For the arrangement of capacitors as shown in | vedclass

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(A) Let the capacitance of each capacitor be $C = 4 \mu F$. Looking at the circuit,we can identify two branches connected in parallel between points $

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$4$

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(A) Let the capacitance of each capacitor be $C = 4 \mu F$. Looking at the circuit,we can identify two branches connected in parallel between points $A$ and $B$. Each branch consists of two capacitors in series. For the upper branch,the two capacitors of $4 \mu F$ each are in series,so their equivalent capacitance $C_1$ is given by: $1/C_1 = 1/4 + 1/4 = 2/4 = 1/2 \implies C_1 = 2 \mu F$. Similarly,for the lower branch,the two capacitors of $4 \mu F$ each are in series,so their equivalent capacitance $C_2$ is given by: $1/C_2 = 1/4 + 1/4 = 2/4 = 1/2 \implies C_2 = 2 \mu F$. Now,these two branches ($C_1$ and $C_2$) are connected in parallel between points $A$ and $B$. Therefore,the effective capacitance $C_{eq}$ is: $C_{eq} = C_1 + C_2 = 2 \mu F + 2 \mu F = 4 \mu F$.

For the arrangement of capacitors as shown in the circuit,the effective capacitance between the points $A$ and $B$ is (capacitance of each capacitor is $4 \mu F$): (in $\mu F$)

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