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Potential difference that is applied across a conductor of resistance 'R' varies with time 't' according to the equation $V (t) = V_{0} e^{- t / z}$ where z is a constant that has dimensions of time. Obviously, the applied potential difference decreases exponentially with time. Charge that passes through any fixed observation point within the conductor between t = 0 and t = z will be

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$\frac{V_{0}}{R}$

$0.368 \frac{V_{0}}{R}$

$0.632 \frac{V_{0} Z}{R}$

$0.368 \frac{V_{0} z}{R}$

answer is C.

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Detailed Solution

$I = \frac{V}{R} = \frac{V_{0}}{R} e^{- t / z}$

$q = \int Idt = \frac{V_{0}}{R} \int_{0}^{z} e^{- t / z} dt$

$= \frac{V_{0}}{R} (- R) {[e^{- t / z}]}_{0}^{z}$

$= 0.632 \frac{V_{0} Z}{R}$

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