5} \times 100\)
= \(\frac{0.795}{0.23} \times 100\) = 345.65°C

Question 5.
A network of resistors is connected to a 16 V battery with internal resistance of 1Ω, a shown in Fig.
a) Compute the equivalent resistance of the network.
b) Obtain the current in each resistor.
c) Obtain the voltage drops V_AB, V_BC and V_CD.

Solution:
a) The network is a simple series and parallel combination of resistors. First the two 4Ω resistors in parallel are equivalent to a resistor
= [(4 × 4)/(4 + 4)]Ω = 2Ω
In the same way, the 12Ω and 6Ω resistors in parallel are equivalent to a resistor of
[(12 × 6)/(12 + 6)]Ω = 4Ω
The equivalent resistance R of the network is obtained by combining these resistors (2Ω and 4Ω) With 1Ω in series, that is,
R = 2Ω + 4Ω + 1Ω = 7Ω

b) The total current I in the circuit is
I = \(\frac{\varepsilon}{R+r}\) = \(\frac{16 \mathrm{~V}}{(7+1) \Omega}\) = 2A
Consider the resistors between A and B. If I₁ is the current in one of the 4 Ω resistors and I₂ the current in the other.
I₁ × 4 = I₂ × 4
That is, I₁ = I₂, which is otherwise obvious from the symmetry of the two arms. But I₁ + I₂ = I = 2A. Thus,
That is, current in each 4Ω resistor is 1 A. Current in lfi resistor between B and C would be 2 A.
Now, consider the resistances between C and D. If I₃ is the current in the 12Ω resistor, and I₄ in the 6Ω resistor,
I₃ × 12 = I₄ × 6 i.e., I₄ = 2I₃
But, I₃ + I₄ = I = 2A
Thus, I₃ = \(\left(\frac{2}{3}\right)\)A, I₄ = \(\left(\frac{4}{3}\right) \mathrm{A}\)
That is, the current in the 12Ω resistor is (2/3)A, while the current in the 6Ω resistor is (4/3) A.

c) The voltage drop across AB is
V_AB = I₁ × 4 = 1 A × 4Ω = 4V
This can also be obtained by multiplying the total current between A and B by the equivalent resistance between A and B, that is,
V_A = 2A × 2Ω = 4V
The voltage drop across BC is
V_BC = 2A × 1Ω = 2V
Finally, the voltage drop across CD is,
V_CD = 12Ω × I₃ = 12Ω × \(\left(\frac{2}{3}\right) \mathrm{A}\) = 8V.
This can alternately be obtained by multiplying total current between C and D by the equivalent resistance between C and D, that is.
V_CD = 2A × 4Ω = 8V
Note that the total voltage drop across AD is 4V + 2V + 8V = 14V. Thus, the terminal voltage of the battery is 14V, while its emf is 16V The loss of the voltage (= 2V) is accounted for by the internal resistance ID of the battery [2A × 1Ω = 2V].

Question 6.
A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1Ω Fig. Determine the equivalent resistance of the network and the current along each edge of the cube.
Solution:

The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network.

The paths AA’. AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further, at the corners A’, B and D, the incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff’s first rule and the symmetry in the problem.
Next take a closed loop, say, ABCC’EA, and apply Kirchhoff’s second rule :
-IR – (1/2)IR – IR + ε = 0
where R is the resistance of each edge and ε the emf of battery. Thus, ε = \(\frac{5}{2}\)IR
The equivalent resistance R_eq of the network is R_eq = \(\frac{\varepsilon}{3 I}\) = \(\frac{5}{6}\)R
For R = IΩ, R,sub>eq = (5/6) Ω and for ε = 10V, the total current (=3I) in the network is 3I = 10V/(5/6) p = 12 A, i.e., I = 4 A
The current flowing in each edge can now be read off from the Fig.

Question 7.
Determine the current in each branch of the network shown in Fig.

Solution:
Each branch of the network is assigned an unknown current to be determined by the application of Krichhoff’s rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknown I₁, I₂ and I₃ which can be found by applying the second rule of Krichhoff to three different closed loops. Kirchhoff s second rule for the closed loop ADCA gives,
10 – 4(I₁ – I₂) + 2 (I₂ + I₃ – I₁) – I₁ = 0
that is, 7I₁ – 6I₂ – 2I₃ = 10 ——-> (1)
For the closed loop ABCA, we get
10 – 4I₂ – 2 (I₂ + I₃) – I₁ = 0
that is, I₁ + 6I₂ + 2I₃ = 10 —-—> (2)
For the closed loop BCDEB, we get
5 – 2 (I₂ + I₃) – 2 (I₂ + I₃ – I₁) = 0
that is, 2I₁ – 4I₂ – 4I₃ = -5 ——–> (3)
Equations (1, 2, 3) are three simultaneous equations in three unknowns. These can be solved by the usual method to give.
I₁ = 2.5A, I₂ = \(\frac{5}{8}\)A, I₃ = 1\(\frac{7}{8}\)A
The currents in the various branches of the network are
AB : \(\frac{5}{8}\) A, CA : 2\(\frac{1}{2}\) A, DEB : 1\(\frac{7}{8}\) A
AD = 1\(\frac{7}{8}\) A, CD : 0 A, BC : 2\(\frac{1}{2}\) A

It is easily verified that Krichhoffs second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage drop over the closed loop BADEB
5V + \(\left(\frac{5}{8} \times 4\right)\) – \(\left(\frac{15}{8} \times 4\right) \mathrm{V}\)
equal to zero, as required by Krichhoffs second rule.

Question 8.
The four arms of a Wheatstone bridge (Fig.) have the following resistances :
AB = 100Ω, BC = 10Ω, CD = 5Ω, and DA = 60Ω

A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.
Solution:
Considering the mesh BADB, we have
100I₁ + 15I_g – 60I₂ = 0
or 20I₁ + 3I_g – 12I₂ = 0 —–> (1)
Considering the mesh BCDB, we have
10(I₁ – I_g) – 15I_g – 5(I₂ + I_g) = 0
10I₁ – 30I_g – 5I₂ = 0
2I₁ – 6I_g – I₂ = 0 ——> (2)
Considering the mesh ADCEA,
60I₂ + 5(I₂ + I_g) = 10
65I₂ + 5I_g = 10
13I₂ + I_g = 2 —–> (3)
Multiplying equation (2) by 10 ‘
20I₁ + 60I_g – 10I₂ = 0
From Equations. (4) and (1) we have  ——> (4)
63I_g + 2I₂ = 0
I₂ = 31.5I_g
Substituting the value of I₂ into Equation (3) we get.
13(31.5I_g) + I_g = 2
410.5 I_g = 2
I_g = 4.87 mA.

Question 9.
In a metre bridge (Fig.), the null point is found at a distance of 36.7 cm from A. If now a resistance of 12Ω is connected in parallel with S, the null point occurs at 51.9 cm. Determine the values of R and S.

Solution:
From the first balance point, we get
\(\frac{\mathrm{R}}{\mathrm{S}}\) = \(\frac{33.7}{66.3}\) —–> (1)
After S is connected in parallel with a resistance of 12Ω, the resistance across the gap changes from S to S_eq, where
S_eq = \(\frac{12 \mathrm{~S}}{\mathrm{~S}+12}\) —–> (2)
and hence the new balance condition now gives
\(\frac{51.9}{48.1}\) = \(\frac{\mathrm{R}}{\mathrm{S}_{\mathrm{eq}}}\) = \(\frac{R(S+12)}{12 S}\)
Substituting the value of R/S from Equation (1), we get
\(\frac{51.9}{48.1}\) = \(\frac{\mathrm{S}+12}{12}\) . \(\frac{33.7}{66.3}\)
Which gives S = 13.5 Ω. Using the value of R/S above, we get R = 6.86 Ω.

