A capacitor discharges through a resistance. The stored energy $U_{0}$ , in one capacitive time constant falls to:

\frac{U_{0}}{e^{2}}

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e U_{0}

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\frac{U_{0}}{e}

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none of these

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Solution

The correct option is A $\frac{U_{0}}{e^{2}}$
For an RC circuit, $q = q_{0} e^{- \frac{t}{R C}}$ and $U_{0} = \frac{q_{0}^{2}}{2 C}$
Now, stored energy at any instant: $U = \frac{q^{2}}{2 C} = \frac{q_{0}^{2} e^{- \frac{2 t}{R C}}}{2 C}$
In one capacitive time constant, stored energy:
$U^{'} = \frac{q_{0}^{2} e^{- 2 \times 1}}{2 C}$ $= U_{0} e^{- 2} = \frac{U_{0}}{e^{2}}$

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Column I	Column II
(A) $U_{1} (x) = \frac{U_{0}}{2} {[1 - {(\frac{x}{a})}^{2}]}^{2}$	(P) The force acting on the particle is zero at x = a.
(B) $U_{2} (x) = \frac{U_{0}}{2} {(\frac{x}{a})}^{2}$	(Q) The force acting on the particle is zero at x = 0.
(C) $U_{3} (x) = \frac{U_{0}}{2} {(\frac{x}{a})}^{2} exp [- {(\frac{x}{a})}^{2}]$	(R) The force acting on the particle is zero at x = – a.
(D) $U_{4} (x) = \frac{U_{0}}{2} [\frac{x}{a} - \frac{1}{3} {(\frac{x}{a})}^{3}]$	(S) The particle experiences an attractive force towards x = 0 in the region $\| x \| < a .$
	(T) The particle with total energy $\frac{U_{0}}{4}$ can oscillate about the point x = – a.

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