The equivalent capacitance between terminals | vedclass

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(D) Let the four capacitors be $C_1, C_2, C_3, C_4$ in a row from left to right. Terminal $A$ is connected to the left plate of $C_1$. The right plate

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$4C$

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(D) Let the four capacitors be $C_1, C_2, C_3, C_4$ in a row from left to right. Terminal $A$ is connected to the left plate of $C_1$. The right plate of $C_1$,left plate of $C_2$,and right plate of $C_4$ are connected together. The right plate of $C_2$ and left plate of $C_3$ are connected. The right plate of $C_3$ and left plate of $C_4$ are connected to terminal $B$. By analyzing the circuit,we see that $C_1, C_2, C_3$ are in series,and this combination is in parallel with $C_4$. However,looking at the diagram,the first three capacitors are in series between $A$ and $B$,and the fourth capacitor is connected between $A$ and $B$ as well. Actually,the circuit shows three capacitors in series between $A$ and $B$,and one capacitor in parallel with the series combination. Equivalent capacitance of three capacitors in series: $\frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \implies C_s = \frac{C}{3}$. Now,this $C_s$ is in parallel with the fourth capacitor $C$. $C_{eq} = C_s + C = \frac{C}{3} + C = \frac{4C}{3}$.

The equivalent capacitance between terminals $A$ and $B$ is:

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