Question

In the following network potential difference across capacitance of 4.5 μF is

• A. 8 V
• B. 4 V
• C. 2 V
• D. 6 V

Answer 1

Answer: (A)

8 V

Answer 2

Equivalent capacitance $C_{eq} = \frac{9 \times 4.5}{9 + 4.5} = 3\mu F$

Charge supplied by battery Q = C_eq × V = 3 × 12 = 36 μC

Hence potential difference across 4.5 μF $= \frac{36}{4.5}$= 8V.