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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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5 V

9 V

10 V

13 V

answer is D.

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Detailed Solution

$q_{1} = q, q_{2} = 2 q$

$\frac{q_{1}}{c_{1}} + \frac{q_{2}}{c_{2}} = 30$

$q = 18 μ c$

$\frac{Q_{1}}{c_{3}} + \frac{Q_{2}}{c_{4}} = 30$

$\frac{2 Q}{5} + \frac{10 Q}{5} = 30$

$Q = 12.5 μ c$

$\begin{array}{l} V_{A} - V_{a} = \frac{18}{1} = 18 \\ V_{A} - V_{b} = \frac{12.5}{2.5} = 5 \end{array}$

Potential difference $V_{a} - V_{b} =$ 13 V

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