AP 10th Class Physics 11th Lesson Questions and Answers Electricity

AP State Board new syllabus AP Board Solutions Class 10 Physics 11th Lesson Electricity Questions and Answers.

10th Class Physics 11th Lesson Electricity Questions and Answers

10th Class Physics 11th Lesson Questions and Answers (Exercise)

Question 1.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is –
a) 1/25
b) 1/5
c) 5
d) 25
Answer:
d) 25

Question 2.
Which of the following terms does not represent electrical power in a circuit ?
a) I²R
b) IR²
c) VI
d) V²/R
Answer:
b) IR²

Question 3.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be
a) 100 W
b) 75 W
c) 50 W
d) 25 W
Answer:
d) 25 W

Question 4.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be-
a) 1 : 2
b) 2 : 1
c) 1 : 4
d) 4 : 1
Answer:
e) 1 : 4

Question 5.
How is a voltmeter connected in the circuit to measure the potential difference . between two points ?
Answer:
A voltmeter is connected in parallel to a circuit with its +ve terminal to the point at higher potential and -ve terminal to the point at lower potential.

Question 6.
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10^-8 Ω m. What will be the length of this wire to make its resistance 10 Ω ? How much does the resistance change if the diameter is doubled ?
Answer:
Radius (r) = \(\frac{0.5}{2}\) = 0.25 mm = 0.025 cm
ρ = 1.6 × 10^-8 Ω m, R = 10 Ω, l = ?
As R = ρ\(\frac{l}{\mathrm{~A}}\) = ρ\(\frac{l}{\mathrm{~A}}\)
∴ l = \(\frac{\pi r^2 R}{\rho}\) = \(\frac{3.14 \times(0.025)^2 \times 10}{1.6 \times 10^{-6}}\) = 12265.625 cm = 122.6 m
Again R = ρ\(\frac{l}{\mathrm{~A}}\) = ρ\(\frac{l}{\pi \frac{\mathrm{~d}^2}{4}}\)
i.e. R ∝ \(\frac{1}{\mathrm{~d}^2}\)
Thus, when the diameter of wire is doubled, the resistance becomes one-fourth of the original value
New resistance = \(\frac{10}{4}\) = – 2.5 Ω.
Decrease in resistance = 10 Ω – 2.5 Ω = 7.5 Ω

Question 7.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –

1 (amperes)	0.5	1.0	2.0	3.0	4.0
V (volts)	1.6	3.4	6.7	10.2	13.2

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:

The slope of V – I graph is resistance
So, Resistance, R = \(\frac{\text { Change in V }}{\text { Change in I }}\) = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
= \(\frac{13.2 – 1.6}{4 – 0.5}\) = \(\frac{11.6}{3.5}\) = 3.314 Ω

Question 8.
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 niA in the circuit. Find the value of the resistance of the resistor.
Answer:
Potential difference = V = 12 V
Current, I = 2.5 mA = 2.5 × 10^-3 A
Resistance, R = ?
From the Ohm’s Law, V = IR
R = \(\frac{\mathrm{V}}{\mathrm{I}}\) = \(\frac{12}{2.5 \times 10^{-3}}\) = 4.8 × 10^-3Ω = 4.8 kΩ
Resistance = 4.8 kΩ

Question 9.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor ?
Answer:
Resistors 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω are connected in series.
Equivalent resistance, R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
Given, potential difference, V = 9.V
Current drawn I = \(\frac{\mathrm{V}}{\mathrm{I}}\) = \(\frac{9}{13.4}\) = 0.67 A
Since all the resistors are in series, the same current, i.e., 0.67 A flows through the 12 Ω resistor.

