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Question

A 120 volt A.C line provides power to 48 Watt bulb (rated at 120v). If a capacitor of capacitance 2.5μF is connected in series with the bulb, then the output power of the bulb is [ω=103rad/s]


A
20 Watt
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B
17.28 Watt
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C
13.3 Watt
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D
35.7 Watt
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Solution

The correct option is B 17.28 Watt
By p=i2R=vi=v2RR=v2p=120248=300Ω
xc=1wc=106103×2.5=400Ω
i=εz=1204002+3002=120500
Pout=i2R=(120500)2×300=17.28 Watt

Physics

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