An electric bulb, marked `40 W` and `200 V`, is used in a circuit of supply voltage `100 V`. Now its power is

Options

A. 100W
B. 20W
C. 40W
D. 10W

Answer

D

Explanation

Actual power of bulb `(P_(1))=40 W `
Actual voltage of bulb ` (V_(1))=200V and ` supply voltage ` (V_(2))=100 V ` .
Power `(P)=(V^(2))/(R) prop V^(2) . ` Therefore ` (P_(1))/(P_(2))=(V_(1)^(2))/(V_(2)^(2))` or `, (40)/(P_(2))=((200)^(2))/((100)^(2))=4 or P_(2)=(40)/(4)=10W ` ( where `P_(2)`=power when voltage is 100 V ).