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(C) The given circuit consists of four capacitors. Let the capacitors be $C_1 = 1\,\mu F$ (top left),$C_2 = 1\,\mu F$ (middle vertical),$C_3 = 1\,\mu

Vedclass · Accepted Answer

$\frac{8}{3}$

Vedclass · Answer

(C) The given circuit consists of four capacitors. Let the capacitors be $C_1 = 1\,\mu F$ (top left),$C_2 = 1\,\mu F$ (middle vertical),$C_3 = 1\,\mu F$ (top right),and $C_4 = 2\,\mu F$ (bottom right).First,notice that the capacitor $C_1$ and $C_2$ are in parallel. Their equivalent capacitance is $C_{12} = C_1 + C_2 = 1 + 1 = 2\,\mu F$.Now,this combination $C_{12}$ is in series with $C_3$. The equivalent capacitance of this branch is $C_{123} = \frac{C_{12} 	imes C_3}{C_{12} + C_3} = \frac{2 	imes 1}{2 + 1} = \frac{2}{3}\,\mu F$.Finally,this branch is in parallel with $C_4$. The total effective capacitance between $x$ and $y$ is $C_{xy} = C_{123} + C_4 = \frac{2}{3} + 2 = \frac{8}{3}\,\mu F$.

Four capacitors are connected as shown in the figure. Their capacities are indicated in the figure. The effective capacitance between points $x$ and $y$ is (in $\mu F$)

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