Question

An electric bulb rated for $500W$ at $100V$ is used in a circuit having a $200V$ supply. The resistance $R$ that must be put in series with the bulb, so that the bulb draws $500W$, is

• A. $700\Omega$
• B. $40\Omega$
• C. $20\Omega$
• D. $18\Omega$

Answer 1

The correct option is C

$20\Omega$

Power consumed by a bulb is

$P=VI,I=\frac{P}{V}\Rightarrow I=\frac{500}{100}=5A$
$\Rightarrow \text{For the bulb to consume}500W\text{it has to draw a current of}5A$

Now, as bulb and resistance $R$ are in series, current in resistance is $5A$
Potential difference across resistance is $100V$
By ohms Law
$V=IR\Rightarrow 5R=100$ or $R=20\Omega$