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Current Electricity

152726 The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

1

20 Ω

2

15 Ω

3

10 Ω

4

30 Ω

Explanation:

C Given,
$V = 10 V$
$P = 30 W$
original image
The equivalent resistance of the given circuit is,
$R_{eq} = \frac{5 R}{5 + R}$
Current Electricity
Power dissipated in the given circuit is,
$P = \frac{V^{2}}{R_{eq}}$
$30 = \frac{(10)^{2}}{(\frac{5 R}{5 + R})} = \frac{100 (5 + R)}{5 R}$
$150 R = 500 + 100 R$
$150 R - 100 R = 500$
$50 R = 500$
$R = \frac{500}{50} = 10 Ω$

CG PET-2007

Current Electricity

152727 An $1800 W$ toaster, a $1.3 kW$ electric fan and $100 W$ lamp are plugged in the same $120 V$ circuit i.e., all the three devices are in parallel. What is the approximate value of the total current (i.e. sum of the current drawn by the three devices) through circuit?

1

40 A

2

120 A

3

18 A

4

27 A

Explanation:

D Given,
$P_{1} = 1800 W$
$P_{2} = 1.3 kW = 1.3 \times 1000 = 1300 W$
$P_{3} = 100 W$
$V = 120 V$
Power consume,
$P = P_{1} + P_{2} + P_{3}$
Power, $P = V \times I$
From equation (i), we get
$VI = P_{1} + P_{2} + P_{3}$
$120 \times I = 1800 + 1300 + 100$
$120 \times I = 3200$
$I = \frac{3200}{120} = 26.66 A = 27 A$

UPSEE - 2016

Current Electricity

152728 A current of $3 A$ flows through the $2 Ω$ resistor shown in the circuit. The power dissipated in the $5 Ω$ resistor is

1

4 W

2

2 W

3

1 W

4

5 W

Explanation:

D Given, current, I = $3 A$
Voltage across $2 Ω$ resistance
$V = 2 \times 3$
$V = 6 V$
Voltage across $4 Ω$ and $(5 Ω + 1 Ω)$ resister is same,
So, current across $5 Ω = \frac{6}{1 + 5} = 1 A (∵ I = \frac{V}{R})$
Thus, power across $5 Ω$
$P = I^{2} R$
$P = (1)^{2} \times 5$
$P = 5 W$

UPSEE - 2014

Current Electricity

152730 A $4 μ F$ condenser is charged to $400 V$ and then its plates are joined through a resistance. The heat produced in the resistance is :

1

0.16 J

2

0.32 J

3

0.64 J

4

1.28 J

Explanation:

B Given, $C = 4 μ F = 4 \times 10^{- 6} F, V = 400 V$
We know that,
Heat produced $=$ energy stored.
$U = H = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 4 \times 10^{- 6} \times (400)^{2}$
$= \frac{1}{2} \times 4 \times 10^{- 6} \times 160000$
$= 2 \times 10^{- 6} \times 160000 = 0.32 J$

UPSEE - 2004

Current Electricity

152731 Two bulbs of equal power are connected in parallel and they totally consume $110 W$ at 220 $V$ . The resistances of each bulb is

1

550 Ω

2

440 Ω

3

330 Ω

4

880 Ω

5

660 Ω

Explanation:

D Given,
We know that,
$V = 220 V$
Bulbs are connected in parallel and they totally consume $110 W$ to $220 V$
$P_{net} = P_{1} + P_{2}$
$P + P = 110 W$
$P = 55 W$
${∵ P_{1} = P_{2} = P}$
$∵ P = \frac{V^{2}}{R}$
$R = \frac{V^{2}}{P}$
$R = \frac{220 \times 220}{55} = 880 Ω$

Kerala CEE - 2016

Current Electricity

152726 The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

1

20 Ω

2

15 Ω

3

10 Ω

4

30 Ω

Explanation:

C Given,
$V = 10 V$
$P = 30 W$
original image
The equivalent resistance of the given circuit is,
$R_{eq} = \frac{5 R}{5 + R}$
Current Electricity
Power dissipated in the given circuit is,
$P = \frac{V^{2}}{R_{eq}}$
$30 = \frac{(10)^{2}}{(\frac{5 R}{5 + R})} = \frac{100 (5 + R)}{5 R}$
$150 R = 500 + 100 R$
$150 R - 100 R = 500$
$50 R = 500$
$R = \frac{500}{50} = 10 Ω$

CG PET-2007

Current Electricity

152727 An $1800 W$ toaster, a $1.3 kW$ electric fan and $100 W$ lamp are plugged in the same $120 V$ circuit i.e., all the three devices are in parallel. What is the approximate value of the total current (i.e. sum of the current drawn by the three devices) through circuit?