Question 10.
A resistance of R Ω draws current from a potentiometer. The potentiometer has a total resistance R₀ Ω. (Figure) A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

Solution:
While the slide is in the middle of, the potentiometer only half of its resistance (R₀/2) will be between the points A and B. Hence, the total resistance between A and B, say, R₁, will be given by the following expression.
\(\frac{1}{\mathrm{R}_1}\) = \(\frac{1}{R}\) + \(\frac{1}{\left(\mathrm{R}_0 / 2\right)}\)
R₁ = \(\frac{\mathrm{R}_0 \mathrm{R}}{\mathrm{R}_0+2 \mathrm{R}}\)
The total resistance between A and C will be süm of resistance between A and B and B and C, i.e., R₁ + R₀/2
∴ The current flowing through the potentiometer will be
I = \(\frac{\mathrm{V}}{\mathrm{R}_1+\mathrm{R}_0 / 2}\) = \(\frac{2 \mathrm{~V}}{2 \mathrm{R}_1+\mathrm{R}_0}\)
The voltage V₁ taken from the potentiometer will be the product of current I and resistance R₁,

Andhra Pradesh BIEAP AP Inter 2nd Year Accountancy Study Material 3rd Lesson Consignment Textbook Questions and Answers.

AP Inter 2nd Year Accountancy Study Material 3rd Lesson Consignment

Essay Questions

Question 1.
What do you mean about consignment? Explain the differences between consignment and sale.
Answer:
Consignment means sending goods to another person. In the case of consignment, goods are sent by the owner of the goods to the agent for the purpose of sale. The ownership of the goods remains with the sender. The agent sells the goods on behalf of the sender according to his instructions. The sender of the goods is known as the consignor and the agent is called the consignee.
Differences between consignment and sale

Basis	Consignment	Sale
1. Ownership of goods	The ownership of goods remains with the consignor and the possession is transferred to the consignee.	Ownership and possession of goods are transferred to the buyer immediately.
2. Parties	Two parties involved are known as consignors and consignees.	Two parties involved are known as the buyer and seller.
3. Relation between parties	The relation between them is that of a principal and agent which continues for long period, till it is ended.	The relationship between them is between buyer and seller, which ends immediately after the delivery and payment of goods.
4. Risk	The risk of loss or damage is of the consignor.	The risk passes with ownership to the buyer.
5. Consideration	The consignee sells goods for consideration.	The goods are sold for profit against the price.
6. Expenses	The expenses are borne by the consignor.	After-sales, the expenses are borne by the buyer.
7. Account sales	Consignee sends consignor account sales from time to time.	The buyer does not need to send any account sales to the seller.
8. Profit/Loss	The profit or loss on the consignment belongs to the consignor.	The profit or loss on the sales belongs to the seller.

Question 2.
What are account sales? Give a specimen copy of account sales.
Answer:
Account sales is a document or statement sent by the consignee to the consignor from time to time. Since the consignee sells the goods on behalf of the consignor, so he has to send a proper statement either on the sale of goods or at the end of a particular period.

In account sales, the consignee shows the details of the gross sale proceeds of the consignment. The various expenses, and charges incurred by him, and the commission due to him are deducted. Any advance payment to the consignor is deducted from the total amount due and the net amount payable is shown. The net amount payable is sent to the consignor by a bank draft or bill of exchange agreed.

Specimen of Account Sales
Account sales sent by Gwaliar Ltd to Sony Ltd regarding 200 T.V.s

Question 3.
What is meant by commission? Explain different types of commission.
Answer:
The consignee is remunerated by a commission which is usually calculated as an agreed percentage of the gross sale proceeds of the sale.
Commission payable to consignee can be divided into 3 types. They are:

• Ordinary commission
• Del credre commission
• Overriding commission

(a) Ordinary commission: Ordinary commission is the commission generally paid by the consignor to the consignee. It is calculated as a fixed percentage of gross sale proceeds. The such commission does not provide any security to the consignor from bad debts.

(b) Del credre commission: The consignee may sell some part of the goods on credit. When goods are sold on credit, there is always a risk of some amount as bad debt. In order to avoid the risk of bad debts, the consignor provides an additional commission known as Del credre commission to the consignee who guarantees the payment in case of credit sale. Del credre commission is paid at a predetermined percentage of gross sale proceeds. However, as regards payment of del credre commission there may be a separate agreement for its payment.

(c) Overriding commission: It is an extra commission allowed over the normal commission. This commission is generally offered when an agent is required to work hard either to introduce a new product in the market or to handle the work of supervising the performance of other agents in a particular area. It is the commission paid by the consignor to the consignee for executing sales on consignment at a price higher than the price fixed by the consignor. In other words, it is the surplus. the commission allowed to consignee, calculated on the surplus price realized by him.

Short Answer Questions

Question 1.
What do you mean by Consignment?
Answer:
The word consignment originated from the French word ‘consigner’ which means to hand over or transmit. To consign means ‘to send’; therefore, consignment means sending goods to another person. In the case of consignment, goods are sent by the owner of the goods to the agent for the purpose of sale. The ownership of the goods remains with the sender. The agent sells the goods on behalf of the sender, according to his instructions. The sender of the goods is known as the consignor and the agent is known as the consignee.

Question 2.
Briefly explain about Consignor and Consignee.
Answer:
Consignor: The person who sends the goods is known as a consignor. In other words, the consignor is the person who consigns the goods. He is the owner of the goods. He sends goods where in the physical delivery is delivered to the receiver without transfer of ownership.

Consignee: The person who receives the goods sent by the consignor is known as the consignee. In other words, the consignee is the person who acts as an agent of the consignor. He receives the goods on behalf of the consignor, stores them, incurs expenses, and sells the goods as per the specifications of the consignor for a consideration called ‘commission’.

Question 3.
What is a Proforma Invoice?
Answer:
Along with the goods, a statement is usually forwarded by the consignor to the consignee, giving a description of the goods consigned, the weight, quantity, price, and other relevant details. The statement is known as a proforma invoice. It resembles a sales invoice in appearance, but its purpose is quite different. It is intended as an evidence record of the consignment and the minimum price at which the consignee is expected to sell the goods sent to him.

Question 4.
What is Account Sales?
Answer:
Account sales is a document sent by the consignee to the consignor showing the details of the gross sale proceeds, the various expenses incurred by him, the commission amount due, any advance payment to the consignor which is deducted from the total amount due, and the net amount payable is shown.

Question 5.
What is Commission?
Answer:
The consignee is remunerated by a commission which is usually calculated as an agreed percentage of the gross proceeds of sale and such commission does not provide any security to the consigner from bad debts.

Question 6.
What is the Del Credre commission?
Answer:
The consignee may sell some part of the goods on credit. When goods are sold on credit, there is always a risk of some amount of bad debt. In order to avoid the risk of bad debts, the consignor provides an additional commission known as the Del-Credre commission to the consignee who guarantees the payment in case of credit sale. Del credre commission is paid at a predetermined percentage of gross sale proceeds.

Question 7.
What is Overriding Commission?
Answer:
It is an extra commission allowed over the normal commission. This commission is generally offered when an agent is required to work hard either to introduce a new product or to supervise the work of other agents in a particular area.

Question 8.
Briefly explain recurring expenses and nonrecurring expenses.
Answer:
All the expenses incurred to bring the goods to the godown of the consignee are treated as nonrecurring or direct expenses. Examples of non-recurring expenses which may be incurred by the consignor or consignee are freight, carriage or cartage insurance, packing, dock dues, loading and unloading charges, customs duty, octroi, etc. All the expenses incurred by the consignee after the goods reach the godown are treated as recurring or indirect expenses.
e.g: Go down rent, godown insurance, salary to salesmen, advertisement selling expenses, commission, bank charges, etc.

Question 9.
Explain the procedure for valuation of unsold stock in consignment.
Answer:
At the end of the accounting period, the unsold goods left with the consignee should be valued properly. Otherwise, true profit cannot be ascertained. Unsold stock is valued at either market price or cost price whichever is less. The cost price of the goods for this purpose does not mean only the cost at which the consignor purchased goods. But the proportion of non-recurring or direct expenses incurred by the consignor as well as the consignee should be added to the cost price.

Question 10.
Explain the term normal loss.
Answer:
In the case of some goods, even after taking all the precautions, some loss of quantity is bound to take place. Therefore, the loss which is unavoidable, natural, and due to the inherent nature of goods is called normal loss. For example, if coal is consigned a small portion of coal is bound to lose while loading and unloading. Similarly, in the case of oil and petroleum products, a portion may last due to evaporation and leakage when they are stored.

Question 11.
Accounting treatment for normal loss.
Answer:
Normal loss is unavoidable. Therefore, it forms part of the cost of consignment. Since this loss is usual, no journal entry is required to be passed but the normal loss is to be considered for calculating the cost of unsold stock left with the consignee, normal loss is spread over the remaining stock. Therefore, for calculating the value of the unsold stock, the following formula can be applied.

