Question

A $100\text{W}$ bulb designed to operate on $100\text{V}$ is to be connected across a $500\text{V}$ source. Find the resistance to be put in series, so that bulb consumes $100\text{W}$ only.

• A. $500\Omega$
• B. $100\Omega$
• C. $400\Omega$
• D. $200\Omega$

Answer 1

The correct option is C $400\Omega$

Given,
Power of the bulb, $P=100\text{W}$,
Voltage drop across bulb, $V=100\text{V},$

Resistance of bulb:

$R_b=\frac{V^2}{P}=\frac{(100{)}^2}{100}=100\Omega$

If bulb consumes $100\text{W}$, then potential difference across the bulb should be $100\text{V}$.

So, current $I=\frac{V}{R_b}=\frac{100}{100}=1\text{A}$.

Let resistance $R$ be put in series, then potential difference across it,

$V_R=500-100=400\text{V}$.


So, $R=\frac{V_R}{I}=\frac{400}{1}=400\Omega$

Hence, option (c) is the correct answer.