1

40 A

2

120 A

3

18 A

4

27 A

Explanation:

D Given,
$P_{1} = 1800 W$
$P_{2} = 1.3 kW = 1.3 \times 1000 = 1300 W$
$P_{3} = 100 W$
$V = 120 V$
Power consume,
$P = P_{1} + P_{2} + P_{3}$
Power, $P = V \times I$
From equation (i), we get
$VI = P_{1} + P_{2} + P_{3}$
$120 \times I = 1800 + 1300 + 100$
$120 \times I = 3200$
$I = \frac{3200}{120} = 26.66 A = 27 A$

UPSEE - 2016

Current Electricity

152728 A current of $3 A$ flows through the $2 Ω$ resistor shown in the circuit. The power dissipated in the $5 Ω$ resistor is

1

4 W

2

2 W

3

1 W

4

5 W

Explanation:

D Given, current, I = $3 A$
Voltage across $2 Ω$ resistance
$V = 2 \times 3$
$V = 6 V$
Voltage across $4 Ω$ and $(5 Ω + 1 Ω)$ resister is same,
So, current across $5 Ω = \frac{6}{1 + 5} = 1 A (∵ I = \frac{V}{R})$
Thus, power across $5 Ω$
$P = I^{2} R$
$P = (1)^{2} \times 5$
$P = 5 W$

UPSEE - 2014

Current Electricity

152730 A $4 μ F$ condenser is charged to $400 V$ and then its plates are joined through a resistance. The heat produced in the resistance is :

1

0.16 J

2

0.32 J

3

0.64 J

4

1.28 J

Explanation:

B Given, $C = 4 μ F = 4 \times 10^{- 6} F, V = 400 V$
We know that,
Heat produced $=$ energy stored.
$U = H = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 4 \times 10^{- 6} \times (400)^{2}$
$= \frac{1}{2} \times 4 \times 10^{- 6} \times 160000$
$= 2 \times 10^{- 6} \times 160000 = 0.32 J$

UPSEE - 2004

Current Electricity

152731 Two bulbs of equal power are connected in parallel and they totally consume $110 W$ at 220 $V$ . The resistances of each bulb is

1

550 Ω

2

440 Ω

3

330 Ω

4

880 Ω

5

660 Ω

Explanation:

D Given,
We know that,
$V = 220 V$
Bulbs are connected in parallel and they totally consume $110 W$ to $220 V$
$P_{net} = P_{1} + P_{2}$
$P + P = 110 W$
$P = 55 W$
${∵ P_{1} = P_{2} = P}$
$∵ P = \frac{V^{2}}{R}$
$R = \frac{V^{2}}{P}$
$R = \frac{220 \times 220}{55} = 880 Ω$

Kerala CEE - 2016

Current Electricity

152726 The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

1

20 Ω

2

15 Ω

3

10 Ω

4

30 Ω

Explanation:

C Given,
$V = 10 V$
$P = 30 W$
original image
The equivalent resistance of the given circuit is,
$R_{eq} = \frac{5 R}{5 + R}$
Current Electricity
Power dissipated in the given circuit is,
$P = \frac{V^{2}}{R_{eq}}$
$30 = \frac{(10)^{2}}{(\frac{5 R}{5 + R})} = \frac{100 (5 + R)}{5 R}$
$150 R = 500 + 100 R$
$150 R - 100 R = 500$
$50 R = 500$
$R = \frac{500}{50} = 10 Ω$

CG PET-2007

Current Electricity

152727 An $1800 W$ toaster, a $1.3 kW$ electric fan and $100 W$ lamp are plugged in the same $120 V$ circuit i.e., all the three devices are in parallel. What is the approximate value of the total current (i.e. sum of the current drawn by the three devices) through circuit?

1

40 A

2

120 A

3

18 A

4

27 A

Explanation:

D Given,
$P_{1} = 1800 W$
$P_{2} = 1.3 kW = 1.3 \times 1000 = 1300 W$
$P_{3} = 100 W$
$V = 120 V$
Power consume,
$P = P_{1} + P_{2} + P_{3}$
Power, $P = V \times I$
From equation (i), we get
$VI = P_{1} + P_{2} + P_{3}$
$120 \times I = 1800 + 1300 + 100$
$120 \times I = 3200$
$I = \frac{3200}{120} = 26.66 A = 27 A$

UPSEE - 2016

Current Electricity

152728 A current of $3 A$ flows through the $2 Ω$ resistor shown in the circuit. The power dissipated in the $5 Ω$ resistor is

1

4 W

2

2 W

3

1 W

4

5 W

Explanation:

