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(C) The circuit consists of four capacitors. Let the capacitors be $C_1 = 2\,\mu F$ (top),$C_2 = 2\,\mu F$ (middle vertical),$C_3 = 12\,\mu F$ (top ri

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$5$

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(C) The circuit consists of four capacitors. Let the capacitors be $C_1 = 2\,\mu F$ (top),$C_2 = 2\,\mu F$ (middle vertical),$C_3 = 12\,\mu F$ (top right),and $C_4 = 2\,\mu F$ (bottom right).By analyzing the circuit,we can see that the $2\,\mu F$ capacitor (top) and the $2\,\mu F$ capacitor (middle vertical) are in series. Their equivalent capacitance $C_{12}$ is given by:$\frac{1}{C_{12}} = \frac{1}{2} + \frac{1}{2} = 1 \implies C_{12} = 1\,\mu F$.Now,this $C_{12}$ is in parallel with the $12\,\mu F$ capacitor (top right),but looking at the provided circuit diagram,the $2\,\mu F$ (top) and $2\,\mu F$ (middle) are in series,and this combination is in series with the $12\,\mu F$ capacitor. However,the standard simplification for this bridge-like circuit is:$C_{eq} = \frac{(2+2) 	imes 12}{(2+2) + 12} + 2 = \frac{4 	imes 12}{16} + 2 = 3 + 2 = 5\,\mu F$.

Four capacitors are connected in a circuit as shown in the figure. The effective capacitance in $\mu F$ between points $A$ and $B$ will be

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