For circuit the equivalent capacitance between $P$ and $Q$ is

\frac{6 C}{11}

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4 C

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\frac{3 C}{2}

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\frac{3 C}{4}

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Solution

The correct option is A $\frac{6 C}{11}$
In this given figure.
$\begin{matrix} C = C_{1} + C_{2} + C_{3} + C_{4} + C_{5} + C_{6} = 1 + 1 + 1 + 1 + 1 + 1 = 6 C \end{matrix}$
Hence,
Option $A$ is correct .

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Similar questions

\frac{{tan}^{2} {(\frac{π}{6})}^{c} + {sin}^{2} {(\frac{π}{6})}^{c} + {cos}^{2} {(\frac{π}{3})}^{c}}{{sec}^{2} {(\frac{π}{4})}^{c} - {cos}^{2} π^{c}} = \frac{1}{\sqrt{3}} sec {(\frac{π}{6})}^{c} + \frac{1}{3} cos {(\frac{π}{3})}^{c}

What is the equivalent capacitance of the system of capacitors between A and B ?

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