QBManager

Current Electricity

152685 Two electric bulbs $A$ and $b$ are rated as $60 W$ and $100 W$ . They are connected in parallel to the same source. Then :

1 B draws more current than A

2 current drawn are in the ratio of their resistances

3 both draw the same current

4 A draws more current than B.

Explanation:

A Given,
Power of electric bulb, $A = 60 W$
Power of electric bulb, $B = 100 W$
We know that,
$Power, P = \frac{V^{2}}{R}$
Now, the resistance of $60 W$ bulb has more resistance as compared to a $100 W$ bulb as $P \propto \frac{1}{R}$ .
Therefore, when bulbs are connected in parallel to the same source, the bulb $B (100 W)$ draws more current than A.

Karnataka CET-2004

Current Electricity

152686 A fuse wire with radius $1 mm$ blows at 1.5 ampere. The radius of the fuse wire of the same material to blow at $3 A$ will be :

1

3^{1 / 4} mm

2

4^{1 / 3} mm

3

3^{1 / 2} mm

4

2^{1 / 3} mm

Explanation:

B Given, $r_{1} = 1 mm, I_{1} = 1.5 A, I_{2} = 3 A$
We know that,
For fusing current,
The relation of current with radius,
$I^{2} \propto r^{3}$
$\frac{I_{1}^{2}}{r_{1}^{3}} = \frac{I_{2}^{2}}{r_{2}^{3}}$
${(\frac{I_{2}}{I_{1}})}^{2} = {(\frac{r_{2}}{r_{1}})}^{3}$
$r_{2} = r_{1} {(\frac{I_{2}}{I_{1}})}^{2 / 3}$
$r_{2} = 1 {(\frac{3}{1.5})}^{2 / 3}$
$r_{2} = 2^{2 / 3} = 4^{1 / 3} mm$

Karnataka CET-2003

Current Electricity

152687 $n$ identical bulbs, each designated to draw a power $P$ from a certain voltage supply. The total power which they will draw is :

1

P / n

2

P / n^{2}

3

nP

4

P

Explanation:

A Resistance of each bulb $= R$
Voltage of source $= V$
We know that,
Power of each bulb, $P = \frac{V^{2}}{R}$
$or R = \frac{V^{2}}{P}$
Now, $n$ bulbs are in series so total resistance,
$R_{t} = nR$
$∴ Current (I) = \frac{V}{R_{t}} = \frac{V}{nR}$
Therefore,
Total power drawn, $P_{t} = I^{2} R_{t}$
$P_{t} = \frac{V^{2}}{n^{2} R^{2}} (n R)$
$P_{t} = \frac{V^{2}}{n R} = \frac{P}{n}$

Karnataka CET-2002

Current Electricity

152688 An electric bulb is designed to draw power $P_{0}$ at voltage $V_{0}$ . If the voltage is $V$ , it draws a power $P$ . Then :

1

P = {(\frac{V_{0}}{V})}^{2} P_{0}

2

P = {(\frac{V}{V_{0}})}^{2} P_{0}

3

P = (\frac{V}{V_{0}}) P_{0}

4

P = (\frac{V_{0}}{V}) P_{0}

Explanation:

B We know that,
Power in resistor, $P = \frac{V^{2}}{R}$
So, $R = \frac{V^{2}}{P}$
Now since the bulb is operating initially at power $P_{0}$ and voltage $V_{0}$ .
So, $P_{0} = \frac{V_{0}^{2}}{R}$
Resistance of bulb $(R) = \frac{V_{0}^{2}}{P_{0}}$
Now the bulb is attached to voltage $V$ and it consumes power P.
$So, P = \frac{V^{2}}{R} = \frac{V^{2} P_{0}}{V_{0}^{2}}$
$P = {(\frac{V}{V_{0}})}^{2} P_{0}$

Karnataka CET-2001

Current Electricity

152689 An electric bulb is rated at $220 V, 200 W$ . Power consumed by it when operated at $110 V$ is

1

25 W

2

50 W

3

75 W

4

90 W

Explanation:

B $V_{1} = 220 V, P_{1} = 200 W, V_{2} = 110 V$
We know that,
$P = VI = \frac{V^{2}}{R}$
$R = \frac{V^{2}}{P}$
$∵ R_{1} = \frac{V_{1}^{2}}{P_{1}} = \frac{(220)^{2}}{200} = 242 Ω$
When, $V_{2} = 110 V$
$P_{2} = \frac{V_{2}^{2}}{R_{2}}$
$= \frac{(110)^{2}}{242} = 50 W$

