JEE MAIN - Physics (2022 - 29th June Evening Shift - No. 6)
A capacitor is discharging through a resistor R. Consider in time t1, the energy stored in the capacitor reduces to half of its initial value and in time t2, the charge stored reduces to one eighth of its initial value. The ratio t1/t2 will be
1/2
1/3
1/4
1/6
Explanation
For a discharging capacitor when energy reduces to half the charge would become $${1 \over {\sqrt 2 }}$$ times the initial value.
$$ \Rightarrow {\left( {{1 \over 2}} \right)^{1/2}} = {e^{ - {t_1}/\tau }}$$
Similarly, $${\left( {{1 \over 2}} \right)^3} = {e^{ - {t_2}/\tau }}$$
$$ \Rightarrow {{{t_1}} \over {{t_2}}} = {1 \over 6}$$
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