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(B) The circuit consists of two parallel branches connected across $AB$. One branch has a capacitor of capacitance $C$. The other branch consists of t

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$15$

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(B) The circuit consists of two parallel branches connected across $AB$. One branch has a capacitor of capacitance $C$. The other branch consists of three capacitors in series: $2C$,$C$,and $2C$. The equivalent capacitance of the series branch is given by $\frac{1}{C_{eq}} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{2C} = \frac{1+2+1}{2C} = \frac{4}{2C} = \frac{2}{C}$. Thus,$C_{eq} = \frac{C}{2}$. Since the capacitors are in series,the charge $Q$ on each capacitor in this branch is the same,given by $Q = C_{eq} 	imes V = (\frac{C}{2}) 	imes 30 = 15C$. The potential difference across the capacitor $C$ (between points $M$ and $N$) is $V_{MN} = \frac{Q}{C} = \frac{15C}{C} = 15\, V$.

In the circuit shown,a potential difference of $30\, V$ is applied across $AB$. The potential difference between the points $M$ and $N$ is....$V$

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