Question 10.
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line ?
Answer:
Suppose n resistances of 176 Ω are connected in parallel.
Answer:
Then, \(\frac{1}{\mathrm{R}}\) = \(\frac{1}{176}\) + \(\frac{1}{176}\) + \(\frac{1}{176}\) + ……….. n factors = \(\frac{n}{176}\)
or R = \(\frac{176}{n}\)Ω
By Ohm’s law R = \(\frac{\mathrm{V}}{\mathrm{I}}\)
\(\frac{176}{n}\) = \(\frac{220}{5}\)
n = \(\frac{176 \times 5}{220}\) = 4

Question 11.
Show how would you connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
Answer:
Here R₁ = R₂ = R₃ = 6 Ω.
i) When we connect R₁ in series with the parallel combination of R₂ and R₃ as shown fig.

The equivalent resistance is R = R₁ + \(\frac{\mathrm{R}_2 \mathrm{R}_3}{\mathrm{R}_2+\mathrm{R}_3}\) = 6 + \(\frac{6 \times 6}{6+6}\) = 6 + 3 = 9 Ω

ii) When we connect a series combination of R₁ and R₂ in parallel with R₃ as shown in fig., the equivalent resistance is R = \(\frac{12 \times 6}{12+6}\) = \(\frac{72}{18}\) = 4 Ω

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5A ?
Answer:
Potential difference, V = 220 V
Power of each bulb P = 10 W
Resistance of each bulb, R = \(\frac{V^2}{\mathrm{P}}\) = \(\frac{220 \times 220}{10}\) = 4840 Ω
Total resistance in the circuit R’ = \(\frac{\mathrm{V}}{\mathrm{I}}\) = \(\frac{220}{5}\) = 14Ω
Let n be the number of bulbs to be connected in parallel to obtain resistance R’
\(\frac{1}{\mathrm{R}^{\prime}}\) = \(\frac{1}{\mathrm{R}}\) + \(\frac{1}{\mathrm{R}}\) + …… n times
\(\frac{1}{\mathrm{R}^{\prime}}\) = \(\frac{\mathrm{n}}{\mathrm{R}}\)
n = \(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}\) = \(\frac{4840}{44}\)
Required number of bulbs = 110

Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Q resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases ?
Answer:
When used individually I = \(\frac{220}{24}\) = 9.16 A in both of them.
When used in series R_s = 24 + 24 = 48 Ω
⇒ I_s = \(\frac{220}{48}\)A = 4.58 A
When used in parallel R_P = \(\) = 12 Ω
⇒ I_P = \(\frac{220}{12}\)A = 18.3 A

Question 14.
Compare the power used in the 2 Ω resistor in each of the following circuits:
i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:
i) I = \(\frac{6}{1 + 2}\) = 2A
Since current flowing is same in both resistors, power used in 2 Ω resistor.
P₁ = I² R = (2)² × 2 = 8 W

ii) Since both 12 Ω and 2 Ω are in parallel to the 4V source.
Power used in 2 Ω resistor P₂ = \(\frac{\mathrm{V}^2}{\mathrm{R}}\) = \(\frac{4^2}{2}\) = \(\frac{16}{12}\) = 8 W
Comparison between the power used in both cases = \(\frac{\mathrm{P}_1}{\mathrm{P}_2}\) = \(\frac{8 \mathrm{~W}}{8 \mathrm{~W}}\) = 1

Question 15.
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V ?
Answer:
R₁₀₀ = \(\frac{220^2}{100}\), R₆₀ = \(\frac{220^2}{60}\)
Current drawn by 100 W bulb \(\frac{220}{\mathrm{R}_{100}}\) = \(\frac{100}{220}\) = 0.45A
Current drawn by 60 W bulb = \(\frac{220}{\mathrm{R}_{60}}\) = \(\frac{60}{220}\)A = 0.27A
Total current drawn from the line = 0.45 A + 0.27 A = 0.72 A

Question 16.
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster In 10 minutes?
Answer:
Energy consumed by 250 W TV set in 1h = 250 × 1 = 250 Wh
Energy consumed by 1200 W toaster ¡n 10 min = 1200 × \(\frac{10}{60}\) 200 Wh
∴ Energy consumed by TV set is more than the energy consumed by toaster in the given timings.