Textual Exercises

Question 1.
On 1st January 2009, Sudha of Srinagar consigned goods valued at ₹ 20,000 to Indira of Warangal. Sudha paid cartage and other expenses ₹ 1,500. On 1st April 2009, Indira sent on account sales with the following information.
(a) 1/2 of the goods sold for ₹ 15,000
(b) Indira incurred expenses of ₹ 750
(c) Indira is entitled to receive commission @ 5% on sales.
A bank draft was enclosed for the balance due. Prepare necessary Ledger accounts in the books of Sudha.
Solution:

Question 2.
On 1st January 2012, Gopi of Hyderabad consigned goods valued at ₹ 30,000 to Sudheer of Madras. Gopi paid cartage and other expenses ₹ 2,000 on 1st April 2012. Sudheer sent the account sales with the following information:
(a) 50% of the goods sold for ₹ 22,000
(b) Sudheer incurred expenses amounting to ₹ 1,200
(c) Sudheer is entitled to receive a commission @ 5% on sales.
A bank draft was enclosed for the balance due. Prepare the necessary ledger accounts in the books of Gopi.
Solution:

Question 3.
Sai and Co., of Chennai, consigned 100 Radios to Deepthi and Co. of Hyderabad. The cost of each Ratio was ₹ 500. Sai and Co. paid insurance ₹ 500; Freight ₹ 800. Account sales were received from Deepthi and Co., showing the sale of 80 Radios at ₹ 600 each. The following expenses were deducted by them.
Carriage – ₹ 20
Selling expenses – ₹ 130
Commission = ₹ 2,400
Sai and Co. received a bank draft for the balance due. Prepare important Ledger accounts in the books of Deepthi and Co.
Solution:

Question 4.
Raj of Bandar sends 200 T V. sets each costing ₹ 15,000 to Rani of Guntur to be sold on a consignment basis. He incurred the following expenses.
Freight ₹ 2,000; Loading and unloading charges ₹ 2,000 and Insurance ₹ 5,000.
Rani sold 185 TVs for ₹ 30,00,000 and paid ₹ 10,000 as shop rent which is to be borne by Raj as per terms and conditions of consignment.
The consignee is entitled to a commission of ₹ 200 per T V. sold. Assuming that Rani settled the account by sending a bank draft to Raj.
Prepare the necessary Ledger Accounts in the books of Raj.
Solution:

Question 5.
Vishnu of Vijayawada consigned goods valued at ₹ 50,000 to Shiva of Secundrabad. Vishnu paid transport charges ₹ 4,000 and drew a bill of two months on Shiva for ₹ 30,000 as advance. The bill was discounted with bankers for ₹ 29,500. Shiva sent the account sales of the consignment stating that the entire stock was sold for ₹ 72,000; Cartage ₹ 2,000; Commission ₹ 3,000 and a Bank Draft for the balance.
Prepare necessary accounts in the books of Vishnu.
Solution:

Question 6.
Laxmi of Vijayawada consigned goods worth ₹ 20,000 to his agent Saraswathi of Kodad on consignment Laxmi spent ₹ 1,000 on transport, ₹ 500 on insurance: Saraswathi sent ₹ 5,000 as advance. After two months, Laxmi received the account sales as follows:
(a) Half of the goods were sold for ₹ 24,000
(b) Selling expenses were ₹ 1,200
(c) 10% commission on sales
Give ledger accounts in the books of Laxmi.
Solution:

Question 7.
On 1st January 2009, Sudha of Srinagar consigned goods valued at ₹ 20,000 to Indira of Warangal. Sudha paid cartage and other expenses ₹ 1500. On 1st April 2009, Indira sent account sales with the following information:
(a) 50% of the goods sold for ₹ 15,000
(b) Indira incurred expenses amounting to ₹ 750
(c) Indira is entitled to receive commission @ 5% on sales.
A bank draft was enclosed for the balance due. Prepare the necessary ledger accounts in the books of Sudha.
Answer:

Question 8.
Robert consigned goods to Rahim valued at ₹ 5,000 to be sold on a 5% commission basis. Robert has paid ₹ 500 freight and ₹ 550 towards insurance.
Robert received account sales and a draft for the balance from Rahim showing the following particulars.
Gross Sales – ₹ 7500
Selling Expenses – ₹ 450
Commission – ₹ 375
Pass necessary entries journal in the and prepare ledger accounts in the books of both parties.
Solution:

Question 9.
Krishna of Mumbai and Gopal of Chennai is in the consignment business. Gopal sent goods to Krishna for ₹ 10,000. Gopal paid freight ₹ 500. Insurance ₹ 1,500 Krishna met sales expenses ₹ 900, Krishna sold the entire stock for ₹ 20,000 and he is entitled to a commission of 5% on sales.
Write the necessary entries in the books of Gopal and Krishna.
Solution:
In the books of Gopal Journal Entries

In the book of Krishna Journal Entries

Question 10.
Manikanta of Vijayawada Consigned goods of value of ₹ 20,000 to Ayyappa of Ahmedabad. Manikanta paid forwarding charges of ₹ 1,000 and drew a bill of two months on Ayyappa for ₹ 10,000. The bill was discounted with bankers for ₹ 9,500. Ayyappa sent received the account sales of the consignment stating that the entire stock was sold for ₹ 28,000 agents commission ₹ 2,000 and a bank draft for the balance.
Prepare necessary accounts.
Solution:

Question 11.
Mrs. Murali sent 50 Bicycles on consignment to Mr. Deepthi invoiced at ₹ 800 each on Jan 1st, 2009. She has paid the following expenses:
₹ 1,350 – freight, ₹ 600 – Insurance, ₹ 1,500 – other expenses.
On 5th January, she received a bill from Deepthi for ₹ 40,000. On Feb 20th Deepthi sent an account sales showing that the bicycles have realized ₹ 1,000 each. He incurred expenditures on carriage ₹ 500, warehousing ₹ 460, and ₹ 300 miscellaneous expenses. He charged a commission of 10% on sales. Prepare the books of the consignor and consignee.
Solution:

Question 12.
M/s. Robert & Co. of Bangalore consigned 100 cases @ 50 each to Mahathi & Co., of Calcutta. M/s. Robert & Co. spent ₹ 700 on Carriage and paid insurance ₹ 250.
In due course account sales were received with the following details:

Pass necessary entries in the books of both parties.
Solution:
In the books of Robert & Co. Journal Entries

In the books of Mahathi & Co. Journal Entries

Question 13.
A & Co., of Hyderabad, consigned 100 Video Games to B & Co., of Delhi to be sold on consignment @ ₹ 500 each. He paid transport ₹ 2,000 and warehouse charges ₹ 3,000. B & Co. sent account sales stating that

Prepare necessary ledger accounts of both books.
Solution:

Question 14.
X of Chirala consigned 200 bales of Tobacco @ 250 per bale to V of Vijayawada. X paid cartage of freight etc., ₹ 1,250. X drew a bill on V for 3 months for ₹ 30,000. V sold the entire consignment and rendered account sales showing that the goods realized ₹ 60,000 out of which he deducted his charges amounting to ₹ 400 and commission at 5% on sales. Make entries in the journal and show necessary ledger accounts in the books of both parties.
Solution:

Question 15.
Amar consigned 100 bales of cloth to Akbar at ₹ 5,000 per bale. Amar incurred the following expenses:
Packing and Forwarding Charges – ₹ 500
Insurance in Transit – ₹ 2,000
Akbar received the consignment and sold 80 bales at ₹ 8,000 per bale.
They incurred the following expenses:
Freight and Cartage – ₹ 3,000
Insurance of godown – ₹ 400
Salesmen’s Salary – ₹ 1,600
Ascertain the value of the consignment.
Solution:

Question 16.
On January 15, 2009, Dharani of Hyderabad sent 400 Bicycles to be sold on consignment to Dheeraj of Warangal. The Bicycles were invoiced at ₹ 1,000 per piece carriage and other expenses amounted to ₹ 6,000.
Dharani received the following account sales.
On 15th March 100 Bicycles were sold at ₹ 1,450 per piece on which 5% commission was charged and ₹ 3,750 were deducted as expenses.
10th April – 150 Bicycles were sold at ₹ 1,400 per piece on which 5% commission was charged and ₹ 2,900 were deducted as expenses incurred after 15 March.
Prepare consignment Accounts and Accounts in the books of Dharani.
Pass the necessary journal entries in the books of Bhagavan and Lakshman.
Solution:

