Question

a light bulb is rated at 100 watt for a 220V supply find the peak voltage of the source

Answer 1

Peak voltage of the source is
220×1.414
i.e. approx 308 V

Answer 2

The peak voltage of the source is 0.45A

Supply of the bulb = 220V (Given)

Power = 100 watt (Given)

Power of the bulb P = VI

P = V x V/R

R = V² / P

= 220 x 220 / 100

= 48400 / 100

= 484

Peak voltage of the source = E/√2

E0 = 220 × √2

= 220√2

= 311.1

Irms = I0/√2 = E0/R√2

= 311.1/484√2

= 311.1/684.4

= 0.45A