Question

A generator in a power plant develops a maximum voltage of 170 V.

a. What is the effective voltage?

b. A 60 W light bulb is placed across the generator. A maximum current of 0.70 A flows
through the bulb. What effective current flows through the bulb?

c. What is the resistance of the light bulb when it is working?

Answer 1

a. Effective voltage is V_max/√2 = 170/√2= 120.2 V
b. Effective current is I_max/√2= 0.7/√2 = 0.49 I
c. The resistance is R = 2W/I_max² from the law W = I²/R = I_max² /2R
R = 2*60 / 0.7² = 244.89 Ohm