JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Current Electricity

An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance '

R

' that must be put in series with the bulb so that the bulb delivers the same power is ____

Ω

Answer: 50

Solution

The bulb gives its rated power only at its rated voltage, so it must carry

I = \frac{P}{V} = \frac{200}{100} = 2 A

with

100

V across it; its own resistance is

R_{b} = \frac{V ^{2}}{P} = \frac{10 0 ^{2}}{200} = 50 Ω

. The series resistor carries the same current and must drop the remaining

200 - 100 = 100

V, so the resistance to be put in series is

R = \frac{100}{2} = 50 Ω

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer