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PHXII03:CURRENT ELECTRICITY

357124 Two resistors of resistances $100 Ω$ and $200 Ω$ are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in $100 Ω$ to that in $200 Ω$ in a given time is

1

2 : 1

2

1 : 4

3

4 : 1

4

1 : 2

Explanation:

As both resistors are in parallel combination so potential drop ( $V$ ) across both are same.
$P = \frac{V^{2}}{R} \Rightarrow P \propto \frac{1}{R}$
We know Power $P = \frac{E n e r g y}{t i m e} = \frac{E}{t}$
Time(t) is same for both $R_{1}, R_{2}$
$\frac{E_{1}}{E_{2}} = \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}} = \frac{200}{100} = \frac{2}{1}$
$= 2 : 1$
Note: We are asked to find ratio of thermal energies
supporting img

NEET - 2022

PHXII03:CURRENT ELECTRICITY

357125 There are $25 W - 220 V$ bulb and a $100 W$ - $220 V$ line. Which electric bulb will glow more brightly?

1

25 W

bulb

2

100 W

bulb

3 Both will have equal incandescence

4 Neither

25 W

nor

100 W

bulb will give light

Explanation:

Since power is given by $P = \frac{V^{2}}{R}$ ,
$so R = \frac{V^{2}}{P}$
For the first bulb,
$R_{1} = (\frac{V^{2}}{P_{1}}) = [\frac{(220)^{2}}{25}] = 1936 Ω$
For the second bulb,
$R_{2} = (\frac{V^{2}}{P_{2}}) = [\frac{(220)^{2}}{100}] = 484 Ω$
Current in series combination is the same in the two bulbs and current $I$ is given by
$I = \frac{V}{R_{1} + R_{2}} = \frac{220}{1936 + 484} = \frac{220}{2420} = \frac{1}{11} A$
If the actual powers in the bulbs be $P_{1}$ and $P_{2}$ then
$P_{1}^{'} = I^{2} R_{1} = {(\frac{1}{11})}^{2} \times 1936 = 16 W$
and $P_{2}^{'} = I^{2} R_{2} = {(\frac{1}{11})}^{2} \times 484 = 4 W$
Since ${P^{'}}_{1} > {P^{'}}_{2}, 25 W$ bulb will glow more brightly.

PHXII03:CURRENT ELECTRICITY

357126 Assertion :
Fuse wire must have high resistance and low melting point.
Reason :
Fuse is used for small current flow only.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII03:CURRENT ELECTRICITY

357127 Two bulbs of 500 $W$ and 200 $W$ are manufactured to operate on 220 $V$ line. The ratio of heat produced in 500 $W$ and 200 $W$ , in two cases, when firstly they are connected in parallel and secondly in series will be

1

\frac{5}{2} : \frac{5}{2}

2

\frac{5}{2} : \frac{2}{5}

3

\frac{2}{5} : \frac{2}{5}

4

\frac{2}{5} : \frac{5}{2}

Explanation:

In parallel, $\frac{H_{1}}{H_{2}} = \frac{P_{1} t}{P_{2} t} = \frac{P_{1}}{P_{2}} = \frac{500}{200} = \frac{5}{2}$
In series
$\frac{H_{1}}{H_{2}} = \frac{I^{2} R_{1} t}{I^{2} R_{2} t} = \frac{R_{1}}{R_{2}} = \frac{V^{2} / P_{1}}{V^{2} / P_{2}} = \frac{P_{2}}{P_{1}} = \frac{200}{500} = \frac{2}{5}$

PHXII03:CURRENT ELECTRICITY

357128 In a large building, there are $12$ bulbs of $40 W$ , $4$ bulbs of $100 W, 7$ fans of $60 W$ and $1$ heater of $1 k W$ . The voltage of the electric mains is $230 V$ . The minimum capacity of the main fuse of the building will be

1

2 A

2

5 A

3

10 A

4

7 A

Explanation:

$P_{t o t a l} = V \times I$
$12 \times 40 + 4 \times 100 + 7 \times 60 + 1 \times 1000 = V \times I$
$480 + 400 + 420 + 1000 = 230 I$
$∴ I = \frac{2300}{230}$
$= 10 A$

PHXII03:CURRENT ELECTRICITY

357124 Two resistors of resistances $100 Ω$ and $200 Ω$ are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in $100 Ω$ to that in $200 Ω$ in a given time is

1

2 : 1

2

1 : 4

3

4 : 1

4

1 : 2

Explanation:

As both resistors are in parallel combination so potential drop ( $V$ ) across both are same.
$P = \frac{V^{2}}{R} \Rightarrow P \propto \frac{1}{R}$
We know Power $P = \frac{E n e r g y}{t i m e} = \frac{E}{t}$
Time(t) is same for both $R_{1}, R_{2}$
$\frac{E_{1}}{E_{2}} = \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}} = \frac{200}{100} = \frac{2}{1}$
$= 2 : 1$
Note: We are asked to find ratio of thermal energies
supporting img

NEET - 2022

PHXII03:CURRENT ELECTRICITY

357125 There are $25 W - 220 V$ bulb and a $100 W$ - $220 V$ line. Which electric bulb will glow more brightly?

1

25 W

bulb

2

100 W

bulb

3 Both will have equal incandescence

4 Neither

25 W

nor

100 W

bulb will give light

Explanation:

Since power is given by $P = \frac{V^{2}}{R}$ ,
$so R = \frac{V^{2}}{P}$
For the first bulb,
$R_{1} = (\frac{V^{2}}{P_{1}}) = [\frac{(220)^{2}}{25}] = 1936 Ω$
For the second bulb,
$R_{2} = (\frac{V^{2}}{P_{2}}) = [\frac{(220)^{2}}{100}] = 484 Ω$
Current in series combination is the same in the two bulbs and current $I$ is given by
$I = \frac{V}{R_{1} + R_{2}} = \frac{220}{1936 + 484} = \frac{220}{2420} = \frac{1}{11} A$
If the actual powers in the bulbs be $P_{1}$ and $P_{2}$ then
$P_{1}^{'} = I^{2} R_{1} = {(\frac{1}{11})}^{2} \times 1936 = 16 W$
and $P_{2}^{'} = I^{2} R_{2} = {(\frac{1}{11})}^{2} \times 484 = 4 W$
Since ${P^{'}}_{1} > {P^{'}}_{2}, 25 W$ bulb will glow more brightly.

PHXII03:CURRENT ELECTRICITY

357126 Assertion :
Fuse wire must have high resistance and low melting point.
Reason :
Fuse is used for small current flow only.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII03:CURRENT ELECTRICITY

357127 Two bulbs of 500 $W$ and 200 $W$ are manufactured to operate on 220 $V$ line. The ratio of heat produced in 500 $W$ and 200 $W$ , in two cases, when firstly they are connected in parallel and secondly in series will be

1

\frac{5}{2} : \frac{5}{2}

2

\frac{5}{2} : \frac{2}{5}

3

\frac{2}{5} : \frac{2}{5}

4

\frac{2}{5} : \frac{5}{2}

Explanation:

In parallel, $\frac{H_{1}}{H_{2}} = \frac{P_{1} t}{P_{2} t} = \frac{P_{1}}{P_{2}} = \frac{500}{200} = \frac{5}{2}$
In series
$\frac{H_{1}}{H_{2}} = \frac{I^{2} R_{1} t}{I^{2} R_{2} t} = \frac{R_{1}}{R_{2}} = \frac{V^{2} / P_{1}}{V^{2} / P_{2}} = \frac{P_{2}}{P_{1}} = \frac{200}{500} = \frac{2}{5}$

PHXII03:CURRENT ELECTRICITY

357128 In a large building, there are $12$ bulbs of $40 W$ , $4$ bulbs of $100 W, 7$ fans of $60 W$ and $1$ heater of $1 k W$ . The voltage of the electric mains is $230 V$ . The minimum capacity of the main fuse of the building will be

1

2 A

2

5 A

3

10 A

4

7 A

Explanation:

$P_{t o t a l} = V \times I$
$12 \times 40 + 4 \times 100 + 7 \times 60 + 1 \times 1000 = V \times I$
$480 + 400 + 420 + 1000 = 230 I$
$∴ I = \frac{2300}{230}$
$= 10 A$

