JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Capacitance

In the circuit shown, find C if the effective capacitance of the whole circuit is to be

0.5 μ F

. All values in the circuit are in

μ F

(A) $\frac{7}{11} μ F$ ✓ Correct
(B) $\frac{6}{5} μ F$
(C) $4 μ F$
(D) $\frac{7}{10} μ F$

Answer: (A)

Solution

Reducing the network, the branches combine to an equivalent capacitance

C_{e q} = \frac{\frac{7}{3} C}{\frac{7}{3} + C}

. Setting this equal to the required

0.5 μ

F gives

\frac{7 C}{3 ( \frac{7}{3} + C )} = \frac{1}{2}

, so

\frac{14 C}{3} = \frac{7}{3} + C

, i.e.

\frac{11 C}{3} = \frac{7}{3}

, and therefore

C = \frac{7}{11} μ

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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