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Q.

Four capacitors with capacitances $C_{1} = 1 μ F, C_{2} = 1.5 μ F, C_{3} = 2.5 μ F$ and $C_{4} = 0.5 μ F$ are connected as shown and are connected to a 30 volt source. The potential difference between points a and b is:

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a

9 V

b

10 V

c

13 V

d

5 V

answer is D.

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Detailed Solution

$q_{1} = q, q_{2} = 2 q \frac{q_{1}}{c_{1}} + \frac{q_{2}}{c_{2}} = 30 q = 18 μ c \frac{Q_{1}}{c_{3}} + \frac{Q_{2}}{c_{4}} = 30 \frac{2 Q}{5} + \frac{10 Q}{5} = 30 Q = 12.5 μ c \begin{array}{l} V_{A} - V_{B} = \frac{18}{1} = 18 - 1 \\ V_{A} - V_{B} = \frac{12.5}{2.5} = 5 - 2 \end{array}$
Potential difference is 13 V

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