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

Answer 1

The correct option is B $20\Omega$

Given that, as bulb draws $500\text{W}$ then voltage across it will be $100\text{V}$.

For a bulb, we know that,
$P=VI,I=\frac{P}{V}I=\frac{500\text{W}}{100\text{V}}=5\text{A}$


As bulb is connected in series to resistance, equal current flows through resistance and bulb.
Voltage across other resistor will be
$200-100=100\text{V}$

Now from Ohm's law,
$IR=V5R=100R=20\Omega$