12th Standard Physics English Medium - Current Electricity 1 Mark Book Back Question Paper and Answer Key 2022 - 2023 | QB365

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12th Standard Physics English Medium - Current Electricity 1 Mark Book Back Question Paper and Answer Key 2022 - 2023

Published on: 01/09/2022

QB365 provides a detailed and simple solution for every Possible Book Back Questions in Class 12 Physics Subject - Current Electricity, English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Std: 12th Standard - Physics

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Questions:

Multiple Choice Question

15 x 1 = 15

In Joule’s heating law, when R and t are constant, if the H is taken along the y axis and I² along the x axis, the graph is ______.

(a)

straight line

(b)

parabola

(c)

circle

(d)

ellipse

A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of ______.

(a)

each of them increases

(b)

each of them decreases

(c)

copper increases and germanium decreases

(d)

copper decreases and germanium increases

The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is ______.

(a)

0.2 Ω

(b)

0.5 Ω

(c)

0.8 Ω

(d)

1.0 Ω

The temperature coefficient of resistance of a wire is 0.00125 per °C. At 20°C, its resistance is 1 Ω. The resistance of the wire will be 2 Ω at ______.

(a)

800 °C

(b)

700 °C

(c)

850 °C

(d)

820 °C

What is the current drawn out from the battery?

(a)

1 A

(b)

2 A

(c)

3 A

(d)

4 A

There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?

(a)

1.5 Ω

(b)

2.5 Ω

(c)

3.5 Ω

(d)

4.5 Ω

In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW are connected. The voltage of electric mains is 220 V. The maximum capacity of the main fuse of the building will be ______.

(a)

14 A

(b)

8 A

(c)

10 A

(d)

12 A

In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60 W bulb for use in India is R, the resistance of a 60 W bulb for use in USA will be ______.

(a)

(b)

(c)

\(\frac{R}{4}\)

(d)

\(\frac{R}{2}\)

A wire connected to a power supply of 230 V has power dissipation P₁. Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case power dissipation is P₂. The ratio \(\frac{P_2}{P_1}\) is ______.

(a)

(b)

(c)

(d)

10.

Two wires of A and B with circular cross section made up of the same material with equal lengths. Suppose R_A = 3 R_B, then what is the ratio of radius of wire A to that of B?

(a)

(b)

\(\sqrt3\)

(c)

\(\frac{1}{\sqrt3}\)

(d)

\(\frac{1}{3}\)

11.

What is the value of resistance of the following resistor?

(a)

100 k Ω

(b)

10 k Ω

(c)

1 k Ω

(d)

1000 k Ω

12.

A carbon resistor of (47 ± 4.7 ) k Ω to be marked with rings of different colours for its identification. The colour code sequence will be ______.

(a)

Yellow – Green – Violet – Gold

(b)

Yellow – Violet – Orange – Silver

(c)

Violet – Yellow – Orange – Silver

(d)

Green – Orange – Violet - Gold

13.

A toaster operating at 240 V has a resistance of 120 Ω. The power is ______.

(a)

400 W

(b)

2 W

(c)

480 W

(d)

240 W

14.

A wire of resistance 2 ohms per meter is bent to form a circle of radius 1m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is

(a)

\(\pi \Omega\)

(b)

\(\frac{\pi}{2}\Omega\)

(c)

2\(\pi \Omega\)

(d)

\(\frac{\pi}{4}\Omega\)

15.

The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?

(a)

2 ohm

(b)

4 ohm

(c)

8 ohm

(d)

1 ohm

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Answers Key:

Multiple Choice Question

15 x 1 = 15

(a)
straight line

Resistivity ∝ temperature for conductor. so, copper → decreases
Resistivity ∝\(\frac{1}{\text {temperature for semiconductor}}\)
so, germanium → increases.

I = 0.2 A, R = 10 Ω, E = 2.1 V
\(I=\frac{ɛ}{R+r}\)
\(0.2=\frac{2.1}{10+r}\)
0.2 x (10 + r) = 2.1
2 + 0.2 r = 2.1
0.2 r = 2.1 - 2 = 0.1
Internal resistance, \(r=\frac{0.1}{0.2}=\frac{1}{2}\)
r = 0.5 Ω

R_t = R_o [1 + α(T₂ - T₁)
2 = 1[1 + 0.00125(T₂ - 293)
I = 5/4 x 10^-3(T₂ - 293)
T₂ = 1093 K
T₂ = 820 ^oC

\(\frac{1}{R_p}=\frac{1}{15}+\frac{1}{15}+\frac{1}{15}.\)
\(=\frac{1+1+1}{15}=\frac{3}{15}=\frac{1}{5}\)
∴ R_p = 5 Ω
\(\therefore I=\frac{V}{R_p}=\frac{5}{5}=1A\)

R_s = 3 + 2.5 + P = 5.5 + P
V = 9 V, I = 1.0 A
R_s = \(\frac{V}{I}=\frac{9}{1}= 9 \Omega\)
∴ 9 = 5.5 + P
∴ P = 9 - 5.5 = 3.5 Ω

Total power = 15 x 40 + 5 x 100 + 5 x 80 + 1000
= 600 + 500 + 400 + 1000
= 2500 W
P = VI, V = 220 V
\(I=\frac{P}{V}=\frac{2500}{220}=11.363 \ A\)
≃ 12 A

\(\mathrm{V}_1 =220 \mathrm{~V}, \quad \mathrm{P}_1=60 \mathrm{~W} \)
\(\mathrm{~V}_{\mathrm{U}} =110 \mathrm{~V}, \mathrm{P}_{\mathrm{U}}=60 \mathrm{~W} \)
\(P =\frac{V^2}{R} \Rightarrow R=\frac{V^2}{P} \)
\(\therefore R_l =\frac{V_I^2}{P_l} \text { Similarly, } \quad \mathrm{R}_U=\frac{V_U^2}{P_U} \)
\(R_l =\frac{220 \times 220}{60} \quad \mathrm{R}_{\mathrm{U}}=\frac{110 \times 110}{60} \)
\(R_I =\frac{48400}{60} \quad R_U=\frac{12100}{60} \)
\(\frac{R_U}{R_l} =\frac{12100}{60} \times \frac{60}{48400}=\frac{1}{4} \)
\(R_U =\frac{R_l}{4}=\frac{R}{4}\)

\(\mathrm{V}=230 \mathrm{~V} \)
\(P=\frac{V^2}{R} \text { since } \mathrm{V} \text { is same } \mathrm{P} \propto \frac{1}{R} \)
\(\frac{1}{R_2}=\frac{1}{\frac{R_1}{2}}+\frac{1}{\frac{R_1}{2}}=\frac{2}{R_1}+\frac{2}{R_1}=\frac{2+2}{R_1} \)
\(\frac{1}{R_2}=\frac{4}{R_1} \)
\(\therefore R_2=\frac{R_1}{4} \)
\(R_1=4 R_2 \)
\(\therefore \frac{P_2}{P_1}=\frac{R_1}{R_2}=\frac{4 R_2}{R_2}=4\)

10.

\(R \propto \frac{1}{A}, R \propto \frac{1}{r^2} \)
\(R_A \propto \frac{1}{r_A^2}, R_B \propto \frac{1}{r_B^2} \)
\(\frac{r_A}{r_B}=\left(\frac{R_B}{R_A}\right)^{1 / 2}=\left(\frac{R_B}{3 R_B}\right)^{1 / 2}=\frac{1}{3^{\frac{1}{2}}}=\frac{1}{\sqrt{3}}\)

11.

Brown - 1
Black - 0
Yellow - 10⁴
(∴ R = 10 x 10⁴ Ω = 100 kΩ)

12.

Yellow - 4
Violet - 7
Orange - 10³
Silver - Tolerance - 10%

13.

\(P=\frac{V^2}{R}=\frac{240 \times 240}{120}=480 \ W\)

14.

Total length = 2πr = 2π
∴ Resistance of each segment = 2π/2 = π Ω

15.

Resistance, \(R=\frac{V}{I}=\frac{4}{2}=2 \ ohm\)