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Question:

A light bulb is rated 150W for 220VAC supply of 60Hz. Calculate (i) the resistance of the bulb (ii) the rms current through the bulb. OR An alternating voltage given by V=70sin100πt is connected across a pure resistor of 25Ω. Find (i) the frequency of the source (ii) the rms current through the resistor.

Solution:

(i) Given -P=150W, Vrms=220V, f=60Hz
by P=Vrms²/R
or R=Vrms²/P=220²/150=322.67Ω
(ii) now Irms=Vrms/R=220/322.67=0.68A
OR
(i) If V=70sin100πt
comparing it with V=V0sinωt
ω=100π
or 2πf=100π
or f=100/2=50Hz
(ii) I0=V0/R=70/25=2.8A
therefore Irms=I0/√2=2.8/√2=1.98A