F \times 6 \mu F}{3 \mu F + 6 \mu F} = \frac{18}{9} = 2 \mu F.$$

With a potential difference of 900 V applied across the series combination, the same charge is stored on each capacitor. The charge is:

$$Q = C_S \times V = 2 \mu F \times 900 V = 1800 \mu C.$$

Voltage across individual capacitors:

Voltage across the 3 μF capacitor:

$$V_1 = \frac{Q}{C_1} = \frac{1800 \mu C}{3 \mu F} = 600 V.$$

Voltage across the 6 μF capacitor:

$$V_2 = \frac{Q}{C_2} = \frac{1800 \mu C}{6 \mu F} = 300 V.$$

Notice that the sum, $$600 V + 300 V = 900 V,$$ matches the applied voltage.

Reconnection in parallel:

After disconnecting from the supply, the charges on the capacitors remain unchanged.

When reconnected in parallel (with like plates connected together), the total charge in the system is the sum of the individual charges:

$$Q_{total} = 1800 \mu C + 1800 \mu C = 3600 \mu C.$$

In a parallel combination, the total capacitance is the sum of the individual capacitances:

$$C_{total} = 3 \mu F + 6 \mu F = 9 \mu F.$$

The common potential difference across the parallel combination is:

$$V = \frac{Q_{total}}{C_{total}} = \frac{3600 \mu C}{9 \mu F} = 400 V.$$

Thus, the potential difference across the combination when reconnected in parallel is 400 V.

Answer: Option D (400 V).

Capacitor

Continue this paper