The capacities and connection of five capacitors are shown in the adjoining figure. The potential difference between the points A and B is 6 volts. Then the equivalent capacity between A and B and the charge on 5 μF capacitance will be respectively.

The circuit can be re drawn as

So equivalent capacitance of the circuit C_AB = 4 $μ$ F

Charge given by the battery Q = C_eqV = 4 × 60 = 240 $μ$ C

Charge in 5 $μ$ F capacitor

$Q^{'} = \frac{5}{(10 + 5 + 9)} \times 240 = 50 μ C$

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The capacities and connection of five capacitors are shown in the adjoining figure. The potential difference between the points A and B is 6 volts. Then the equivalent capacity between A and B and the charge on 5 $μF$ capacitance will be respectively.