Textual Examples

Question 1.
When the consignee sold total goods.
Sri Manikanta of Guntur consigned goods of the value of ₹ 1,00,000 to their agent Sri Rama of Hyderabad. Sri Manikanta paid loading and insurance in transit ₹ 5,000. On receiving the consignment Sri Rama sent ₹ 50,000 worth of Bank draft as advance.
Sri Rama sent account sales which show the following particulars.
Gross Sales – ₹ 2,00,000
Godown Rent – ₹ 1,000
Advertisements – ₹ 2,000
Commission 10% on sales
Sri Rama attached a bank draft for the balance due to Sri Manikanta your required to pass journal entries and prepare necessary ledger accounts in the books of Sri Manikanta and Sri Rama.
Solution:

Question 2.
Bhaskar of Rajahmundry consign 500 radio sets each at ₹ 600 to Prasad to Tenali on consignment Bhaskar paid ₹ 12,000 as freight and insurance in transit Bhaskar drawn a bill on Prasad for 3 months for ₹ 1,00,000. Prasad sends account sales which show the following particulars.
(1) Gross sale proceeds are ₹ 4,50,000.
(2) Unloading and godown rent ₹ 10,000.
(3) Commission 5% on Gross sales.
Prasad sends a bank draft for the balance due to Bhaskar.
You are required to prepare necessary ledger accounts in the books of the consignor and consignee.
Solution:

Question 3.
Kishore of Guntur sends 200 bicycles costing ₹ 1,20,000 to Pavan of Vijayawada on consignment. Kishore spends ₹ 6,000 on freight and insurance in transit. Pavan spent unloading charges ₹ 1,200 godown rent ₹ 800, consignee sold 180 bicycles at ₹ 2,00,000. Calculate the value of the closing stock.
Solution:

Note: Godown rent paid by the consignee is a recurring expense. Hence godown rent is not included in the valuation of closing stock.

Question 4.
X Send 500 radios costing ₹ 1,000 each to Y on consignment. X spends ₹ 50,000 on expenses. Y spent ₹ 12,000 as an advertisement. Consignee sold 400 radios each at ₹ 1,200. Calculate the value of the closing stock.
Solution:

Question 5.
Murali and Co of Warangal consign 500 radio sets to Han and Co of Hyderabad. The cost of each radio is ₹ 500. Murali and Co paid insurance ₹ 10,000 and freights ₹ 15,000. Account sales were received from Hari and Co showing the following particulars.
1. 400 radio sets sold each at ₹ 600.
2. Advertisement expenses ₹ 20,000.
3. Commission 10% on sales.
Hari and Co send a bank draft for the balance due to the consignor.
Show journal entries and ledger accounts in the books of both parties.
Solution:

Question 6.
A dealer is apple consign 1,000 tonnes of apples at a cost of ₹ 10,000 and paid ₹ 2,000 towards freight and insurance. The consignee received 950 tonnes of apples. Consignee sold 500 tonnes of apples. 50 tonnes of apples were treated as an unavoidable loss. Calculate the value of the unsold stock.
Solution:
Cost of 1,000 tonnes of apples = ₹ 10,000
Add: Consignor expenses = ₹ 2,000
Total = ₹ 12,000
Total quantity of goods less normal loss in quantity = 1,000 – 50 = 950 tonnes
Stock of unsold goods = 950 – 500 = 450 tonnes

Value of unsold stock = 12000 × \(\frac{450}{950}\)
= ₹ 5,684.21
= ₹ 5684

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 16th Lesson Communication Systems Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 16th Lesson Communication Systems

Very Short Answer Questions

Question 1.
What are the basic blocks of a communication system ?
Answer:
Basic blocks in communication system are

1. Transmitter
2. Receiver
3. Channel.

Question 2.
What is “World Wide Web” (WWW) ?
Answer:
Tern Berners -Lee invented the World Wide Web.
It is an encyclopedia of knowledge accessible to every one round the clock through out the year.

Question 3.
Mention the frequency range of speech signals.
Answer:
Speech signals frequency range is 300 Hz to 3100 Hz.

Question 4.
What is sky wave propagation ?
Answer:
In the frequency range from a MHz upto about 30 MHz, long distance communication can be achieved by ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave propagation.

Question 5.
Mention the various parts of the ionosphere ?
Answer:
Parts of ionosphere are

1. D (Part of stratosphere (65-70 km day only),
2. E (Part of stratosphere (100 km day only),
3. F₁ (Part of mesosphere (170 km – 190 km),
4. F₂ (Part of thermosphere [300 km at night 250 – 400 km during day time]).

Question 6.
Define modulation. Why is it necessary ? (T.S. Mar.’19, 16, 15; A.P. Mar. 16, 15) (Mar. ’14)
Answer:
Modulation : The process of combining low frequency audio signal with high frequency carrier wave is called modulation.

Necessary : Low frequency signals cannot transmit directly. To reduce size of the antenna and to avoid mixing up of signal from different transmitters modulation is necessary.

Question 7.
Mention the basic methods of modulation. (A.P. Mar. ’19, ’16, T.S. Mar. ’15)
Answer:
The basic methods of modulation are :

1. Amplitude modulation (AM)
2. Frequency modulation (FM)
3. Phase modulation (PM)

Question 8.
Which type of communication is employed in Mobile Phones ? (A.P. Mar. ’15)
Answer:
Space wave mode of propagation is employed in mobile phones.

Short Answer Questions

Question 1.
Draw the block diagram of a generalized communication system and explain it briefly.
Answer:
Every communication system has three essential elements :

1. Transmitter
2. Medium / Channel
3. Receiver.

The block diagram is shown in Fig.

In Communication system, the transmitter and the receiver are located at two different places separate from the transmitter.

The channel is the physical medium that connects. The purpose of transmitter is to convert the message signal produced by the source of information into a form, suitable for transmission through the channel. If the output of the information source is a non-electrical signal like a voice signal, a transducer converts it to electrical form before giving it as an input to the transmitter. When a transmitted signal propagates along the channel it may get distorted due to channel imperfection. Moreover, noise adds to the transmitted signal and the receiver receives a corrupted version of the transmitted signal. The receiver has the task of operating on the received signal. It reconstructs a recognizable form of the original message signal for delivering it to the user of information.

Question 2.
What is a Ground wave ? When is it used for communication ?
Answer:
Ground Wave : To radiate signals with high efficiency, The antennas should have a size comparable to the wavelength λ of the signal (at least ~ λ/4). At longer wavelengths (i.e., at lower frequencies), the antennas have large physical size and they are located on or very near to the ground. In standard AM broadcast, ground based vertical towers are generally used as transmitting antennas. For such antennas, ground has a strong influence on the propagation of the signal. The mode of propagation is called surface wave propagation and the wave glides over the surface of the earth. A wave induces current in the ground over which it passes and it is attenuated as a result of absorption of energy by the earth. The attenuation of surface waves increases very rapidly with increase in frequency. The maximum range of coverage depends on the transmitted power and frequency (less than 2 MHz). Ground waves will propagate long distances over sea water due to its high conductivity.

Question 3.
What are Sky Waves ? Explain Sky Wave propagation, briefly.
Answer:
Sky Waves : Long distance communication between two points on the earth is achieved through reflection of electromagnetic waves by Ionosphere, Such waves are called sky waves.

This mode of propagation is used by short wave broadcast services. The Ionosphere is so called because of the presence of a large number of ions or charged particles. It extends from a height of ~ 65 Km to about 400 km above the earth’s surface.

The degree of ionisation varies with the height. The density of atmosphere decreases with height. At greater heights the solar radiation is intense but there are few molecules to be ionized. Close to the earth, the radiation intensity is low so that the ionization is again low. However at some intermediate heights, there occurs a peak of ionization density. The ionospheric layer acts as a reflector for a certain range of frequencies (3 to 30 MHz).
Electromagnetic waves of frequencies higher than 30 MHz penetrate ionosphere and escape. This phenomenon is shown in the Fig.

The phenomenon of bending of em waves is so that they are diverted towards the earth which is similar to Total Internal Reflection in optics.

Question 4.
What is Space Wave Communication ? Explain.
Answer:
A spcae wave travels in a straight line from transmitting antenna to the receiving antenna. Space waves are used for line – of – sight (LOS) communication as well as satellite communication. At frequencies above 40 MHz, communication is essentially limited to line-of-sight paths. At these frequencies, the antennas are relatively smaller and can be placed at heights of many wavelengths above the ground. Because of line-of-sight nature of propagation, direct waves get blocked at some point by the curvature of the earth as illustrated in Fig. If the signal is to be received beyond the horizon then the receiving antenna must be high enough to intercept the line-of-sight waves.

If the transmitting antenna is at a height hT then it can be shown that the distance to the horizon d_T is given as d_T = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{T}}\right)}\) where R is the radius of the earth (approximately 6400 km). Similarly if the receiving antenna is at a height hR, the distance to the horizon d_R is d_R = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{R}}\right)}\). With reference to Fig. the maximum line-of sight distance d_M between the two antennas having heights h_T and h_R above the earth is given by d_M = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{T}}\right)}\) = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{R}}\right)}\)

Television broadcast, microwave links and satellite communication are some examples of communication systems that use space wave mode of propogation.

Question 5.
What do you understand by modulation ? Explain the need for modulation.
Answer:
The process of combining audio frequency signal with high frequency signal is called modulation.
To transmit an electronic signal in the audio frequency range (20 Hz to 20 KHz) over a long distance directly the following constraints limit the possibility :

1. Size of the antenna
2. Effective power radiated by the antenna
3. Mixing up of signals from different transmitters.

For 20 KHz signal the height of antenna is about 4km, still a large height. This makes antenna length impractical. Even if we transmit the signal, they may combine with low frequency signals present in the atmosphere and it is impossible to distinguish the signals at the receiving end.

In order to avoid these problems a low frequency audio signal is combined with high frequency signal to translate the audio signal to high frequencies.

Question 6.
What should be the size of the antenna or aerial ? How the power radiated is related to length of the antenna and wavelength ?
Answer:
Size of antenna (or) aerial: For trasmitting a signal, we need an antenna. The size of the antenna comparable to the wavelength of the signal (at least λ/4). So that the antenna properly senses the time variation of the signal. For an e.m waves of frequency 20 kHz, the wavelength λ is 15 km. Obviously such a long antenna is not possible to construct and operate. There is a need of translating the information contained in our original low frequency base band signal into high frequencies before transmission.

Effective power radiated by an antenna : A linear antenna (length l) show that the power radiated is proportional to \(\frac{l}{\lambda^2}\). For the same antenna length, the power radiated increases with decreasing λ. i.e., increasing frequency. Hence the effective power radiated by a long wave length base band signal would be small.

Question 7.
Explain amplitude modulation.
Answer:
Amplitude modulation (AM.) : In amplitude modulation, the amplitude of carrier wave varies, but frequency and phase remains constant.
Here we can explain AM using a sinusoidal signal as a modulating signal.

Let C(t) = A_c sin ω_c t represent carrier wave
m(t) = A_m sin ω_m t represent modulating signal.
The modulating C_m(t) can be written as
C_m(t) = (A_c + A_m sin ω_mt) sin ω_Ct
C_m(t) = A_c (1 + \(\frac{A_m}{A_c}\) sin ω_mt) sin ω_ct —— (1)
Where ω_m = 2πf_m is the angular frequency of message signai
Note that the modulated signal now contains the message signal.
C_m(t) = A_c sin ω_ct + µ A_c sin ω_mt sin ω_c t —– (2)
Where µ = \(\frac{A_m}{A_c}\) = Modulation index.
To avoid distortion keep . µ ≤ 1
C_m(t) = A_c sin ωt + \(\frac{\mu A_c}{2}\) cos (ω_c – ω_m) t – \(\frac{\mu \mathrm{A}_{\mathrm{c}}}{2}\) cos (ω_c + ω_m)t —— (3)
Here (ω_c – ω_m) and (ω_c + ω_m) are lower side and upper side frequencies.
As long as the broadcast frequencies (camer wave) are sufficiently spaced out the side bands donot over lap.

Question 8.
How can an amplitude modulated wave be generated?
Answer:

The modulated signal A_m sin ω_mt is added to the camer wave signal A_c sin ω_ct to produce the signal x(t).
x(t) = A_m sin ω_m t + A sin ω_ct is passed through a square law device produces an output.
y(t) = B x(t) + Cx² (t)
where B and C are constant.
This signal is passed through a band pass filter. The output of band pass filter produces AM wave, In band pass filter dc and the sinusoidal frequencies ω_m, 2ω_m and 2ω_c rejects and retains the frequency ω_c, (ω_c – ω_m) and (ω_c + ω_m)

Question 9.
How can an amplitude modulated wave be detected ?
Answer:

A block diagram of a typical receiver is shown in figure. Detection is the process of recovering the modulating signal from the modulated carrier wave.
We just saw that the modulated carrier wave contains the frequencies ω_c and ω_c ± ω_m.. In order to obtain message signal m(t) of angular frequency ω_m, a simple method is shown in figure.

The modulated signal is passed through a rectifier produces the output message signal. This message signal is passed through envelope detector (RC circuit).

Textual Exercises

Question 1.
Which of the following frequencies will be suitable for beyond the horizon communication using sky waves ?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Solution:
(b) 10 kHz frequencies cannot be radiated due to large antenna size, 1GHz and 1000 GHz will be generated. So option (b) is correct.

Question 2.
Frequencies in the UHF range normally propagate by means of:
(a) Ground waves
(b) Sky waves
(c) Surface waves
(d) Space waves
Answer:
(d). The frequencies in UHF range normally propagate by means.of space waves. The high frequency space does not bend with ground but are ideal for frequency modulation.

Question 3.
Digital signals
(i) do not provide a continuous set of values
(ii) represent values are discrete steps
(iii) can utilize binary system and
(iv) can utilize decimal as well as binary systems.
Which of the above statements are true ?
(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (i), (ii) and (iii) but not (iv)
(d) All of (i), (ii), (iii) and (iv).
Answer:
(c). A digital signal is a discontinuous function of time in contrast to an analogue signal. The digital signals can be stored as digital data and cannot be transmitted along the telephone lines. Digital signal cannot utilize decimal signals.

Question 4.
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication ? A TV transmitting antenna is 81m tall. How much service area can it cover if the receiving antenna is at the ground level ?
Answer:
No, it is not necessary for line of sight communication, the two antennas may not be at the same height.
Given, height of antenna h = 81 m.
Radius of earth R = 6.4 × 10⁶ m.
Area = πd²; Range, d = \(\sqrt{2 \mathrm{hR}}\)
∴ Service area = π × 2πR = \(\frac{22}{7}\) × 81 × 2 × 6.4 × 10⁶ = 3258.5 km².

Question 5.
A carrier wave of peak voltage 12V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75% ?
Answer:
Given, Peak voltage V₀ = 12V
Modulation index μ = 75% = \(\frac{75}{100}\)
We know that Modulation index (μ)

So, peak voltage of modulating signal V_m = μ × Peak voltage = \(\frac{75}{100}\) × 12 = 9V.

Question 6.
A modulating signal is a square wave, as shown in fig.
The carrier wave is given by c(t) = 2 sin (8πt) volts.
(i) Sketch the amplitude modulated waveform.
(ii) What is the modulation index ?

Answer:
Given, equation of carrier wave c(t) = 2 sin (8πt) —– (1)

(i) According to the diagram
Amplitude of modulating signal A_m = 1V
Amplitude of carrier wave A_c = 2V
T_M = 1s; Ωm = \(\frac{2 \pi}{T_m}\) = \(\frac{2 \pi}{1}\) = 2π rad/s —– (2)
From equation (1)
c(t) = 2 sin 8πt = A_c sin ω_c
From equation (2)
so, ω_c = 4ω_m
Amplitude of modulated wave A = A_m + A_c = 2 + 1 = 3V.
The sketch of amplitude modulated waveform is shown below.

