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=vi⇒i=100200=12=irms for an AC source. p=v2R⇒R=400Ω i.e., resistance of bulb. irms=i0√2=ε0√2z=ε0√2√(ωL)2+R2 12=400√2√(ωL)2+4002⇒ωL=400⇒L=0.4H