PHXII03:CURRENT ELECTRICITY

357124 Two resistors of resistances $100 Ω$ and $200 Ω$ are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in $100 Ω$ to that in $200 Ω$ in a given time is

1

2 : 1

2

1 : 4

3

4 : 1

4

1 : 2

Explanation:

As both resistors are in parallel combination so potential drop ( $V$ ) across both are same.
$P = \frac{V^{2}}{R} \Rightarrow P \propto \frac{1}{R}$
We know Power $P = \frac{E n e r g y}{t i m e} = \frac{E}{t}$
Time(t) is same for both $R_{1}, R_{2}$
$\frac{E_{1}}{E_{2}} = \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}} = \frac{200}{100} = \frac{2}{1}$
$= 2 : 1$
Note: We are asked to find ratio of thermal energies
supporting img

NEET - 2022

PHXII03:CURRENT ELECTRICITY

357125 There are $25 W - 220 V$ bulb and a $100 W$ - $220 V$ line. Which electric bulb will glow more brightly?

1

25 W

bulb

2

100 W

bulb

3 Both will have equal incandescence

4 Neither

25 W

nor

100 W

bulb will give light

Explanation:

Since power is given by $P = \frac{V^{2}}{R}$ ,
$so R = \frac{V^{2}}{P}$
For the first bulb,
$R_{1} = (\frac{V^{2}}{P_{1}}) = [\frac{(220)^{2}}{25}] = 1936 Ω$
For the second bulb,
$R_{2} = (\frac{V^{2}}{P_{2}}) = [\frac{(220)^{2}}{100}] = 484 Ω$
Current in series combination is the same in the two bulbs and current $I$ is given by
$I = \frac{V}{R_{1} + R_{2}} = \frac{220}{1936 + 484} = \frac{220}{2420} = \frac{1}{11} A$
If the actual powers in the bulbs be $P_{1}$ and $P_{2}$ then
$P_{1}^{'} = I^{2} R_{1} = {(\frac{1}{11})}^{2} \times 1936 = 16 W$
and $P_{2}^{'} = I^{2} R_{2} = {(\frac{1}{11})}^{2} \times 484 = 4 W$
Since ${P^{'}}_{1} > {P^{'}}_{2}, 25 W$ bulb will glow more brightly.

PHXII03:CURRENT ELECTRICITY

357126 Assertion :
Fuse wire must have high resistance and low melting point.
Reason :
Fuse is used for small current flow only.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII03:CURRENT ELECTRICITY

357127 Two bulbs of 500 $W$ and 200 $W$ are manufactured to operate on 220 $V$ line. The ratio of heat produced in 500 $W$ and 200 $W$ , in two cases, when firstly they are connected in parallel and secondly in series will be

1

\frac{5}{2} : \frac{5}{2}

2

\frac{5}{2} : \frac{2}{5}

3

\frac{2}{5} : \frac{2}{5}

4

\frac{2}{5} : \frac{5}{2}

Explanation:

In parallel, $\frac{H_{1}}{H_{2}} = \frac{P_{1} t}{P_{2} t} = \frac{P_{1}}{P_{2}} = \frac{500}{200} = \frac{5}{2}$
In series
$\frac{H_{1}}{H_{2}} = \frac{I^{2} R_{1} t}{I^{2} R_{2} t} = \frac{R_{1}}{R_{2}} = \frac{V^{2} / P_{1}}{V^{2} / P_{2}} = \frac{P_{2}}{P_{1}} = \frac{200}{500} = \frac{2}{5}$

PHXII03:CURRENT ELECTRICITY

357128 In a large building, there are $12$ bulbs of $40 W$ , $4$ bulbs of $100 W, 7$ fans of $60 W$ and $1$ heater of $1 k W$ . The voltage of the electric mains is $230 V$ . The minimum capacity of the main fuse of the building will be

1

2 A

2

5 A

3

10 A

4

7 A

Explanation:

$P_{t o t a l} = V \times I$
$12 \times 40 + 4 \times 100 + 7 \times 60 + 1 \times 1000 = V \times I$
$480 + 400 + 420 + 1000 = 230 I$
$∴ I = \frac{2300}{230}$
$= 10 A$