(ii) Modulation index μ = \(\frac{A_m}{A_c}\) = 0.5

Question 7.
For an amplitude modulated wave, the maximum amplitude is found to be 10V while the ‘ minimum amplitude is found to be 2V. Determine the modulation index, μ. What would be the value of m if the minimum amplitude is zero volt ?
Answer:
Given, maximum amplitude A_max = 10V
Minimum amplitude A_min = 2V
Let A_c and A_m be amplitudes of carrier wave and signal wave
A_max = A_c + A_m = 10 —— (1)
and A_min = A_c – A_m = 2 —– (2)
Adding the equations (1) & (2) we get 2 A_c = 12; A_c = 6V; A_m = 10 – 6 = 4V
Modulation index μ = \(\frac{A_m}{A_c}\) = \(\frac{4}{6}\) = \(\frac{2}{3}\).
When the minimum amplitude is zero, then i.e., A_min = 0
A_c + A_m = 10 —– (3)
A_c – A_m = 0 —- (4)
By solving (3) & (4) we get 2 A_m = 10; A_m = 5; A_c = 5
Modulation index μ = \(\frac{A_m}{A_c}\) = \(\frac{5}{5}\) = 1.

Question 8.
Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.
Answer:
Let ω_c be the angular frequency of carrier waves & ω_m be the angular frequency of signal waves.
Let the signal received at the receiving station be e = E₁. cos (ω_c + ω_m) t
Let the instantaneous voltage of carrier wave e_c = E₀ cos ω_c t is available at receiving station.
Multiplying these two signals, we get
e × e_c = E₁E_c cos ω_ct. cos (ω_c + ω_m) t
E = \(\frac{E_1 E_c}{2}\) 2. cos ω_ct. cos (ω_c + ω_m) t. (Let e × e_c = E)
= \(\frac{\mathrm{E}_1 \mathrm{E}_{\mathrm{c}}}{2}\) [cos (ω_c + ω_c + ω_m) t + cos (ω_c + ω_m – ω_c) t]
∵ [2 cos A cos B = cos(A + B) + cos (A – B)]
\(\frac{E_1 E_c}{2}\) = [cos (2 (ω_c + ω_m)t + cos ω_mt]

Now, at the receiving end as the signal passes through filter, it will pass the high frequency (2ω_c + ω_m)t + cos ω_m) but obstruct the frequency ω_m. So we can record the modulating signal \(\frac{\mathrm{E}_1 \mathrm{E}_{\mathrm{c}}}{2}\) cos ω_mt which is a signal of angular frequency to ω_m.

Andhra Pradesh BIEAP AP Inter 2nd Year Accountancy Study Material 2nd Lesson Depreciation Textbook Questions and Answers.

AP Inter 2nd Year Accountancy Study Material 2nd Lesson Depreciation

Essay Questions

Question 1.
Define depreciation. What are the main causes of depreciation?
Answer:
Spicer and Pegler define depreciation as follows ‘Depreciation is the measure of exhaustion of the effective life of an asset from any cause during a given period. Depreciation means a decline in the value of fixed assets due to use, the passage of time, obsolescence, or any other cause.
Causes of depreciation: The main causes of depreciation include the following:

• Wear and tear: When fixed assets are put to use in business operations for earning revenue, the value of such assets may decrease. Such a decrease in the value of assets is said to be due to wear and tear.
• Physical forces: When the assets are exposed to the forces of nature like weather, winds, rains, etc. The value of such assets may decrease even if they are not being put to use.
• Expiration of legal rights: When the use of the assets like patents, copyrights, leases, etc., is governed by a time-bound agreement, the value of such assets may decrease with the passage of time.
• Obsolescence: Obsolescence implies an existing asset becoming out of date on account of the availability of a better type of asset due to technological changes or improvements in production methods.
• Accidents: A decline in the usefulness of the asset may be caused by accidents due to fire, earthquakes, floods, etc. Accidental loss is permanent.
• Depletion: Assets of wasting character such as mines, quarries, oil wells, etc., get depleted with the extraction of raw materials them.

Question 2.
Define depreciation. Explain the need of providing depreciation.
Answer:
“Depreciation is the diminution in the intrinsic value of the asset due to use and or the lapse of time”.
So, depreciation is a permanent, continuous, and gradual decrease in the book value of an asset due to various causes.
Need for Depreciation:

• To ascertain true profit or loss: The true profit or loss can be ascertained only when the depreciation is debited to the profit and loss account along with other revenue expenses like salaries, postage, etc. Which are incurred for the purpose of earning revenue.
• To disclose true and fair financial position: If reasonable depreciation is not deducted from the value of assets, the balance sheet will not reflect the true and fair financial position of business concerns. Hence depreciation should be provided every year to present a true and fair financial position.
• To have funds for the replacement of assets: A portion of profits is set aside in the form of depreciation every year. The amount accumulates and is available for replacement of the asset at the end of its life or when it is discarded.
• To ascertain the true cost of production: For ascertaining the cost of production, it is necessary to charge depreciation as an item of cost of production. If the depreciation is not charged, the cost records would not present the true cost of production.
• To fulfill legal requirements: In the case of joint stock companies, it is compulsory to provide depreciation on fixed assets. Without providing depreciation dividends cannot be declared.

Question 3.
Explain the meaning, merits, and demerits of the Straight Line Method.
Answer:
The straight-line method is the simplest and one of the most widely used methods of providing depreciation. This method is called as Fixed Instalment Method or Equal Instalment Method or Original Cost Method. Under this method, depreciation is calculated at a fixed percentage of the original value of the asset every year. Thus the amount of annual depreciation is uniform throughout the year.
The annual depreciation amount and rate of depreciation is calculated as follows.

Merits of straight-line method:

• It is veiy easy to understand.
• It is very simple to calculate depreciation.
• Assets can be depreciated up to net scrap value or zero value.
• This method is suitable for those assets whose estimated life can be estimated accurately.
• Depreciation will remain the same throughout the life of the asset.

Demerits of straight-line method:

• In this method, the depreciation amount will remain the same throughout the life of the asset.
• But in reality depreciation and repairs will be less in the earlier years and gradually increase in the later part of the life of the asset.
• It becomes difficult to ascertain the amount of depreciation if additions are made during the year.
• This method is not recognized by the income tax authorities.
• No provision is made for interest on the amount invested in the asset.
• It is very difficult to estimate the life of the asset.

Question 4.
Explain the meaning, merits, and demerits of the Reducing Balance Method.
Answer:
This method is known as Diminishing Balance Method or Written Down Value Method. Under this method, depreciation is charged at a fixed percentage on the book value of the asset. Since the book value keeps on reducing by the annual charge of depreciation, it is known as the reducing balance method.
The amount of depreciation decreases year after year. Depreciation is calculated on the diminishing value of the asset. Under this method, the amount of depreciation is larger in the earlier years than in the later years.
Merits:

• The total charge i.e., depreciation and repairs remain uniform year after year. In the earlier years, the amount of depreciation is more and the amount of repairs is less, whereas in later years, the depreciation amount is less and the amount of repairs is more.
• This method is logical in the sense that as an asset grows older, the amount of depreciation also goes on decreasing.
• Income Tax Act, 1961 accept this method for tax purpose.
• This method can be used where obsolescence is high.
• This method is suitable for those assets which have a longer life.

Demerits:

• It is difficult to calculate depreciation.
• The book value of the asset doesn’t become zero.
• It doesn’t take into account the interest on the amount invested in the asset.
• It doesn’t provide for the replacement of assets on the expiry of its life.

Question 5.
What are the differences between the Straight Line Method and the Reducing balance method?
Answer:

Basis of difference	Straight line method	Reducing balance method
1. Basis of calculation	Depreciation is calculated on the original cost.	Depreciation is calculated on the book value (i.e., original cost – depreciation charged till date).
2. Amount of depreciation	The amount of depreciation remains constant.	The amount of depreciation decreases year after year.
3. Total change against Profit and Loss a/c in respect of depreciation and repairs	The total charge against the Profit and Loss account in respect of depreciation and repairs expenses in later years under this method.	The depreciation charge declined in later years. Therefore total depreciation and repair expenses remain similar or equal every year.
4. Recognition by Income Tax Law	This method is not recognized by Income Tax Act.	This method is recognized by Income Tax Act.
5. Suitability	This method is suitable for assets in which repair charges are more, and the possibility of obsolescence is low.	This method is suitable for assets that are affected by technological changes and more repair expenses with the passage of time.

Short Answer Questions

Question 1.
What is depreciation?
Answer:
Depreciation is the gradual reduction or loss in the value of fixed assets like building plants, furniture, etc. If a company purchases a plant for ₹ 1,00,000 with an estimated life of 10 years, the plant will lose a value of ₹ 10,000 every year. This reduction loss is called depreciation.

