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Question

A 200 V, 100 Watt bulb is to be used across an A.C source of peak voltage 400 V and an angular frequency of 103 rad/s. The inductance of a choke coil to be used in series with the bulb so that the bulb won't blow off is

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Solution

By p=vii=100200=12=irms for an AC source.
p=v2RR=400Ω i.e., resistance of bulb.
irms=i02=ε02z=ε02(ωL)2+R2
12=4002(ωL)2+4002ωL=400L=0.4 H

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