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Question

A light bulb is rated 150W for 220V AC 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

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Solution

(i) Given - P=150W,Vrms=220V,f=60Hz
by P=V2rms/R
or R=V2rms/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

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