Current Electricity

152685 Two electric bulbs $A$ and $b$ are rated as $60 W$ and $100 W$ . They are connected in parallel to the same source. Then :

1 B draws more current than A

2 current drawn are in the ratio of their resistances

3 both draw the same current

4 A draws more current than B.

Explanation:

A Given,
Power of electric bulb, $A = 60 W$
Power of electric bulb, $B = 100 W$
We know that,
$Power, P = \frac{V^{2}}{R}$
Now, the resistance of $60 W$ bulb has more resistance as compared to a $100 W$ bulb as $P \propto \frac{1}{R}$ .
Therefore, when bulbs are connected in parallel to the same source, the bulb $B (100 W)$ draws more current than A.

Karnataka CET-2004

Current Electricity

152686 A fuse wire with radius $1 mm$ blows at 1.5 ampere. The radius of the fuse wire of the same material to blow at $3 A$ will be :

1

3^{1 / 4} mm

2

4^{1 / 3} mm

3

3^{1 / 2} mm

4

2^{1 / 3} mm

Explanation:

B Given, $r_{1} = 1 mm, I_{1} = 1.5 A, I_{2} = 3 A$
We know that,
For fusing current,
The relation of current with radius,
$I^{2} \propto r^{3}$
$\frac{I_{1}^{2}}{r_{1}^{3}} = \frac{I_{2}^{2}}{r_{2}^{3}}$
${(\frac{I_{2}}{I_{1}})}^{2} = {(\frac{r_{2}}{r_{1}})}^{3}$
$r_{2} = r_{1} {(\frac{I_{2}}{I_{1}})}^{2 / 3}$
$r_{2} = 1 {(\frac{3}{1.5})}^{2 / 3}$
$r_{2} = 2^{2 / 3} = 4^{1 / 3} mm$

Karnataka CET-2003

Current Electricity

152687 $n$ identical bulbs, each designated to draw a power $P$ from a certain voltage supply. The total power which they will draw is :

1

P / n

2

P / n^{2}

3

nP

4

P

Explanation:

A Resistance of each bulb $= R$
Voltage of source $= V$
We know that,
Power of each bulb, $P = \frac{V^{2}}{R}$
$or R = \frac{V^{2}}{P}$
Now, $n$ bulbs are in series so total resistance,
$R_{t} = nR$
$∴ Current (I) = \frac{V}{R_{t}} = \frac{V}{nR}$
Therefore,
Total power drawn, $P_{t} = I^{2} R_{t}$
$P_{t} = \frac{V^{2}}{n^{2} R^{2}} (n R)$
$P_{t} = \frac{V^{2}}{n R} = \frac{P}{n}$

Karnataka CET-2002

Current Electricity

152688 An electric bulb is designed to draw power $P_{0}$ at voltage $V_{0}$ . If the voltage is $V$ , it draws a power $P$ . Then :

1

P = {(\frac{V_{0}}{V})}^{2} P_{0}

2

P = {(\frac{V}{V_{0}})}^{2} P_{0}

3

P = (\frac{V}{V_{0}}) P_{0}

4

P = (\frac{V_{0}}{V}) P_{0}

Explanation:

B We know that,
Power in resistor, $P = \frac{V^{2}}{R}$
So, $R = \frac{V^{2}}{P}$
Now since the bulb is operating initially at power $P_{0}$ and voltage $V_{0}$ .
So, $P_{0} = \frac{V_{0}^{2}}{R}$
Resistance of bulb $(R) = \frac{V_{0}^{2}}{P_{0}}$
Now the bulb is attached to voltage $V$ and it consumes power P.
$So, P = \frac{V^{2}}{R} = \frac{V^{2} P_{0}}{V_{0}^{2}}$
$P = {(\frac{V}{V_{0}})}^{2} P_{0}$

Karnataka CET-2001

Current Electricity

152689 An electric bulb is rated at $220 V, 200 W$ . Power consumed by it when operated at $110 V$ is

1

25 W

2

50 W

3

75 W

4

90 W

Explanation:

B $V_{1} = 220 V, P_{1} = 200 W, V_{2} = 110 V$
We know that,
$P = VI = \frac{V^{2}}{R}$
$R = \frac{V^{2}}{P}$
$∵ R_{1} = \frac{V_{1}^{2}}{P_{1}} = \frac{(220)^{2}}{200} = 242 Ω$
When, $V_{2} = 110 V$
$P_{2} = \frac{V_{2}^{2}}{R_{2}}$
$= \frac{(110)^{2}}{242} = 50 W$

