Question

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

• A. 50W
• B. 100W
• C. 25W
• D. 0.25W

Answer 1

The correct option is B

100W

Given, resistance of the bulb, R = 1 ohm; Power, P = 400 W

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