PHXII03:CURRENT ELECTRICITY

357124 Two resistors of resistances $100 Ω$ and $200 Ω$ are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in $100 Ω$ to that in $200 Ω$ in a given time is

1

2 : 1

2

1 : 4

3

4 : 1

4

1 : 2

Explanation:

As both resistors are in parallel combination so potential drop ( $V$ ) across both are same.
$P = \frac{V^{2}}{R} \Rightarrow P \propto \frac{1}{R}$
We know Power $P = \frac{E n e r g y}{t i m e} = \frac{E}{t}$
Time(t) is same for both $R_{1}, R_{2}$
$\frac{E_{1}}{E_{2}} = \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}} = \frac{200}{100} = \frac{2}{1}$
$= 2 : 1$
Note: We are asked to find ratio of thermal energies
supporting img

NEET - 2022

PHXII03:CURRENT ELECTRICITY

357125 There are $25 W - 220 V$ bulb and a $100 W$ - $220 V$ line. Which electric bulb will glow more brightly?

1

25 W

bulb

2

100 W

bulb

3 Both will have equal incandescence

4 Neither

25 W

nor

100 W

bulb will give light

Explanation:

Since power is given by $P = \frac{V^{2}}{R}$ ,
$so R = \frac{V^{2}}{P}$
For the first bulb,
$R_{1} = (\frac{V^{2}}{P_{1}}) = [\frac{(220)^{2}}{25}] = 1936 Ω$
For the second bulb,
$R_{2} = (\frac{V^{2}}{P_{2}}) = [\frac{(220)^{2}}{100}] = 484 Ω$
Current in series combination is the same in the two bulbs and current $I$ is given by
$I = \frac{V}{R_{1} + R_{2}} = \frac{220}{1936 + 484} = \frac{220}{2420} = \frac{1}{11} A$
If the actual powers in the bulbs be $P_{1}$ and $P_{2}$ then
$P_{1}^{'} = I^{2} R_{1} = {(\frac{1}{11})}^{2} \times 1936 = 16 W$
and $P_{2}^{'} = I^{2} R_{2} = {(\frac{1}{11})}^{2} \times 484 = 4 W$
Since ${P^{'}}_{1} > {P^{'}}_{2}, 25 W$ bulb will glow more brightly.

PHXII03:CURRENT ELECTRICITY

357126 Assertion :
Fuse wire must have high resistance and low melting point.
Reason :
Fuse is used for small current flow only.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII03:CURRENT ELECTRICITY

357127 Two bulbs of 500 $W$ and 200 $W$ are manufactured to operate on 220 $V$ line. The ratio of heat produced in 500 $W$ and 200 $W$ , in two cases, when firstly they are connected in parallel and secondly in series will be

1

\frac{5}{2} : \frac{5}{2}

2

\frac{5}{2} : \frac{2}{5}

3

\frac{2}{5} : \frac{2}{5}

4

\frac{2}{5} : \frac{5}{2}

Explanation:

In parallel, $\frac{H_{1}}{H_{2}} = \frac{P_{1} t}{P_{2} t} = \frac{P_{1}}{P_{2}} = \frac{500}{200} = \frac{5}{2}$
In series
$\frac{H_{1}}{H_{2}} = \frac{I^{2} R_{1} t}{I^{2} R_{2} t} = \frac{R_{1}}{R_{2}} = \frac{V^{2} / P_{1}}{V^{2} / P_{2}} = \frac{P_{2}}{P_{1}} = \frac{200}{500} = \frac{2}{5}$

PHXII03:CURRENT ELECTRICITY

357128 In a large building, there are $12$ bulbs of $40 W$ , $4$ bulbs of $100 W, 7$ fans of $60 W$ and $1$ heater of $1 k W$ . The voltage of the electric mains is $230 V$ . The minimum capacity of the main fuse of the building will be

1

2 A

2

5 A

3

10 A

4

7 A

Explanation:

$P_{t o t a l} = V \times I$
$12 \times 40 + 4 \times 100 + 7 \times 60 + 1 \times 1000 = V \times I$
$480 + 400 + 420 + 1000 = 230 I$
$∴ I = \frac{2300}{230}$
$= 10 A$

