Solveeit Logo

Question

Physics Question on Electromagnetic induction

An electric bulb is rated at 220V,220\,V, 100W;100\,W; Power consumed by it when operated at 110V110\,V is

A

25W25\,W

B

50W50\,W

C

75W75\,W

D

90W90\,W

Answer

25W25\,W

Explanation

Solution

The electric bulb is rated 220V220\, V and 100W100 \,W
\therefore Resistance of the bulb
R=V2P=(220)2100=484ΩR=\frac{V^{2}}{P}=\frac{(220)^{2}}{100}=484\, \Omega
Power consumed when operated on 110V110\, V is
P=V2R=(110)2484=25WP=\frac{V^{2}}{R}=\frac{(110)^{2}}{484}=25\, W