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(D) The circuit consists of a capacitor $C$ connected in parallel with a series combination of three capacitors: $2C$,$C$,and $2C$.Since the capacitor

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$30$

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(D) The circuit consists of a capacitor $C$ connected in parallel with a series combination of three capacitors: $2C$,$C$,and $2C$.Since the capacitor $C$ is connected directly across $AB$,the potential difference across it is $60\,V$.The series branch containing capacitors $2C$,$C$,and $2C$ is also connected across $AB$,so the total potential difference across this branch is $60\,V$.In a series combination,the potential difference $V$ across a capacitor is inversely proportional to its capacitance $(V = Q/C)$. Since the charge $Q$ is the same for all capacitors in series,$V \propto 1/C$.Let the potential differences across the capacitors $2C$,$C$,and $2C$ be $V_1$,$V_2$,and $V_3$ respectively.$V_1 : V_2 : V_3 = \frac{1}{2C} : \frac{1}{C} : \frac{1}{2C} = \frac{1}{2} : 1 : \frac{1}{2} = 1 : 2 : 1$.The sum of these potential differences is $V_1 + V_2 + V_3 = 60\,V$.Using the ratio $1:2:1$,the sum of parts is $1+2+1 = 4$.Thus,$V_2 = \frac{2}{4} 	imes 60\,V = 30\,V$.The points $M$ and $N$ are the terminals of the middle capacitor $C$ in the series branch. Therefore,the potential difference between $M$ and $N$ is $V_2 = 30\,V$.

In the circuit shown,a potential difference of $60\,V$ is applied across $AB$. The potential difference between the points $M$ and $N$ is.....$V$.

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