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Question

Two capacitors, C1=2μF and C2=8μF are connected in series across a 300V source. Then

A
the charge on each capacitor is 4.8×104C
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B
the potential difference across C1 is 60V
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C
the potential difference across C2 is 240V
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D
the energy stored in the system is 5.2×102J
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Solution

The correct option is A the charge on each capacitor is 4.8×104C
The equivalent capacitance is Ceq=C1C2C1+C2=2×82+8=1.6μF.
Equivalent charge is Qeq=CeqV=1.6×106×300=4.8×104C
As the capacitors are in series so charge on each capacitor is Qeq=4.8×104C.
The potential difference across C1 is V1=QeqC1=4.8×1042×106=240V
The potential difference across C2 is V2=QeqC2=4.8×1048×106=60V
Energy stored in the system is U=12CeqV2=12×1.6×106×(300)2=7.2×102J

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