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Question

Calculate the potential difference and the energy stored in the capacitor C2 in the circuit shown in the figure. Given potential at A in 90 V, C1=20μF, C2=30μF and C3=15μF.


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Solution

The given capacitors are connected in series,

1C=1C1+1C2+1C3.

=120+130+115

=3+2+460=960=320

C=203μF.

As the capacitirs are connected in series, so the charge remains same, and is given by:

Q=CV=203×90=600μF

Potential difference across C2,

V2=QC2

=60030=20V

Energy stored in capacitor C2,

U=12C2V2

=12×30×106×(20)2

=12×30×106×400

=6×103J

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