In the circuit shown below, the charge on the 60 μF capacitor is

Step-by-step solution

The circuit shown in the figure is a balanced Wheatstone bridge, so the 80 μF doesn't have any effect on the circuit and it can be removed. After removing the 80 μF , we see that the 30 μF and 60 μF are in series. Similarly, the 15 μF and 30 μF are also in series. C 1 = 30 × 60 30 + 60 = 20 μF , C 2 = 15 × 30 15 + 30 = 10 μF Hence, the equivalent capacitance of the entire circuit is C = C 1 + C 2 = 30 μF C eq = C × 30 C + 30 = 15 μF The charge that flows through the battery = 10 × 15 = 150 μC This charge will be di