Question

An electric bulb is rated 500 W and 100 V is used in 200 V supply. Calculate the resistance R which is connected in series to the bulb so that the bulb delivers 500 W.

• A. 10 ohm
• B. 15 ohm
• C. 25 ohm
• D. 20 ohm

Answer 1

Answer: (D)

20 ohm

Resistance of the bulb $R_b=\frac{V_b^2}{P_b}=\frac{{100}^2}{500}=20\Omega$

Let the resistance connected in series be $R$.

$\therefore$ Current through the circuit $I=\frac{V}{R+R_b}=\frac{200}{R+20}$

Power delivered by bulb $P_b^{\prime }=I^2{R}_b$

$\therefore$ $500=\frac{{200}^2}{(R+20{)}^2}\times 20$ $⟹R=20\Omega$