Question

An electric bulb rated for 500 watt at 100 volt is used in a circuit having a 200 volt supply. Resistance R that must be connected in series with the bulb so that the bulb delivers 500 watts is

• A. 10$\Omega$
• B. 20$\Omega$
• C. 60$\Omega$
• D. 80$\Omega$

Answer 1

The correct option is B 20$\Omega$

$Power=\frac{V^2}{R}$ therefore, $R_{Bulb}=\frac{V^2}{Power}=\frac{{100}^2}{500}=20\Omega$

for the bulb to provide $500$Watts

$I^2{R}_{Bulb}=500$

since $R_{Bulb}=20\Omega$

therefore, we get $I=5Amperes$

now, using KVL $V=I(R_{Bulb}+R)$

$200=5(20+R)$

$R=20\Omega$