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Question

A capacitor of C1 = 1.0μF withstands maximum voltage V1 = 6.0kV. Another capacitor C2 = 2.0μF withstands maximum voltage V2 = 4.0 kV. What voltage will the system of these capacitors withstand if they are connected in series?

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Solution

Given,
A capacitor C1=1.0μ=1×106F
Withstand maximum voltageV1=6.0KV=6×103V
Another capacitor,
C2=20×106F
Withstand maximum voltage
V2=4.0KV=4×103V
What voltage will the system of the capacitor withstand if they are connected in series
Now, Charge on first capacitor,
q1=C1V1=0.1×106×6×103=6×104C
Charge on second capacitor,
q2=C2V2=2.0×106×4×103=8×103C
In series combination, the magnitude of charge on each capacitor must be the same. As maximum charge on C1 is
6×104C
and therefore maximum charge on C2 must also be 6×104C
Hence, maximum voltage for the combination is,
V1=V11+V12=6×1040.1+6×1042.0=6×103+3×104=61000+310000=60+310000V=6.310000V=0.0063V

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