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Q. What is the potential difference across $ 2\mu F $ capacitor in the circuit shown?Physics Question Image

VMMC MedicalVMMC Medical 2012

Solution:

Net emf in the circuit here $ E={{E}_{2}}-{{E}_{1}}=16-6=10\,V $ While the equivalent capacitance $ C=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{2\times 3}{2+3}=\frac{6}{5}\mu F $ Charge on each capacitor $ q=CV=\frac{6}{5}\times 10=12\mu C $ $ \therefore $ Potential difference across $ 2\mu F $ capacitor $ {{V}_{1}}=\frac{q}{{{C}_{1}}}=\frac{12}{2}=6V $

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