D Given, current, I = $3 A$
Voltage across $2 Ω$ resistance
$V = 2 \times 3$
$V = 6 V$
Voltage across $4 Ω$ and $(5 Ω + 1 Ω)$ resister is same,
So, current across $5 Ω = \frac{6}{1 + 5} = 1 A (∵ I = \frac{V}{R})$
Thus, power across $5 Ω$
$P = I^{2} R$
$P = (1)^{2} \times 5$
$P = 5 W$

UPSEE - 2014

Current Electricity

152730 A $4 μ F$ condenser is charged to $400 V$ and then its plates are joined through a resistance. The heat produced in the resistance is :

1

0.16 J

2

0.32 J

3

0.64 J

4

1.28 J

Explanation:

B Given, $C = 4 μ F = 4 \times 10^{- 6} F, V = 400 V$
We know that,
Heat produced $=$ energy stored.
$U = H = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 4 \times 10^{- 6} \times (400)^{2}$
$= \frac{1}{2} \times 4 \times 10^{- 6} \times 160000$
$= 2 \times 10^{- 6} \times 160000 = 0.32 J$

UPSEE - 2004

Current Electricity

152731 Two bulbs of equal power are connected in parallel and they totally consume $110 W$ at 220 $V$ . The resistances of each bulb is

1

550 Ω

2

440 Ω

3

330 Ω

4

880 Ω

5

660 Ω

Explanation:

D Given,
We know that,
$V = 220 V$
Bulbs are connected in parallel and they totally consume $110 W$ to $220 V$
$P_{net} = P_{1} + P_{2}$
$P + P = 110 W$
$P = 55 W$
${∵ P_{1} = P_{2} = P}$
$∵ P = \frac{V^{2}}{R}$
$R = \frac{V^{2}}{P}$
$R = \frac{220 \times 220}{55} = 880 Ω$

Kerala CEE - 2016

Current Electricity

152726 The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

1

20 Ω

2

15 Ω

3

10 Ω

4

30 Ω

Explanation:

C Given,
$V = 10 V$
$P = 30 W$
original image
The equivalent resistance of the given circuit is,
$R_{eq} = \frac{5 R}{5 + R}$
Current Electricity
Power dissipated in the given circuit is,
$P = \frac{V^{2}}{R_{eq}}$
$30 = \frac{(10)^{2}}{(\frac{5 R}{5 + R})} = \frac{100 (5 + R)}{5 R}$
$150 R = 500 + 100 R$
$150 R - 100 R = 500$
$50 R = 500$
$R = \frac{500}{50} = 10 Ω$

CG PET-2007

Current Electricity

152727 An $1800 W$ toaster, a $1.3 kW$ electric fan and $100 W$ lamp are plugged in the same $120 V$ circuit i.e., all the three devices are in parallel. What is the approximate value of the total current (i.e. sum of the current drawn by the three devices) through circuit?

1

40 A

2

120 A

3

18 A

4

27 A

Explanation:

D Given,
$P_{1} = 1800 W$
$P_{2} = 1.3 kW = 1.3 \times 1000 = 1300 W$
$P_{3} = 100 W$
$V = 120 V$
Power consume,
$P = P_{1} + P_{2} + P_{3}$
Power, $P = V \times I$
From equation (i), we get
$VI = P_{1} + P_{2} + P_{3}$
$120 \times I = 1800 + 1300 + 100$
$120 \times I = 3200$
$I = \frac{3200}{120} = 26.66 A = 27 A$

UPSEE - 2016

Current Electricity

152728 A current of $3 A$ flows through the $2 Ω$ resistor shown in the circuit. The power dissipated in the $5 Ω$ resistor is

1

4 W

2

2 W

3

1 W

4

5 W

Explanation:

D Given, current, I = $3 A$
Voltage across $2 Ω$ resistance
$V = 2 \times 3$
$V = 6 V$
Voltage across $4 Ω$ and $(5 Ω + 1 Ω)$ resister is same,
So, current across $5 Ω = \frac{6}{1 + 5} = 1 A (∵ I = \frac{V}{R})$
Thus, power across $5 Ω$
$P = I^{2} R$
$P = (1)^{2} \times 5$
$P = 5 W$

UPSEE - 2014

Current Electricity

152730 A $4 μ F$ condenser is charged to $400 V$ and then its plates are joined through a resistance. The heat produced in the resistance is :

1

0.16 J

2

0.32 J

3

0.64 J

4

1.28 J

Explanation:

B Given, $C = 4 μ F = 4 \times 10^{- 6} F, V = 400 V$
We know that,
Heat produced $=$ energy stored.
$U = H = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 4 \times 10^{- 6} \times (400)^{2}$
$= \frac{1}{2} \times 4 \times 10^{- 6} \times 160000$
$= 2 \times 10^{- 6} \times 160000 = 0.32 J$

