JEE MAIN - Physics (2012 - No. 10)
The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant $$\tau $$ of this circuit lies between
100 sec and 150 sec
0 and 50 sec
50 sec and 100 sec
150 sec and 200 sec
Explanation
During discharging of a capacitor, V = V0$${e^{ - {t \over \tau }}}$$
At t = $$\tau $$,
V = $${{{V_0}} \over e}$$ = 0.37V0
From the graph, t = 0, V0 = 25 V
$$ \therefore $$ V = 0.37 $$ \times $$ 25 = 9.25 V
This voltage will occur at time between 100 sec and 150 sec. Hence, time constant $$\tau $$ of this circuit lies between 100 sec and 150 sec.
At t = $$\tau $$,
V = $${{{V_0}} \over e}$$ = 0.37V0
From the graph, t = 0, V0 = 25 V
$$ \therefore $$ V = 0.37 $$ \times $$ 25 = 9.25 V
This voltage will occur at time between 100 sec and 150 sec. Hence, time constant $$\tau $$ of this circuit lies between 100 sec and 150 sec.
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