In the figure shown,the effective capacitance | vedclass

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(A) The circuit consists of five capacitors,each of capacitance $C$. Looking at the circuit,the two capacitors on the left side are in series,and thei

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$2\,C$

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(A) The circuit consists of five capacitors,each of capacitance $C$. Looking at the circuit,the two capacitors on the left side are in series,and their equivalent capacitance is $C_{s1} = \frac{C 	imes C}{C + C} = \frac{C}{2}$. Similarly,the two capacitors on the right side are in series,and their equivalent capacitance is $C_{s2} = \frac{C 	imes C}{C + C} = \frac{C}{2}$. Now,these two equivalent capacitors ($C_{s1}$ and $C_{s2}$) are in parallel with the central capacitor of capacitance $C$. Therefore,the total effective capacitance between points $A$ and $B$ is $C_{AB} = C_{s1} + C_{s2} + C = \frac{C}{2} + \frac{C}{2} + C = C + C = 2\,C$.

In the figure shown,the effective capacitance between the points $A$ and $B$,if each capacitor has capacitance $C$,is

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