Question 17.
An electric heater of resistance 8 £2 draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Answer:
R = 8 Ω, I = 15 A, t = 2h
Rate of the heat developed = \(\frac{\mathrm{H}}{\mathrm{t}}=\frac{i^2 \mathrm{Rt}}{\mathrm{t}}\) = 15² × 8 = 225 × 8 = 1800 Js^-1

Question 18.
Explain the following.
a) Why is the tungsten used almost exclusively for filament of electric lamps ?
b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal ?
c) Why is the series arrangement not used for domestic circuits ?
d) How does the resistance of a wire vary with its area of cross-section ?
e) Why are copper and aluminium wires usually employed for electricity transmission ?
Answer:
a) It has high metling point and emits light at a high temperature.

b) It has more resistivity and less temperature coefficient of resistance.

c) i) All appliances do not get same potential in series arrangement,
ii) All appliances cannot be individually aparted.

d) R ∝ \(\frac{1}{\text { Area of cross-section }}\)

e) They are very good conductors of electricity.

10th Class PS 11th Lesson Questions and Answers (InText)

Page No. 242

Question 1.
What does an electric circuit mean ?
Answer:
A continuous and closed path along which an electric current flows is called an electric circuit.

Question 2.
Define the unit of current.
Answer:
If one coulomb of charge flows through any section of a conductor in one second, then current through it is said to be one ampere.

Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Answer:
Charge on one electron e = 1.6 × 10^-19 C
Total charge Q = 1 C
Number of electrons, n = \(\frac{\mathrm{Q}}{\mathrm{e}}\) = \(\frac{1 \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}\) = 6.25 × 10¹⁸

Page No. 244

Question 4.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
A battery.

Question 5.
What is meant by saying that the potential difference between two points is 1 V ?
Answer:
The potential difference between two points is 1 volt if one joule of work is done in – moving a positive charge of one coulomb from one point to the other point.

Question 6.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
Energy given by battery = charge x potential difference
= 1 C × 6 V = 6 J

Page No. 256

Question 7.
On what factors does the resistance of a conductor depend ? .
Answer:
The resistance of a conductor depends (i) on its length (ii) on its area of cross section (iii) on the nature of its material.

Question 8.
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
The current will flow more easily through a thick wire than a thin wire of the same material.
Larger the area of cross-section of a conductor, more is the case with which the electrons can move through the conductor. Hence smaller is the resistance of the conductor.

Question 9.
Let the resistance of, an electrical compound remains constant while the potential difference across the two ends of the compound decreases to half of its former value. What change will occur in the current through it ?
Answer:
When potential difference is halved, the current through the component also decreases to half of its intial value. This is in accordance with ohm’s law i.e. V ∝ I

Question 10.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal ?
Answer:
The coils of electric toasters and electric irons are made of alloys instead of pure metal due to the following reasons :

1. Alloys have higher resistivity than that of their constituent metals.
2. Alloys do not oxidies (or burn) readily at high temperatures.

Question 11.
Use the data in Table 11.2 to answer the following –
a) Which among.iron and mercury is a better conductor ?
b) Which material is the best conductor?
Answer:
a) Resistivity of iron = 10.0 × 10^-8 Ωm
Resistivity of mercury = 94.0 × 10^-8 Ωm
Thus iron is a better conductor because it has lower resistivity than mercury.
b) As silver has the lowest resistivity (= 1.60 × 10^-8 Ωm), so silver is the best conductor.

Question 12.
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer:
The required circuit diagram is given below.

Question 13.
Redraw the circuit of Question 12, putting in an ammeter to measure the çurrent through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer:
The required circuit diagram is given below.

Total voltage V = 3 × 2 = 6 V
Total resistance R = 5 + 8 + 12 = 25 Ω
Reading of ammeter, I = \(\frac{\mathrm{V}}{\mathrm{R}}\) = \(\frac{6}{25}\) = 0.24 A
Reading of voltmeter V = I × R = 0.24 × 12 = 2.88 V
The voltmeter will measure the potential difference across the 12 Ω resistor, which is 2.88 V.

Page No. 266

Question 14.
Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10⁶ Ω, (b) 1 Ω and 10³ Ω and 10⁶ Ω.
Answer:
When the resistances are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance.
a) Equivalent resistance R = \(\frac{1 \times 10^6}{1+10^6}\) < 1 Ω
b) Equivalent resistance R = \(\frac{1 \times 10^3 \times 10^6}{10^3 \times 10^6+1 \times 10^6+1 \times 10^3}\) < 1 Ω

Question 15.
An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances and what is the current through it ?
Answer:
Resistance of electrical lamp R₁ = 100 Ω
Resistance of toaster R₂ = 50 Ω
Resistance of water filter R₃ = 500 Ω
Equivalent resistance R_p of the three appliances connected in parallel is given by
\(\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3}\)
= \(\frac{1}{100}\) + \(\frac{1}{50}\) + \(\frac{1}{500}\) = \(\frac{16}{500}\)
R_p = \(\frac{500}{16}\) Ω = 31.25 Ω
Or
Resistance of electric iron = Equivalent resistance of the three appliances connected
in parallel = 31.25 Ω
Applied voltage V = 220 V
Current I = \(\frac{\mathrm{V}}{\mathrm{R}}\) = \(\frac{220 \mathrm{~V}}{31.25 \Omega}\) = 7.04 A

Page No. 268

Question 16.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ?
Answer:
The advantages of connecting electrical devices in parallel with the battery are as follows :

1. Each device gets the full battery voltage.
2. The parallel circuit divides the current through the electrical devices. Each device gets proper current depending on its resistance.
3. If one device is switched OFF / ON others are not affected.

Question 17.
How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω by connected to give a total resistance of (a) 4 Ω, (b) 1 Ω ?
Answer:
a) We can obtain a total resistance of 4 Ω by connecting the 2 Ω resistance in series with the parallel combination of 3 Ω and 6 Ω
R = R + \(\frac{\mathrm{R}_2 \mathrm{R}_3}{\mathrm{R}_2+\mathrm{R}_3}\) = 2 + \(\frac{3 \times 6}{3+6}\) 4 Ω

b) We can obtain a total resistance of 1 Ω by connecting resistance of 2 Ω, 3 Ω and 6 Ω in parallel
\(\frac{1}{R}\) = \(\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3}\) + \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1}{1}\)
or R = 1 Ω

Question 18.
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω,12 Ω, 24 Ω ?
Answer:
a) Highest resistance can be obtained by connecting the four coils in series.
Then, R = 4 + 8 + 12 + 24 = 48 Ω

b) Lowest resistance can be obtained by connecting the four coils in parallel.

Then, \(\frac{1}{R}\) = \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{12}\) + \(\frac{1}{24}\) = \(\frac{12}{24}\) = \(\frac{1}{2}\)
∴ R = 2 Ω

Page No. 270

Question 19.
Why does the cord of an electric heater not glow white the learning element does ?
Answer:
Both the cord and the heating element of an electric heater carry the same current. But the heating element becomes hot due to its high resistance (H ∝ I² Rt) and begins to glow. The cord remains cold due to its low resistance and does not glow.

Question 20.
Compare the heat generated while transferring 96,000 coulomb of charge in one hour through a potential difference of 50 V.
Answer:
Here (Q) = 96,000 C; (t) = 1 hour = 3600 sec.; V = 50 V;
Heat generated is H = VQ = 50 V × 96,000 C = 48,00,000 J

Question 21.
An electric iron of resistance 20 £2 takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:
Here R = 20 Ω; I = 5A; t = 30 sec.
Heat developed is H = I² Rt = 25 × 20 × 30 = 15,000J

Page No. 274

Question 22.
What determines the rate at which energy is delivered by a current ?
Answer:
Resistance of the circuit determines the rate at which energy is delivered by a current.

Question 23.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Here, I = 5 A; V = 220 V; t = 2 hrs = 7200 s
Power P = VI = 220 × 5 = 1100 W
Energy consumed = power × time
= 1100 W × 7200 S = 7920000 J

Example Problems [Textbook)

Question 1.
A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit. (T.B. Page No. 242)
Solution:
We are given, I = 0.5 A; t = 10 min = 600 s.
From Eq. 1 = Q/t we have
Q = It = 0.5 A × 600 s = 300 C

Question 2.
How much work is done in moving a charge of 2 C across two points having a potential difference 12 V? (T.B. Page No. 242)
Solution:
The amount of charge Q, that flows between two points at potential difference V (= 12 V) is 2 C. Thus, the amount of work W, done in moving the charge [from Eq. V = \(\frac{\mathrm{W}}{\mathrm{Q}}\) ] is
W = VQ = 12 V × 2 C = 24 J.

Question 3.
a) How much current will an electric bulb draw from a 220 V source, if the resistance
of the bulb fiiament is 1200 Ω ? (T.B. Page No. 254)
b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 Ω ?
Solution:
a) We are given V = 220 V; R = 1200 Ω.
From Eq. R = \(\frac{V}{I}\), we have the current I = \(\frac{220 \mathrm{~V}}{1200 \Omega}\) = 0.18 A

b) We are given,. V = 220 V, R = 100 Ω.
From Eq. R = \(\frac{V}{I}\), we have the current I = \(\frac{220 \mathrm{~V}}{100 \Omega}\) = 2.2 A.
Note the difference of current drawn by an electric bulb and electric heater from the same 220 V source!

Question 4.
The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V ? (T.B. Page No. 254)
Solution:
We are given, potential difference V = 60 V, current I = 4 A.
According to Ohm’s law, R = \(\frac{V}{I}\) = \(\frac{60 \mathrm{~V}}{4 \mathrm{~A}}\) = 15 Ω
When the potential difference is increased to 120 V the current is given by current \(\frac{V}{I}\) = \(\frac{120 \mathrm{~V}}{15 \Omega}\) = 8 A

Question 5.
Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature ? .Using Table 11.2, predict the material of the wire. (T.B. Page No. 256)
Solution:
We are given the resistance R of the wire = 26 Ω,
the diameter d = 0.3 mm = 3 × 10^-4 m, and
the length l of the wire = 1 m.
Therefore, from Eq. (R = ρ\(\frac{l}{\mathrm{~A}}\)),
the resistivity of the given metallic wire is ρ = (\(\frac{\mathrm{RA}}{l}\)) = (\(\frac{\mathrm{R} \pi \mathrm{~d}^2}{4 l}\))
Substitution of values in this gives ρ = 1.84 × 10^-6 Ω m
The resistivity of the metal at 20°C is 1.84 × 10^-6 Ω m. From Table 11.2, we see that this
is the resistivity of manganese.

Question 6.
A wire of given material having length 1 and area of cross-section A has a resistance of 4 Ω. What would be the resistance of another wire of the same material having length 1/2 and area of cross-section 2A? (T.B. Page No. 256)
Solution:
For first wire R₁ ρ\(\frac{l}{A}\) = 4Ω
Now for second wire R₂ = ρ\(\frac{l / 2}{A}\) = \(\frac{1}{4}\)ρ\(\frac{l}{A}\)
R₂ = \(\frac{1}{4}\)R₂
R₂ = 1 Ω
The resistance of the new wire is 1 Ω.

Question 7.
An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are con-nected to a 6 V battery (Figure). Calculate
(a) the total resistance of the circuit,
(b) the current through the circuit, and
(c) the potential difference across the electric lamp and conductor. (T.B. Page No. 260)

Solution:
The resistance of electric lamp, R₁ = 20 Ω,

The resistance of the conductor connected in series, R₂ = 4 Ω.
a) Then the total resistance in the circuit R = R₁ + R₂
R_S = 20 Ω + 4 Ω = 24 Ω.
The total potential difference across the two terminals of the battery V = 6 V.

b) Now by Ohm’s law, the current through the circuit is given by
I = \(\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{s}}}\) = \(\frac{6 \mathrm{~V}}{24 \Omega}\) = 0.25 V

c) Applying Ohm’s law to the electric lamp and conductor separately, we get potential
difference across the electric lamp, V₁ = 20 Ω × 0.25 A = 5 V;
and,
that across the conductor, V₂ = 4 Ω × 0.25 A = 1 V.

Suppose that we like to replace the series combination of electric lamp and conductor by a single and equivalent resistor. Its resistance must be such that a potential difference of 6 V across the battery terminals will cause a current of 0.25 A in the circuit. The resistance R of this equivalent resistor would be
R = \(\frac{V}{I}\) = \(\frac{6 \mathrm{~V}}{0.25 \mathrm{~A}}\) = 24 Ω
This is the total resistance of the series circuit it is equal to the sum of the two resistances.

Question 8.
In the circuit diagram given in Figure, suppose the resistors R₁, R₂ and R₃ have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have been connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance. (T.B. Page No. 264)

Solution:
R₁ = 5 Ω, R₂ = 10 Ω, and R₃ = 30 Ω.
Potential difference across the battery, V = 12 V.
This is also the potential difference across each of the individual resistor; therefore, to . calculate the current in the resistors, we use Ohm’s law.
The current I₁, through R₁ = [ltaex]\frac{\mathrm{V}}{\mathrm{R}_1}[/latex]
I₁ = \(\frac{12 \mathrm{~V}}{5 \Omega}\) = 2.4 A.

Thus, R_p = 3 Ω.

Question 9.
If in Figure, R₁ = 10 Ω, R₂ = 40 Ω, R₃ = 30 Ω, R₄ = 20 Ω, R₅ = 60 Ω, and a 12 V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit. (T.B. Page No. 266)
Solution:

Suppose we replace the parallel resistors R₁ and R₂ by an equivalent resistor of resistance, R’. Similarly we replace the parallel resistors R₃, R₄ and R₅ by an equivalent single resistor of resistance R”. Then using Eq. \(\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
we have
\(\frac{1}{\mathrm{R}^{\prime}}=\frac{1}{10}+\frac{1}{40}=\frac{5}{40}\) ; ’that is R’ = 8 Ω
Similarly, \(\frac{1}{\mathrm{R}^n}=\frac{1}{30}+\frac{1}{20}+\frac{1}{60}=\frac{6}{60}\)
that is, R” = 10 Ω.
a) Thus, the total resistance, R = R’ + R” = 18 Ω.

b) To calculate the current, we use Ohm’s law, and get I = \(\frac{V}{R}\) = \(\frac{12 \mathrm{~V}}{18 \Omega}\) = 0.67 A

Question 10.
An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case ? (T.B. Page No. 270)
Solution:
From Eq. P = \(\frac{V Q}{t}\) = VI, we know that the power input is P = VI
Thus the current I = \(\frac{P}{V}\)

a) When heating is at the maximum rate, I = \(\frac{840 \mathrm{~W}}{220 \mathrm{~V}}\) = 3.82 A;
and the resistance of the electric iron is R = \(\frac{V}{I}\) = \(\frac{220 \mathrm{~V}}{3.82 \mathrm{~A}}\) = 57.60 Ω.

b) When heating is at the minimum rate, I = \(\frac{360 \mathrm{~W}}{220 \mathrm{~V}}\) = 1.64 A;
and the resistance of the electric iron is R = \(\frac{V}{I}\) = \(\frac{220 \mathrm{~V}}{1.64 \mathrm{~A}}\) = 134.15 Ω

Question 11.
100 J of heat is produced each second in a 4 Ω resistance. Find the potential difference across the resistor. (T.B. Page No. 270)
Solution:
H = 100 J, R = 4 Ω t = 1 s, V = ?
From Eq. H = I² Rt we have the current through the resistor as
I = \(\sqrt{\left(\frac{\mathrm{H}}{\mathrm{Rt}}\right)}\) = \(\sqrt{\frac{100 \mathrm{~J}}{4 \Omega \times 1 \mathrm{~s}}}\) = 5 A
Thus the potential difference across the resistor, V [from Eq. V = IR ] is
V = IR = 5A × 4Ω = 20V.

Question 12.
An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb ? (T.B. Page No. 272)
Solution:
P = VI = 220 V × 0.50 A = 110 J/s = 110 W.

Question 13.
An electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kW h ? (T.B. Page No. 272)
Solution:
The total energy consumed by the refrigerator in 30 days would be 400 W × 8.0 hour/ day × 30 days = 96000 W h = 96 kW h
Thus the cost of energy to operate the refrigerator for 30 days is
96 kW h × Rs 3.00 per kW h = Rs 288.00

AP 10th Class Physical Science Chapter 11 Questions and Answers (Lab Activities)

Activity – 11.1 (Page. No. 246)

Question 1.
Write an activity to verify the Ohm’s Law.
Answer:
Aim : To study the potential difference across a conductor and current through it.
Apparatus required : Nichrome wire 0.5 m length, an ammeter, a voltmeter and three cells each of 1.5 V.
Procedure :

1. Set up a circuit as shown in Figure, consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.)
2. First use only one cell as the source in the circuit. Note the reading in the ammeter I, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given.
3. Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.
4. Repeat the above steps using three cells and then four cells in the circuit separately.
5. Calculate the ratio of V to I for each pair of potential difference V and current I.

S. No.	Number of cells used in the circuit (ampere)	Current through the nichrome wire, I (amplere)	Potential difference across the nichrome wire, V (volt)	V/I (volt/ampere)
1.	1
2.	2
3.	3
4.	4

Observation :

1. Voltmeter and ammeter reading increases as the number of cells increase in series.
2. Same value of V/I is obtained in each case.
3. V – I graph is a straight line passing through the origin of the graph, as shown in Figure.

Conclusion :

1. V/I is a constant ratio and equal to the resistance of the nichrome wire i.e., = V/I = R
2. Straight line nature of graph shows that the current is proportional to the potential difference.

Activity – 11.2 (Page. No. 248)

Question 2.
Write an activity to understand/study about electrical resistance of a conductor.
Answer:
Aim : To make a study of electrical conductivity of different materials or components.
Apparatus required : Nichrome wire, a torch bulb, a 10W bulb and an ammeter (0 – 5 A range), a plug key and some connecting wires.

Procedure :

1. Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in Figure.
2. Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. Take out the key from the plug.
3. Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.
4. Now repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.

Observation : With different components connected in the gap XY, we get different ammeter readings. These observations show that some components provide ah easy path to the flow of current, while other components opposed by its resistance.

Conclusion : A component of a given size which offers a low resistance is called a good conductor. A component of identical size which offers a relatively higher resistance is called a poor conductor. A component of the same size which offers a much higher resistance is called an insulator.

Question 3.
Write an activity to verify that the resistance of the conductor depends (i) on its length, (ii) on its area of cross-section, and (iii) on the nature of its material.
Answer:
Aim : To study the factors on which the resistance of a conducting wire depends.
Apparatus required : Cell, ammeter, nichrome wire and a plug key.
Procedure :

1. Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length l [say, marked (1)] and a plug key, as shown in Figure.
2. Now, plug the key. Note the current in the ammeter.
3. Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is 2l [marked (2) in the Figure].
4. Note the ammeter reading.
5. Now replace the wire by a thicker nichrome wire, of the same length l[marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
6. Instead of taking a nichrome wire, connect a copper wire [marked (4) in Figure] in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)]. Note the value of the current.
7. Notice the difference in the current in all cases.

Observations:

1. When the length of the wire is doubled, the ammeter reading decreases to one half of its initial value.
2. When we use a, thicker wire of the same material and of the same length, the current in the circuit increases.
3. When we use copper wire of similar dimensions in place of nichrome wire, the current in the circuit increases.

Conclusion : This activity helps us to conclude that the resistance of a conductor depends (i) on its length (ii) on its area of cross-section and (iii) on the nature of its material.

Activity – 11.4 (Page. No. 258)

Question 4.
What happens to the value of current when a number of resistors are connected in series in a circuit ? What would be their equivalent resistance ?Explain with an activity.
(OR)
How do you prove that in a series combination of resistors, the current is the same in every part of the circuit or the same current through each resistor ?
Answer:
Aim : To measure current in a series circuit by connecting ammeter in its different positions.
Apparatus required : Three resistors, battery, ammeter and plug key.
Procedure :

1. Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in Figure. You may use the resistors of values like 1 Ω, 2 Ω, 3 Ω etc., and a battery of 6 V for performing this Activity.
2. Plug the key. Note the ammeter reading.
3. Change the position of ammeter to anywhere in between the resistors. Note the ammeter reading each time.

Observations and Conclusions : It is observed that the value of the current in the ammeter is the same, independent of its position in the electric circuit. It means that in a series combination of resistors the current is the same in every part of the circuit or the same current flows through each resistor.

Activity – 11.5 (Page. No. 258)

Question 5.
Deduce a relationship between V, V₁, V₂ and V₃ in a series combination of three resistors through an activity.
(OR)
Prove that the total potential difference across a combination of resistors in series is equal to theisum of potential difference across the individual resistors.
Answer:
Aim : To measure potential difference across a series combination of resistors and across individual resistance and hence to deduce a relationship between them
Apparatus required : Three resistors, voltmeter, battery, ammeter and plug key.
Procedure :

1. Set up a circuit as shown in figure.
2. Plug the key in the circuit and note the voltmeter reading. It gives the potential difference across the series combination of resistors. Let it be V.
3. Now measure the potential difference across the two terminals of the battery. Compare the two values.
4. Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ends X and P of the first resistor, as shown in Figure.
5. Plug the key and measure the potential difference across the first resistor. Let it be V₁.
6. Similarly, measure the potential difference across the other two resistors, separately.
   Let these values be V₂ and V₃, respectively.
7. Deduce a relationship between V, V₁, V₂ and V₃.

Observation : It is observed that the potential difference V is equal to the sum of potential differences V₁, V₂ and V₃.

Conclusion : That is the total potential differences across a combination of resistors in series is equal to the sum of potential difference across the individual resistors. That is
V = V₁ + V₂ + V₃.

Activity – 11.6 (Page. No. 262)

Question 6.
Deduce I = I₁ + I₂ + I₃ for the parallel combination of resistors.
(OR)
How do you verify that the total current I, is equal to.the sum of the separate currents
through each branch of the combination, I = I₁ + I₂ + I₃ ?
Answer:
Aim : To measure potential difference and current across a parallel combination of resistors and through individual resistors and establish relationship between them.
Apparatus tequired : Three resistors, battery, voltmeter, plug key and ammeter.
Procedure :

• Make a parallel combination, XY of three resistors having resistances R₁, R₂ and R₃ respectively.
• Connect it with a battery, a plug key and an ammeter, as shown in Figure (a). A13o connect a voltmeter in parallel with the combination of resistors (Figure b).

• Plug the key and note the ammeter reading. Let the current be 1. Also take the voltmeter reading. It gives the potential difference V, across the combination.
• The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor.
• Take out the plug from the key. Remove the ammeter and voltmeter from the circuit.
  Insert the ammeter in series with the resistor R₁. Note the ammeter reading I₁.
• Similarly measure the currents through R₂ and R₃. Let these be I₂ and I₃, respectively.

Observation and Conclusion : It is observed that the total current 1 is equal to the sum of the separate currents through each branch of the combination. That is
I = I₁ + I₂ + I₃