A capacitor is discharging through a resistor R. Consider in time t 1, 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 :

Question

A capacitor is discharging through a resistor R. Consider in time t 1, 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 : (A) 1 / 2 (B) 1 / 3 (C) 1 / 4 (D) 1 / 6

Tardigrade Admin · Accepted Answer

In t 1 time energy becomes half so charge will become (1/√2) time q = Q 0 e -( t 1/ RC )=( Q 0/√2) and q = Q 0 e -( t 1/ RC )=( Q 0/8)=(( Q 0/√2))6 t 2=6 t 1 ( t 1/ t 2)=(1/6)

Q. A capacitor is discharging through a resistor $R$ . Consider in time $t_{1}$ , the energy stored in the capacitor reduces to half of its initial value and in time $t_{2}$ , the charge stored reduces to one eighth of its initial value. The ratio $t_{1} / t_{2}$ will be :

$1/2$

$1/3$

$1/4$

$1/6$

In $t_{1}$ time energy becomes half so charge will become $\frac{1}{2}$ time
$q = Q_{0} e^{- \frac{t _{1}}{RC}} = \frac{Q _{0}}{2}$
and $q = Q_{0} e^{- \frac{t _{1}}{RC}} = \frac{Q _{0}}{8} = (\frac{Q _{0}}{2})^{6}$
$t_{2} = 6 t_{1}$
$\frac{t _{1}}{t _{2}} = \frac{1}{6}$

Q. 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