Current Electricity

152685 Two electric bulbs $A$ and $b$ are rated as $60 W$ and $100 W$ . They are connected in parallel to the same source. Then :

1 B draws more current than A

2 current drawn are in the ratio of their resistances

3 both draw the same current

4 A draws more current than B.

Explanation:

A Given,
Power of electric bulb, $A = 60 W$
Power of electric bulb, $B = 100 W$
We know that,
$Power, P = \frac{V^{2}}{R}$
Now, the resistance of $60 W$ bulb has more resistance as compared to a $100 W$ bulb as $P \propto \frac{1}{R}$ .
Therefore, when bulbs are connected in parallel to the same source, the bulb $B (100 W)$ draws more current than A.

Karnataka CET-2004

Current Electricity

152686 A fuse wire with radius $1 mm$ blows at 1.5 ampere. The radius of the fuse wire of the same material to blow at $3 A$ will be :

1

3^{1 / 4} mm

2

4^{1 / 3} mm

3

3^{1 / 2} mm

4

2^{1 / 3} mm

Explanation:

B Given, $r_{1} = 1 mm, I_{1} = 1.5 A, I_{2} = 3 A$
We know that,
For fusing current,
The relation of current with radius,
$I^{2} \propto r^{3}$
$\frac{I_{1}^{2}}{r_{1}^{3}} = \frac{I_{2}^{2}}{r_{2}^{3}}$
${(\frac{I_{2}}{I_{1}})}^{2} = {(\frac{r_{2}}{r_{1}})}^{3}$
$r_{2} = r_{1} {(\frac{I_{2}}{I_{1}})}^{2 / 3}$
$r_{2} = 1 {(\frac{3}{1.5})}^{2 / 3}$
$r_{2} = 2^{2 / 3} = 4^{1 / 3} mm$

Karnataka CET-2003

Current Electricity

152687 $n$ identical bulbs, each designated to draw a power $P$ from a certain voltage supply. The total power which they will draw is :

1

P / n

2

P / n^{2}

3

nP

4

P

Explanation:

A Resistance of each bulb $= R$
Voltage of source $= V$
We know that,
Power of each bulb, $P = \frac{V^{2}}{R}$
$or R = \frac{V^{2}}{P}$
Now, $n$ bulbs are in series so total resistance,
$R_{t} = nR$
$∴ Current (I) = \frac{V}{R_{t}} = \frac{V}{nR}$
Therefore,
Total power drawn, $P_{t} = I^{2} R_{t}$
$P_{t} = \frac{V^{2}}{n^{2} R^{2}} (n R)$
$P_{t} = \frac{V^{2}}{n R} = \frac{P}{n}$

Karnataka CET-2002

Current Electricity

152688 An electric bulb is designed to draw power $P_{0}$ at voltage $V_{0}$ . If the voltage is $V$ , it draws a power $P$ . Then :

1

P = {(\frac{V_{0}}{V})}^{2} P_{0}

2

P = {(\frac{V}{V_{0}})}^{2} P_{0}

3

P = (\frac{V}{V_{0}}) P_{0}

4

P = (\frac{V_{0}}{V}) P_{0}

Explanation:

B We know that,
Power in resistor, $P = \frac{V^{2}}{R}$
So, $R = \frac{V^{2}}{P}$
Now since the bulb is operating initially at power $P_{0}$ and voltage $V_{0}$ .
So, $P_{0} = \frac{V_{0}^{2}}{R}$
Resistance of bulb $(R) = \frac{V_{0}^{2}}{P_{0}}$
Now the bulb is attached to voltage $V$ and it consumes power P.
$So, P = \frac{V^{2}}{R} = \frac{V^{2} P_{0}}{V_{0}^{2}}$
$P = {(\frac{V}{V_{0}})}^{2} P_{0}$

Karnataka CET-2001

Current Electricity

152689 An electric bulb is rated at $220 V, 200 W$ . Power consumed by it when operated at $110 V$ is

1

25 W

2

50 W

3

75 W

4

90 W

Explanation:

B $V_{1} = 220 V, P_{1} = 200 W, V_{2} = 110 V$
We know that,
$P = VI = \frac{V^{2}}{R}$
$R = \frac{V^{2}}{P}$
$∵ R_{1} = \frac{V_{1}^{2}}{P_{1}} = \frac{(220)^{2}}{200} = 242 Ω$
When, $V_{2} = 110 V$
$P_{2} = \frac{V_{2}^{2}}{R_{2}}$
$= \frac{(110)^{2}}{242} = 50 W$