PHXII03:CURRENT ELECTRICITY

357124 Two resistors of resistances $100 Ω$ and $200 Ω$ are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in $100 Ω$ to that in $200 Ω$ in a given time is

1

2 : 1

2

1 : 4

3

4 : 1

4

1 : 2

Explanation:

As both resistors are in parallel combination so potential drop ( $V$ ) across both are same.
$P = \frac{V^{2}}{R} \Rightarrow P \propto \frac{1}{R}$
We know Power $P = \frac{E n e r g y}{t i m e} = \frac{E}{t}$
Time(t) is same for both $R_{1}, R_{2}$
$\frac{E_{1}}{E_{2}} = \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}} = \frac{200}{100} = \frac{2}{1}$
$= 2 : 1$
Note: We are asked to find ratio of thermal energies
supporting img

NEET - 2022

PHXII03:CURRENT ELECTRICITY

357125 There are $25 W - 220 V$ bulb and a $100 W$ - $220 V$ line. Which electric bulb will glow more brightly?

1

25 W

bulb

2

100 W

bulb

3 Both will have equal incandescence

4 Neither

25 W

nor

100 W

bulb will give light

Explanation:

Since power is given by $P = \frac{V^{2}}{R}$ ,
$so R = \frac{V^{2}}{P}$
For the first bulb,
$R_{1} = (\frac{V^{2}}{P_{1}}) = [\frac{(220)^{2}}{25}] = 1936 Ω$
For the second bulb,
$R_{2} = (\frac{V^{2}}{P_{2}}) = [\frac{(220)^{2}}{100}] = 484 Ω$
Current in series combination is the same in the two bulbs and current $I$ is given by
$I = \frac{V}{R_{1} + R_{2}} = \frac{220}{1936 + 484} = \frac{220}{2420} = \frac{1}{11} A$
If the actual powers in the bulbs be $P_{1}$ and $P_{2}$ then
$P_{1}^{'} = I^{2} R_{1} = {(\frac{1}{11})}^{2} \times 1936 = 16 W$
and $P_{2}^{'} = I^{2} R_{2} = {(\frac{1}{11})}^{2} \times 484 = 4 W$
Since ${P^{'}}_{1} > {P^{'}}_{2}, 25 W$ bulb will glow more brightly.

PHXII03:CURRENT ELECTRICITY

357126 Assertion :
Fuse wire must have high resistance and low melting point.
Reason :
Fuse is used for small current flow only.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII03:CURRENT ELECTRICITY

357127 Two bulbs of 500 $W$ and 200 $W$ are manufactured to operate on 220 $V$ line. The ratio of heat produced in 500 $W$ and 200 $W$ , in two cases, when firstly they are connected in parallel and secondly in series will be

1

\frac{5}{2} : \frac{5}{2}

2

\frac{5}{2} : \frac{2}{5}

3

\frac{2}{5} : \frac{2}{5}

4

\frac{2}{5} : \frac{5}{2}

Explanation:

In parallel, $\frac{H_{1}}{H_{2}} = \frac{P_{1} t}{P_{2} t} = \frac{P_{1}}{P_{2}} = \frac{500}{200} = \frac{5}{2}$
In series
$\frac{H_{1}}{H_{2}} = \frac{I^{2} R_{1} t}{I^{2} R_{2} t} = \frac{R_{1}}{R_{2}} = \frac{V^{2} / P_{1}}{V^{2} / P_{2}} = \frac{P_{2}}{P_{1}} = \frac{200}{500} = \frac{2}{5}$

PHXII03:CURRENT ELECTRICITY

357128 In a large building, there are $12$ bulbs of $40 W$ , $4$ bulbs of $100 W, 7$ fans of $60 W$ and $1$ heater of $1 k W$ . The voltage of the electric mains is $230 V$ . The minimum capacity of the main fuse of the building will be

1

2 A

2

5 A

3

10 A

4

7 A

Explanation:

$P_{t o t a l} = V \times I$
$12 \times 40 + 4 \times 100 + 7 \times 60 + 1 \times 1000 = V \times I$
$480 + 400 + 420 + 1000 = 230 I$
$∴ I = \frac{2300}{230}$
$= 10 A$