UPSEE - 2004

Current Electricity

152731 Two bulbs of equal power are connected in parallel and they totally consume $110 W$ at 220 $V$ . The resistances of each bulb is

1

550 Ω

2

440 Ω

3

330 Ω

4

880 Ω

5

660 Ω

Explanation:

D Given,
We know that,
$V = 220 V$
Bulbs are connected in parallel and they totally consume $110 W$ to $220 V$
$P_{net} = P_{1} + P_{2}$
$P + P = 110 W$
$P = 55 W$
${∵ P_{1} = P_{2} = P}$
$∵ P = \frac{V^{2}}{R}$
$R = \frac{V^{2}}{P}$
$R = \frac{220 \times 220}{55} = 880 Ω$

Kerala CEE - 2016

Current Electricity

152726 The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

1

20 Ω

2

15 Ω

3

10 Ω

4

30 Ω

Explanation:

C Given,
$V = 10 V$
$P = 30 W$
original image
The equivalent resistance of the given circuit is,
$R_{eq} = \frac{5 R}{5 + R}$
Current Electricity
Power dissipated in the given circuit is,
$P = \frac{V^{2}}{R_{eq}}$
$30 = \frac{(10)^{2}}{(\frac{5 R}{5 + R})} = \frac{100 (5 + R)}{5 R}$
$150 R = 500 + 100 R$
$150 R - 100 R = 500$
$50 R = 500$
$R = \frac{500}{50} = 10 Ω$

CG PET-2007

Current Electricity

152727 An $1800 W$ toaster, a $1.3 kW$ electric fan and $100 W$ lamp are plugged in the same $120 V$ circuit i.e., all the three devices are in parallel. What is the approximate value of the total current (i.e. sum of the current drawn by the three devices) through circuit?

1

40 A

2

120 A

3

18 A

4

27 A

Explanation:

D Given,
$P_{1} = 1800 W$
$P_{2} = 1.3 kW = 1.3 \times 1000 = 1300 W$
$P_{3} = 100 W$
$V = 120 V$
Power consume,
$P = P_{1} + P_{2} + P_{3}$
Power, $P = V \times I$
From equation (i), we get
$VI = P_{1} + P_{2} + P_{3}$
$120 \times I = 1800 + 1300 + 100$
$120 \times I = 3200$
$I = \frac{3200}{120} = 26.66 A = 27 A$

UPSEE - 2016

Current Electricity

152728 A current of $3 A$ flows through the $2 Ω$ resistor shown in the circuit. The power dissipated in the $5 Ω$ resistor is

1

4 W

2

2 W

3

1 W

4

5 W

Explanation:

D Given, current, I = $3 A$
Voltage across $2 Ω$ resistance
$V = 2 \times 3$
$V = 6 V$
Voltage across $4 Ω$ and $(5 Ω + 1 Ω)$ resister is same,
So, current across $5 Ω = \frac{6}{1 + 5} = 1 A (∵ I = \frac{V}{R})$
Thus, power across $5 Ω$
$P = I^{2} R$
$P = (1)^{2} \times 5$
$P = 5 W$

UPSEE - 2014

Current Electricity

152730 A $4 μ F$ condenser is charged to $400 V$ and then its plates are joined through a resistance. The heat produced in the resistance is :

1

0.16 J

2

0.32 J

3

0.64 J

4

1.28 J

Explanation:

B Given, $C = 4 μ F = 4 \times 10^{- 6} F, V = 400 V$
We know that,
Heat produced $=$ energy stored.
$U = H = \frac{1}{2} {CV}^{2}$
$= \frac{1}{2} \times 4 \times 10^{- 6} \times (400)^{2}$
$= \frac{1}{2} \times 4 \times 10^{- 6} \times 160000$
$= 2 \times 10^{- 6} \times 160000 = 0.32 J$

UPSEE - 2004

Current Electricity

152731 Two bulbs of equal power are connected in parallel and they totally consume $110 W$ at 220 $V$ . The resistances of each bulb is

1

550 Ω

2

440 Ω

3

330 Ω

4

880 Ω

5

660 Ω

Explanation:

D Given,
We know that,
$V = 220 V$
Bulbs are connected in parallel and they totally consume $110 W$ to $220 V$
$P_{net} = P_{1} + P_{2}$
$P + P = 110 W$
$P = 55 W$
${∵ P_{1} = P_{2} = P}$
$∵ P = \frac{V^{2}}{R}$
$R = \frac{V^{2}}{P}$
$R = \frac{220 \times 220}{55} = 880 Ω$

Kerala CEE - 2016

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