When a bulb of resistance $1 Ω$ is connected across another battery, it draws a constant power of 400W. What is the total power drawn when 4 such bulbs are connected in series across the second battery?

50W

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100W

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25W

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0.25W

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Solution

The correct option is B 100W
We know that
$P = I^{2} R = \frac{V^{2}}{R} 400 = \frac{V^{2}}{R} = \frac{V^{2}}{1} V = 20 V$
Now, when four bulbs are connected in series across the second battery, the equivalent resistance is
$I \times 4 = 4 Ω$
The power drawn when the resistance is $4 Ω i s \frac{V^{2}}{R} = \frac{20^{2}}{4} = \frac{400}{4} = 100 W$

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When a bulb of resistance 1Ω is connected across another battery, it draws a constant power of 400W. What is the total power drawn when 4 such bulbs are connected in series across the second battery?

The correct option is B 100WWe know that P=I2R=V2R400=V2R=V21V=20V Now, when four bulbs are connected in series across the second battery, the equivalent resistance is I×4=4Ω The power drawn when the resistance is 4Ω is V2R=2024=4004=100W

When a bulb of resistance $1 Ω$ is connected across another battery, it draws a constant power of 400W. What is the total power drawn when 4 such bulbs are connected in series across the second battery?