Q.
In the circuit shown in figure, the charge on the 5μF capacitor will be
516
177
Electrostatic Potential and Capacitance
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Solution:
Potential difference across upper branch is 6V. V2=V(C1+C2C1)=10[3+73]=3V
Hence, potential difference across 5μF capacitor is 3V.
Thus, charge in 5μF capacitor is Q5μF=5×3=15μC