Current Electricity

152685 Two electric bulbs $A$ and $b$ are rated as $60 W$ and $100 W$ . They are connected in parallel to the same source. Then :

1 B draws more current than A

2 current drawn are in the ratio of their resistances

3 both draw the same current

4 A draws more current than B.

Explanation:

A Given,
Power of electric bulb, $A = 60 W$
Power of electric bulb, $B = 100 W$
We know that,
$Power, P = \frac{V^{2}}{R}$
Now, the resistance of $60 W$ bulb has more resistance as compared to a $100 W$ bulb as $P \propto \frac{1}{R}$ .
Therefore, when bulbs are connected in parallel to the same source, the bulb $B (100 W)$ draws more current than A.

Karnataka CET-2004

Current Electricity

152686 A fuse wire with radius $1 mm$ blows at 1.5 ampere. The radius of the fuse wire of the same material to blow at $3 A$ will be :

1

3^{1 / 4} mm

2

4^{1 / 3} mm

3

3^{1 / 2} mm

4

2^{1 / 3} mm

Explanation:

B Given, $r_{1} = 1 mm, I_{1} = 1.5 A, I_{2} = 3 A$
We know that,
For fusing current,
The relation of current with radius,
$I^{2} \propto r^{3}$
$\frac{I_{1}^{2}}{r_{1}^{3}} = \frac{I_{2}^{2}}{r_{2}^{3}}$
${(\frac{I_{2}}{I_{1}})}^{2} = {(\frac{r_{2}}{r_{1}})}^{3}$
$r_{2} = r_{1} {(\frac{I_{2}}{I_{1}})}^{2 / 3}$
$r_{2} = 1 {(\frac{3}{1.5})}^{2 / 3}$
$r_{2} = 2^{2 / 3} = 4^{1 / 3} mm$

Karnataka CET-2003

Current Electricity

152687 $n$ identical bulbs, each designated to draw a power $P$ from a certain voltage supply. The total power which they will draw is :

1

P / n

2

P / n^{2}

3

nP

4

P

Explanation:

A Resistance of each bulb $= R$
Voltage of source $= V$
We know that,
Power of each bulb, $P = \frac{V^{2}}{R}$
$or R = \frac{V^{2}}{P}$
Now, $n$ bulbs are in series so total resistance,
$R_{t} = nR$
$∴ Current (I) = \frac{V}{R_{t}} = \frac{V}{nR}$
Therefore,
Total power drawn, $P_{t} = I^{2} R_{t}$
$P_{t} = \frac{V^{2}}{n^{2} R^{2}} (n R)$
$P_{t} = \frac{V^{2}}{n R} = \frac{P}{n}$

Karnataka CET-2002

Current Electricity

152688 An electric bulb is designed to draw power $P_{0}$ at voltage $V_{0}$ . If the voltage is $V$ , it draws a power $P$ . Then :

1

P = {(\frac{V_{0}}{V})}^{2} P_{0}

2

P = {(\frac{V}{V_{0}})}^{2} P_{0}

3

P = (\frac{V}{V_{0}}) P_{0}

4

P = (\frac{V_{0}}{V}) P_{0}

Explanation:

B We know that,
Power in resistor, $P = \frac{V^{2}}{R}$
So, $R = \frac{V^{2}}{P}$
Now since the bulb is operating initially at power $P_{0}$ and voltage $V_{0}$ .
So, $P_{0} = \frac{V_{0}^{2}}{R}$
Resistance of bulb $(R) = \frac{V_{0}^{2}}{P_{0}}$
Now the bulb is attached to voltage $V$ and it consumes power P.
$So, P = \frac{V^{2}}{R} = \frac{V^{2} P_{0}}{V_{0}^{2}}$
$P = {(\frac{V}{V_{0}})}^{2} P_{0}$

Karnataka CET-2001

Current Electricity

152689 An electric bulb is rated at $220 V, 200 W$ . Power consumed by it when operated at $110 V$ is

1

25 W

2

50 W

3

75 W

4

90 W

Explanation:

B $V_{1} = 220 V, P_{1} = 200 W, V_{2} = 110 V$
We know that,
$P = VI = \frac{V^{2}}{R}$
$R = \frac{V^{2}}{P}$
$∵ R_{1} = \frac{V_{1}^{2}}{P_{1}} = \frac{(220)^{2}}{200} = 242 Ω$
When, $V_{2} = 110 V$
$P_{2} = \frac{V_{2}^{2}}{R_{2}}$
$= \frac{(110)^{2}}{242} = 50 W$

