For the circuit below, the input is a full wave rectified 60 Hz ac signal with p
ID: 2080351 • Letter: F
Question
Explanation / Answer
2C) V=10v, C=0.1uf at f=1kHz, R=5k ohms
Xc=1/(j2fC)
Xc=-j1591.55 ohms
Z=R+jXc=(5000-j1591.55) ohms
I=V/Z=10/(5000-j1591.55)
=0.0019L17.65 Amps
Power dissipated across 5k resister=i^2*R
=(0.0019L17.65)^2*5000
=0.018L35.30 watts
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