Two capacitors of 2 μF and 4 μF are connected in parallel. A third capacitor of 6 μF is connected in series. The combination is connected across a 12 V battery. The voltage across 2 μF capacitor is:
2 V
6 V
1 V
8 V
A
2 V
B
8 V
C
1 V
D
6 V
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Solution
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CP=2+4=6μF 1C=16+16=26=13orC=3μF Total charge Q=CV=3×12=36μC voltage across 6μF capacitor =36μC6μF=6V ∴ Voltage across each of 2μF and 4μF capacitors =12V−6V V=6V
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