Question 2.
What are the causes of depreciation?
Answer:
The main causes of depreciation are:

• Wear and tear
• Physical forces
• Expiration of legal rights
• Obsolescence
• Accidents
• Depletion

Question 3.
What is obsolescence?
Answer:
Obsolescence implies an existing asset becoming out of date on account of the availability of a better type of asset due to technological changes or improvements in production methods.

Question 4.
What is depletion?
Answer:
Assets of wasting in nature such as mines, quarries, etc., get depleted with the extraction of raw materials them.

Question 5.
Write different methods of providing depreciation.
Answer:
The following are the different methods of providing depreciation.

• Straight line method
• Reducing balance method
• Annuity method
• Depreciation fund method
• Insurance policy method
• Revaluation method
• Depletion method
• Machine hour rate method

Question 6.
What is the Straight Line Method?
Answer:
The straight-line method is the simple and widely used method of providing depreciation. Under this method, depreciation is calculated at a fixed percentage of the original cost of the asset every year.

Question 7.
What is the Reducing Balance Method?
Answer:
This method is also known as the written down value method. Under this method, depreciation is charged at a fixed percentage on the book value of the asset. Since the book value keeps on reducing by the annual charge of depreciation, it is known as the reducing balance method.

Textual Exercises

Question 1.
Praveen traders purchased a machine for ₹ 80,000. The life of the machine is estimated at 10 years and the residual value is ₹ 10,000. Calculate the annual amount of depreciation according to the Straight Line Method.
Solution:

Question 2.
A machine is purchased for ₹ 40,000. It is estimated that the useful life of the machine is 9 years and the residual value is ₹ 4,000. You are required to find out the annual amount of depreciation and the rate of depreciation under the Straight Line Method.
Solution:

Question 3.
A truck is purchased for ₹ 50,000. It is estimated that the useful life of the truck is 10 years and the residual value is ₹ 5,000. Calculate the annual amount of depreciation and the rate of depreciation under the Straight Line method.
Solution:

Question 4.
On 1st April 2010, Anand traders purchased a machine for ₹ 2,60,000 and spent ₹ 40,000 on its installation. It is estimated that working life is 10 years and after 10 years its scrap value will be ₹ 20,000. Books are closed on 31st March every year.
Write necessary journal entries and prepare machine accounts for the first three years in the books of Anand traders according to the Straight Line Method.
Solution:
Original cost = Purchase price + Installation expenses
= 2,60,000 + 40,000
= ₹ 3,00,000

Question 5.
On 1st July 2011, Neeharika & Co purchased a printing machine for ₹ 2,16,000 and spent ₹ 24,000 on its installation. It was estimated that the effective useful life of the printing machine will be 12 years and its scrap value will be ₹ 24,000. The books are closed on 31st December every year.
Prepare printing machine account and depreciation account for the first three years according to the Straight Line Method.
Solution:

Question 6.
Madan & Company purchased machinery on 1st January 2011 for ₹ 80,000 and spent ₹ 4,000 for its installation. The estimated life of the machinery is 10 years with a scrap value of ₹ 4,000. Books are closed on 31st December every year.
Calculate the amount of annual depreciation under the Straight Line Method and prepare the machinery account for the first three years.
Solution:



Note: 2 and 3 entries are to be passed for another three years.

Question 7.
On 1st January 2011, Raghavendra traders purchased Furniture for ₹ 60,000. Depreciation is to be calculated at the rate of 10% p.a. on Straight Line method. The books are closed on 31st December every year. Write necessary journal entries and prepare Furniture Account for the first four years.
Solution:

Question 8.
On 1st October 2011 Jagannadham & Sons purchased a machine for ₹ 90,000 and spent ₹ 10,000 for its installation. The books are closed on 31st March every year. The firm writes off depreciation at the rate of 10% on the original cost every year.
Prepare Machine Account and Depreciation Account for the first three years.
Solution:

Question 9.
Venugopal traders limited purchased machinery on 1st July 2010 for ₹ 50,000 and spent ₹ 2,000 on its installation. Depreciation is to be provided at @10% p.a. under the Straight Line Method. Books of account are closed on 31st December every year.
Show the machinery account for the first three years.
Solution:

Question 10.
On 1st January 2011 Suma purchased Furniture for ₹ 80,000. Depreciation is to be provided annually at 10% under the Straight Line Method. On 31st December 2013 furniture was sold for ₹ 40,000.
Show the Furniture Account assuming that the books are closed on 31st December every year.
Solution:

Question 11.
Suneetha traders purchased a second-hand machine for ₹ 72,000 on 1st January 2011 and spent ₹ 8,000 on repairs and installed the same. Depreciation is written off at 10% p.a.on the Straight Line Method. On 30th June 2013, the machine was sold for ₹ 50,000.
Prepare machinery accounts assuming that the accounts are closed on 31st December every year.
Solution:

Question 12.
Ranadheer & Co purchased a machine for ₹ 60,000 on 1st January 2011. Depreciation is calculated at @10% on the Straight Line Method. On 1st April 2013, the company sold the machine for ₹ 36,000.
Prepare machine accounts assuming that the accounts are closed on 31st December every year.
Solution:

Question 13.
On 1st January 2011, Siva traders purchased a second-hand machine for ₹ 40,000 and spent ₹ 5,000 on repairs and installed the same. It is estimated that the working life of the machine is 10 years and the scrap value is ₹ 2,500. On 31st December 2013, the machine was sold for ₹ 25,000. Prepare machine account assuming that the books are closed on 31st December every year according to the Straight Line Method.
Solution:

Question 14.
Manoj & Company purchased a second-hand machine for ₹ 18,000 on 1st April 2011 and spent ₹ 2,000 on repairs and installed the same. Depreciation is written off at 10% p.a. on the Straight Line Method. On 30th June 2013, it was sold for ₹ 13,000.
Prepare machine accounts assuming that the accounts are closed on 31st December every year.
Solution:

Question 15.
Ramesh & Co purchased machinery on 1st January 2011 for ₹ 3,00,000. On 1st September 2011, another machine was purchased for ₹ 4,20,000. Depreciation is provided on machinery at 10% p.a. on the Straight Line Method. Books are closed on 31st December every year.
Prepare machinery account for three years.
Solution:

Question 16.
Andhra sugars Ltd. purchased a plant for ₹ 1,00,000 on 1st January 2011. On 1st July of the same year, an additional plant was purchased for ₹ 50,000. On 1st October 2013, the plant purchased on 1st January 2011 has become obsolete and was sold for ₹ 60,000. On the same date, a fresh plant was purchased for ₹ 1,25,000. Depreciation is provided at 10% p.a. on the Straight Line Method.
Prepare plant account for the years assuming that the accounts are closed on 31st December every year.
Solution:

Question 17.
On 1st July 2010 Ganga & Co purchased a secondhand machine for ₹ 40,000 and spent ₹ 6,000 on repairs. On 1st January 2011, a new machine was purchased for ₹ 24,000. On 30th June 2012 the machine purchased on 1st January 2011 was sold for ₹ 16,000 and another machine was installed at a cost of ₹ 30,000. The company writes off depreciation @ 10% p.a. on the original cost every year on 31st March.
Show the machinery account for three years.
Solution:

Question 18.
Rama transport company purchased 6 trucks at ₹ 5,00,000 each on 1st January 2011. The company writes off the depreciation of @10% p.a. on the original cost. The books of account are closed on 31st December every year. On 1st July 2013, one of the trucks is involved in an accident and completely destroyed. A sum of ₹ 2,50,000 is received from the insurance company in full settlement. Prepare Trucks Account for the first three years.
Solution:

Reducing Balance Method

Question 19.
Kushal textile mills purchased machinery on 1st April 2011 for ₹ 4,00,000 and spent ₹ 20,000 for its installation. Depreciation is provided at @ 10% p.a. on the Reducing Balance Method. Books are closed on 31st March every year.
Write necessary journal entries and prepare Machinery Account and Depreciation Account for the first three years.
Solution:

Question 20.
On 1st July 2010, Pradeep & Co purchased machinery for ₹ 50,000. Depreciation is written off at the rate of 10% p.a. under the Reducing Balance Method. Show the Machinery Account for 3 years assuming that the books are closed on 31st December every year.
Solution:

Question 21.
On 1st January 2012, Siva & Co purchased second-hand machinery for ₹ 34,000 and spent ₹ 6,000 on its repairs and installed the same. On 31st December 2014, the machinery was sold for ₹ 26,000. The books are closed on 31st December every year. Depreciation is provided @ 10% p.a. on the Reducing Balance Method. Show the Machinery Account.
Solution:

Question 22.
On 1st January 2011, Geetha traders purchased a printing machine for ₹ 3,00,000. On 1st July 2013, the printing machine was sold for ₹ 1,30,000. Depreciation is provided @ 10% p.a. on the Reducing Balance Method. The books are closed on 31st December every year.
Prepare Printing Machine Account.
Solution:

Question 23.
Sravanthi enterprises purchased a machine for ₹ 40,000 on 1st July 2011 and spent ₹ 5,000 on its installation. Another Machine for ₹ 35,000 was purchased on 1st January 2013. Depreciation is charged @ 20% p.a. on the Reducing Balance Method. Books are closed on 31st March every year. Prepare Machinery Account for three years.
Solution:

Question 24.
On 1st January 2012, Swathi & Co purchased the plant for ₹ 3,00,000. On 1st October 2012, another plant was purchased for ₹ 1,00,000. Depreciation is charged @10% p.a. on the Reducing Balance Method. On 1st October 2013, the first plant was Sold for ₹ 2,20,000.
Prepare plant account for three years assuming that the accounts are closed on 31st December every year.
Solution:

Textual Examples

Illustrations of Straight Line Method

Question 1.
An Asset is purchased for ₹ 40,000. The useful life of the asset is 10 years and the residual value is ₹ 4,000. Find out the annual depreciation and the rate of depreciation under the straight-line method.
Solution:

Question 2.
Radha & Company purchased machinery for ₹ 45,000 on 1st Jan 2010. The estimated life of the machinery is 8 years and the residual value at the end of its life period is ₹ 5,000. The books are closed on 31st December every year.
Write journal entries and show the Machinery Account and Depreciation Account for 3 years on the Straight Line Method.
Solution:

Question 3.
Vasavi & Co purchased machinery for ₹ 80,000 on 1st January 2011. Depreciation is provided annually at 10% on the original cost every year. Die books are closed on 31st December every year.
Prepare Machinery Account for the first 3 years.
Solution:

Question 4.
Narayana and Bros purchased a plant for ₹ 2,00,000 on 1st January 2010 and spent ₹ 50,000 for its installation. The salvage value of the plant after its useful life of 10 years is estimated to be ₹ 20,000. The Books are closed on 31st December every year.
Prepare Plant account and Depreciation Account for the first 3 years using the Straight Line Method.
Solution:
The original cost of Plant = Purchase price + Installation expenses
= 2,00,000 + 50,000
= ₹ 2,50,000

Question 5.
Rama Rao and his sons purchased a machine for ₹ 1,40,000 on 1st July 2011 and spent ₹ 10,000 for its installation. The firm writes off depreciation at the rate of 10% on the original cost every year. The books are closed on December 31st every year.
Prepare Machine Account and Depreciation Account for the three years.
Solution:
1. Original cost of Machine = Purchase price + Installation expenses
= 1,40,000 + 10,000
= ₹ 1,50,000
2. Annual depreciation = 1,50,000 × \(\frac{10}{100}\) = ₹ 15,000
3. Depreciation for the year 2011 = the Asset is used only 6 months from 01-07-2011.
Depreciation shall be provided only for 6 months i.e. ₹ 7,500

Question 6.
On 1st January 2010, Sahithi & Co purchased a second-hand machine for ₹ 80,000 and spent ₹ 4,000 on carriage inwards, ₹ 40,000 as repair charges, and ₹ 2,000 on installation expenses. It is estimated that the machine will have a scrap value of ₹ 5,000 at end of its useful life which is 10 years. On 31st December 2012, the machine was sold for ₹ 50,000. The books are closed on 31st December every year.
Prepare Machine Account.
Solution:

Question 7.
Nagaraju & Co purchased a second-hand machine on 1st January 2011, for ₹ 45,000 and spent ₹ 5,000 on repairs and installed the same. Depreciation is provided at the rate of 10% p.a. under the Straight Line Method. On 1st July 2014, the machine was sold for ₹ 30,000. The books are closed on 31st December every year.
Prepare the Machine Account.
Solution:

Question 8.
Bhavani traders purchased a machine for ₹ 50,000 on 01-01-2010. Another machine was bought on 01-01-2011 for ₹ 60,000 and used from 1st July 2011 onwards. The depreciation is provided at 10% per annum under the Straight Line Method. The books are closed on 31st December every year.
Prepare the machinery account for 3 years.
Solution:

Working Notes:
1. Annual depreciation is calculated as under.
1st Machine = 50,000 × \(\frac{10}{100}\) = ₹ 5,000
2nd Machine = 60,000 × \(\frac{10}{100}\) = ₹ 6,000
2. The Second machine was purchased on 1st January 2011, but started working from 1st July 2011. Hence the depreciation is calculated for 6 months only, i.e. from the date of use to the closing date of the year 2011.
Depreciation on 2nd machine for 6 months = 60,000 × \(\frac{6}{12}\) = ₹ 3,000

Question 9.
On 1st July 2011, Anupama Traders purchased a machine for ₹ 80,000. On 1st April 2012, the firm purchased another machine for ₹ 40,000. on 31st March 2014, the machine which was purchased on 1st April 2012 was sold for ₹ 29,000. The firm writes off 10% depreciation on the original cost. The books are closed on 31st March every year.
Show the machinery account for three years.
Solution:

Question 10.
On 1st April 2011 Rajesh transport company purchased 4 trucks at ₹ 6,00,000 each. The company writes off depreciation @ 10% per annum on the original cost. On 1st July 2013, one of the trucks is involved in an accident and completely destroyed. Insurance company paid ₹ 3,00,000 in full settlement of the claim. The books are closed on 31st December every year.
Prepare trucks to account for three years.
Solution:

Difference between Straight Line Method and Reducing Balance Method

Question 11.
Nagarjuna & Co purchased plant and machinery for ₹ 70,000 on 1st January 2011 and spent ₹ 10,000. for installation expenses. Depreciation is to be provided at 10% on the Reducing Balance Method. Books are closed on 31st December every year.
Write the necessary journal entries and prepare the Plant and Machinery Account and Depreciation Account for three years.
Solution:
Journal Entries in the Books of Nagarjuna & Co

Question 12.
Sujatha enterprises purchased machinery on 1st April 2011 for ₹ 4,00,000. Depreciation is calculated @ 10% per annum on the Reducing Balance Method. Books are closed on 31st December every year.
Prepare Machinery Account for the first three years.
Solution:

Question 13.
Kiran enterprises purchased a printing machine for ₹ 80,000 on 1st July 2011 and spent ₹ 10,000 on its transport and installation expenses. Another machine for ₹ 70,000 was purchased on 1st January 2013. Depreciation is charged at the rate of 20% on the Reducing Balance Method. Books are closed on 31st March every year.
Prepare printing Machine Account for three years.
Solution:

Question 14.
On 1st July 2010, Venkatesh & Co purchased a machine for ₹ 40,000. On 30th June 2013, the machine was disposed of for ₹ 26,000. The books are closed on 31st December every year. Depreciation is to be calculated @ 10% per annum on the Reducing Balance Method.
Show the Machine Account.
Solution:

Question 15.
On 1st January 2011, Bhargava traders purchased machinery for ₹ 40,000. On 1st July of the same year, the firm purchased additional machinery for ₹ 20,000. On 1st July 2013, the machinery purchased on 1st January 2011 has become obsolete, it was sold for ₹ 32,000. The books are closed on 31st December every year.
Prepare Machinery Account for three years providing depreciation @ 10% p.a. on Reducing Balance Method.
Solution:

Question 16.
On 1st January 2010, Manjula & Co purchased a plant for ₹ 30,000. The company purchased another plant on 1st January 2011 for ₹ 28,000 and spent ₹ 2,000 on installation expenses. The books are closed on 31st December every year.
Show the plant account for the first three years providing depreciation at 10% on the first plant and 15% on the second plant on the Reducing Balance Method.
Solution: