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PHXII03:CURRENT ELECTRICITY

357137 Two bulbs marked $60 W, 220 V$ and $100 W, 220 V$ are connected in series and the series combination is now connected across a $220 V$ main supply. The power dissipiated in the circuit is

1

37.5 W

2

75 W

3

80 W

4

40 W

Explanation:

When connected in series, power dissipated is
$P_{S} = \frac{P_{1} P_{2}}{P_{1} + P_{2}} = \frac{100 \times 60}{100 + 60}$
$= \frac{6000}{160} = 37.5 W$

PHXII03:CURRENT ELECTRICITY

357138 An electric heater of resistance $6 Ω$ is run for $10 \min$ on a $120 V$ line. The energy liberated in this period of time is

1

7.2 \times 10^{3} J

2

14.4 \times 10^{5} J

3

43.2 \times 10^{4} J

4

28.8 \times 10^{4} J

Explanation:

Given, $R = 6 Ω, t = 10 min = 600 s$
$V = 120$ volt
Energy liberated $= \frac{V^{2}}{R} \cdot t = \frac{120 \times 120 \times 600}{6}$
$= 144 \times 10^{4} J = 14.4 \times 10^{5} J$

PHXII03:CURRENT ELECTRICITY

357139 Heat produced in a resistance if 1.5 $A$ current is passed through it is 500 $J$ in 20 sec. If the current is 3 $A$ , the heat produced in the same resistor in the same duration will be

1 500

J

2 1000

J

3 1500

J

4 2000

J

Explanation:

Heat Produced $= H = I^{2} R t$
$∴ \frac{H_{1}}{H_{2}} = {(\frac{I_{1}}{I_{2}})}^{2}$ ( $∵$ time $t$ is same in both cases)
$\Rightarrow H_{2} = 500 \times {(\frac{3}{1.5})}^{2} = 2000 J$

JEE - 2021

PHXII03:CURRENT ELECTRICITY

357140 Find the power supplied by 10 $V$ battery in the circuit shown in figure
supporting img

1

8 W

2

15 W

3

20 W

4

3 W

Explanation:

Let us distribute the potentials at different junctions of the circuit as shown,
supporting img
Writing KCL equation for $x, \frac{x}{5} + \frac{x - 10}{10} + \frac{x + 30}{10} + \frac{x + 5}{5} = 0$
$\frac{2 x + x - 10 + x + 30 + 2 x + 10}{10} = 0$
$\Rightarrow x = - 5 V$
Current through the batteries and power supplied by batteries, $I_{10 V} = \frac{x}{5} + \frac{x + 30}{10} = - 1 + 2.5 = 1.5 A$
and $P_{supplied 10 V} = ε I = 10 \times 1.5 = 15 W$

PHXII03:CURRENT ELECTRICITY

357137 Two bulbs marked $60 W, 220 V$ and $100 W, 220 V$ are connected in series and the series combination is now connected across a $220 V$ main supply. The power dissipiated in the circuit is

1

37.5 W

2

75 W

3

80 W

4

40 W

Explanation:

When connected in series, power dissipated is
$P_{S} = \frac{P_{1} P_{2}}{P_{1} + P_{2}} = \frac{100 \times 60}{100 + 60}$
$= \frac{6000}{160} = 37.5 W$

PHXII03:CURRENT ELECTRICITY

357138 An electric heater of resistance $6 Ω$ is run for $10 \min$ on a $120 V$ line. The energy liberated in this period of time is

1

7.2 \times 10^{3} J

2

14.4 \times 10^{5} J

3

43.2 \times 10^{4} J

4

28.8 \times 10^{4} J

Explanation:

Given, $R = 6 Ω, t = 10 min = 600 s$
$V = 120$ volt
Energy liberated $= \frac{V^{2}}{R} \cdot t = \frac{120 \times 120 \times 600}{6}$
$= 144 \times 10^{4} J = 14.4 \times 10^{5} J$

PHXII03:CURRENT ELECTRICITY

357139 Heat produced in a resistance if 1.5 $A$ current is passed through it is 500 $J$ in 20 sec. If the current is 3 $A$ , the heat produced in the same resistor in the same duration will be

1 500

J

2 1000

J

3 1500

J

4 2000

J

Explanation:

Heat Produced $= H = I^{2} R t$
$∴ \frac{H_{1}}{H_{2}} = {(\frac{I_{1}}{I_{2}})}^{2}$ ( $∵$ time $t$ is same in both cases)
$\Rightarrow H_{2} = 500 \times {(\frac{3}{1.5})}^{2} = 2000 J$

JEE - 2021

PHXII03:CURRENT ELECTRICITY

357140 Find the power supplied by 10 $V$ battery in the circuit shown in figure
supporting img

1

8 W

2

15 W

3

20 W

4

3 W

Explanation:

Let us distribute the potentials at different junctions of the circuit as shown,
supporting img
Writing KCL equation for $x, \frac{x}{5} + \frac{x - 10}{10} + \frac{x + 30}{10} + \frac{x + 5}{5} = 0$
$\frac{2 x + x - 10 + x + 30 + 2 x + 10}{10} = 0$
$\Rightarrow x = - 5 V$
Current through the batteries and power supplied by batteries, $I_{10 V} = \frac{x}{5} + \frac{x + 30}{10} = - 1 + 2.5 = 1.5 A$
and $P_{supplied 10 V} = ε I = 10 \times 1.5 = 15 W$

PHXII03:CURRENT ELECTRICITY

357137 Two bulbs marked $60 W, 220 V$ and $100 W, 220 V$ are connected in series and the series combination is now connected across a $220 V$ main supply. The power dissipiated in the circuit is

1

37.5 W

2

75 W

3

80 W

4

40 W

Explanation:

When connected in series, power dissipated is
$P_{S} = \frac{P_{1} P_{2}}{P_{1} + P_{2}} = \frac{100 \times 60}{100 + 60}$
$= \frac{6000}{160} = 37.5 W$

PHXII03:CURRENT ELECTRICITY

357138 An electric heater of resistance $6 Ω$ is run for $10 \min$ on a $120 V$ line. The energy liberated in this period of time is

1

7.2 \times 10^{3} J

2

14.4 \times 10^{5} J

3

43.2 \times 10^{4} J

4

28.8 \times 10^{4} J

Explanation:

Given, $R = 6 Ω, t = 10 min = 600 s$
$V = 120$ volt
Energy liberated $= \frac{V^{2}}{R} \cdot t = \frac{120 \times 120 \times 600}{6}$
$= 144 \times 10^{4} J = 14.4 \times 10^{5} J$

PHXII03:CURRENT ELECTRICITY

357139 Heat produced in a resistance if 1.5 $A$ current is passed through it is 500 $J$ in 20 sec. If the current is 3 $A$ , the heat produced in the same resistor in the same duration will be

1 500

J

2 1000

J

3 1500

J

4 2000

J

Explanation:

Heat Produced $= H = I^{2} R t$
$∴ \frac{H_{1}}{H_{2}} = {(\frac{I_{1}}{I_{2}})}^{2}$ ( $∵$ time $t$ is same in both cases)
$\Rightarrow H_{2} = 500 \times {(\frac{3}{1.5})}^{2} = 2000 J$

JEE - 2021

PHXII03:CURRENT ELECTRICITY

357140 Find the power supplied by 10 $V$ battery in the circuit shown in figure
supporting img

1

8 W

2

15 W

3

20 W

4

3 W

Explanation:

Let us distribute the potentials at different junctions of the circuit as shown,
supporting img
Writing KCL equation for $x, \frac{x}{5} + \frac{x - 10}{10} + \frac{x + 30}{10} + \frac{x + 5}{5} = 0$
$\frac{2 x + x - 10 + x + 30 + 2 x + 10}{10} = 0$
$\Rightarrow x = - 5 V$
Current through the batteries and power supplied by batteries, $I_{10 V} = \frac{x}{5} + \frac{x + 30}{10} = - 1 + 2.5 = 1.5 A$
and $P_{supplied 10 V} = ε I = 10 \times 1.5 = 15 W$

PHXII03:CURRENT ELECTRICITY

357137 Two bulbs marked $60 W, 220 V$ and $100 W, 220 V$ are connected in series and the series combination is now connected across a $220 V$ main supply. The power dissipiated in the circuit is

1

37.5 W

2

75 W

3

80 W

4

40 W

Explanation:

When connected in series, power dissipated is
$P_{S} = \frac{P_{1} P_{2}}{P_{1} + P_{2}} = \frac{100 \times 60}{100 + 60}$
$= \frac{6000}{160} = 37.5 W$

PHXII03:CURRENT ELECTRICITY

357138 An electric heater of resistance $6 Ω$ is run for $10 \min$ on a $120 V$ line. The energy liberated in this period of time is

1

7.2 \times 10^{3} J

2

14.4 \times 10^{5} J

3

43.2 \times 10^{4} J

4

28.8 \times 10^{4} J

Explanation:

Given, $R = 6 Ω, t = 10 min = 600 s$
$V = 120$ volt
Energy liberated $= \frac{V^{2}}{R} \cdot t = \frac{120 \times 120 \times 600}{6}$
$= 144 \times 10^{4} J = 14.4 \times 10^{5} J$

PHXII03:CURRENT ELECTRICITY

357139 Heat produced in a resistance if 1.5 $A$ current is passed through it is 500 $J$ in 20 sec. If the current is 3 $A$ , the heat produced in the same resistor in the same duration will be

1 500

J

2 1000

J

3 1500

J

4 2000

J

Explanation:

Heat Produced $= H = I^{2} R t$
$∴ \frac{H_{1}}{H_{2}} = {(\frac{I_{1}}{I_{2}})}^{2}$ ( $∵$ time $t$ is same in both cases)
$\Rightarrow H_{2} = 500 \times {(\frac{3}{1.5})}^{2} = 2000 J$

JEE - 2021

PHXII03:CURRENT ELECTRICITY

357140 Find the power supplied by 10 $V$ battery in the circuit shown in figure
supporting img

1

8 W

2

15 W

3

20 W

4

3 W

Explanation:

Let us distribute the potentials at different junctions of the circuit as shown,
supporting img
Writing KCL equation for $x, \frac{x}{5} + \frac{x - 10}{10} + \frac{x + 30}{10} + \frac{x + 5}{5} = 0$
$\frac{2 x + x - 10 + x + 30 + 2 x + 10}{10} = 0$
$\Rightarrow x = - 5 V$
Current through the batteries and power supplied by batteries, $I_{10 V} = \frac{x}{5} + \frac{x + 30}{10} = - 1 + 2.5 = 1.5 A$
and $P_{supplied 10 V} = ε I = 10 \times 1.5 = 15 W$

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