Two capacitors of 2μF and 4μF are connected in parallel. A third capacitor of 6μF is connected in series. The combination is connected across a 12 V battery. The voltage across 2μF capacitor is:
A
2 V
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B
8 V
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C
6 V
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D
1 V
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Solution
The correct option is C 6 V
Cp=2+4=6μF 1C=16+16=26=13orC=3μF Total charge, Q=CV=3×12=36μC Voltage across 6μF capacitor = 36μF6μF=6V ∴Voltage across each of 2μF and 4μF capacitors =12V−6V V=6V