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Motional EMF in an Open Conductor

The numerical...

Question

The numerical value of charge on any plate of the capacitor of value $C$ is

C E

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\frac{C E R_{1}}{(R_{1} + R_{3})}

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\frac{C E R_{2}}{(R_{2} + R_{3})}

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\frac{C E R_{1}}{(R_{2} + R_{3})}

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Solution

The correct option is A $\frac{C E R_{1}}{(R_{1} + R_{3})}$
Capacitance offer infinite resistance for d.c. Hence, there will not be any current through the loop containing $C$ .
$∴ I_{2} = 0$
Apply kirchhoff's first law at junction $N$
$I = I_{1} + I_{2}$
$I = I_{1} (∵ s i n c e I_{2} = 0)$
Apply Kirchhoff's second law to the closed loop $M N Q R M$ , we get
$- I_{1} R_{1} - (I_{1} + I_{2}) R_{3} + E = 0$
or $I_{1} = \frac{E}{R_{1} + R_{3}} (∵ I_{2} = 0)$
Potential difference between $N$ and $Q$
$V_{1} = I_{1} R_{1} = \frac{E R_{1}}{R_{1} + R_{3}}$
Charge on plate of capacitor is $Q = V_{1} C$
$= \frac{E R_{1} C}{R_{1} + R_{3}}$ .

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