Current Electricity

152685 Two electric bulbs $A$ and $b$ are rated as $60 W$ and $100 W$ . They are connected in parallel to the same source. Then :

1 B draws more current than A

2 current drawn are in the ratio of their resistances

3 both draw the same current

4 A draws more current than B.

Explanation:

A Given,
Power of electric bulb, $A = 60 W$
Power of electric bulb, $B = 100 W$
We know that,
$Power, P = \frac{V^{2}}{R}$
Now, the resistance of $60 W$ bulb has more resistance as compared to a $100 W$ bulb as $P \propto \frac{1}{R}$ .
Therefore, when bulbs are connected in parallel to the same source, the bulb $B (100 W)$ draws more current than A.

Karnataka CET-2004

Current Electricity

152686 A fuse wire with radius $1 mm$ blows at 1.5 ampere. The radius of the fuse wire of the same material to blow at $3 A$ will be :

1

3^{1 / 4} mm

2

4^{1 / 3} mm

3

3^{1 / 2} mm

4

2^{1 / 3} mm

Explanation:

B Given, $r_{1} = 1 mm, I_{1} = 1.5 A, I_{2} = 3 A$
We know that,
For fusing current,
The relation of current with radius,
$I^{2} \propto r^{3}$
$\frac{I_{1}^{2}}{r_{1}^{3}} = \frac{I_{2}^{2}}{r_{2}^{3}}$
${(\frac{I_{2}}{I_{1}})}^{2} = {(\frac{r_{2}}{r_{1}})}^{3}$
$r_{2} = r_{1} {(\frac{I_{2}}{I_{1}})}^{2 / 3}$
$r_{2} = 1 {(\frac{3}{1.5})}^{2 / 3}$
$r_{2} = 2^{2 / 3} = 4^{1 / 3} mm$

Karnataka CET-2003

Current Electricity

152687 $n$ identical bulbs, each designated to draw a power $P$ from a certain voltage supply. The total power which they will draw is :

1

P / n

2

P / n^{2}

3

nP

4

P

Explanation:

A Resistance of each bulb $= R$
Voltage of source $= V$
We know that,
Power of each bulb, $P = \frac{V^{2}}{R}$
$or R = \frac{V^{2}}{P}$
Now, $n$ bulbs are in series so total resistance,
$R_{t} = nR$
$∴ Current (I) = \frac{V}{R_{t}} = \frac{V}{nR}$
Therefore,
Total power drawn, $P_{t} = I^{2} R_{t}$
$P_{t} = \frac{V^{2}}{n^{2} R^{2}} (n R)$
$P_{t} = \frac{V^{2}}{n R} = \frac{P}{n}$

Karnataka CET-2002

Current Electricity

152688 An electric bulb is designed to draw power $P_{0}$ at voltage $V_{0}$ . If the voltage is $V$ , it draws a power $P$ . Then :

1

P = {(\frac{V_{0}}{V})}^{2} P_{0}

2

P = {(\frac{V}{V_{0}})}^{2} P_{0}

3

P = (\frac{V}{V_{0}}) P_{0}

4

P = (\frac{V_{0}}{V}) P_{0}

Explanation:

B We know that,
Power in resistor, $P = \frac{V^{2}}{R}$
So, $R = \frac{V^{2}}{P}$
Now since the bulb is operating initially at power $P_{0}$ and voltage $V_{0}$ .
So, $P_{0} = \frac{V_{0}^{2}}{R}$
Resistance of bulb $(R) = \frac{V_{0}^{2}}{P_{0}}$
Now the bulb is attached to voltage $V$ and it consumes power P.
$So, P = \frac{V^{2}}{R} = \frac{V^{2} P_{0}}{V_{0}^{2}}$
$P = {(\frac{V}{V_{0}})}^{2} P_{0}$

Karnataka CET-2001

Current Electricity

152689 An electric bulb is rated at $220 V, 200 W$ . Power consumed by it when operated at $110 V$ is

1

25 W

2

50 W

3

75 W

4

90 W

Explanation:

B $V_{1} = 220 V, P_{1} = 200 W, V_{2} = 110 V$
We know that,
$P = VI = \frac{V^{2}}{R}$
$R = \frac{V^{2}}{P}$
$∵ R_{1} = \frac{V_{1}^{2}}{P_{1}} = \frac{(220)^{2}}{200} = 242 Ω$
When, $V_{2} = 110 V$
$P_{2} = \frac{V_{2}^{2}}{R_{2}}$
$= \frac{(110)^{2}}{242} = 50 W$

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