The shown circuit comprises of two identical capacitors of capacitance C Farad and resistor of resistance $R Ω .$ The key K is initially open. At time t = 0 the charge on left capacitor is $Q_{0}$ Coloumbs and the right capacitor is uncharged as shown. The key K is closed at time t = 0. Then the magnitude of current in amperes through the resistor at any later time t (in sec.) is:

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$\frac{Q_{0}}{R C} e^{\frac{- t}{R C}}$

$\frac{Q_{0}}{R C} e^{\frac{- 2 t}{R C}}$

$\frac{Q_{0}}{2 R C} e^{\frac{- t}{R C}}$

$\frac{Q_{0}}{2 R C} e^{\frac{- 2 t}{R C}}$

answer is B.

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Detailed Solution

At any time t, the charge on right capacitor be q. Applying Kirchoff's law
$\frac{Q_{0} - q}{C} = i R + \frac{q}{C} ∴ \frac{Q_{0} - 2 q}{2 C R} = \frac{d q}{d t}$
integrating and evaluating the constant we get
Question Image
Hence $q = \frac{Q_{0}}{2} (1 - e^{- \frac{2 t}{2 R C}})$
or $i = \frac{d q}{d t} = \frac{Q_{0}}{R C} e^{- \frac{2 t}{2 R C}}$

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