A capacitor is discharging through a resistor | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.In time $t_{1}$,the energy reduces to half,so $U(t_{1}) = \frac{U_{0}}{2}$.This

Vedclass · Accepted Answer

$1/6$

Vedclass · Answer

(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.In time $t_{1}$,the energy reduces to half,so $U(t_{1}) = \frac{U_{0}}{2}$.This implies $\frac{q(t_{1})^2}{2C} = \frac{1}{2} \frac{Q_{0}^2}{2C}$,which simplifies to $q(t_{1}) = \frac{Q_{0}}{\sqrt{2}}$.Using the discharging equation $q(t) = Q_{0} e^{-t/RC}$,we have $\frac{Q_{0}}{\sqrt{2}} = Q_{0} e^{-t_{1}/RC}$,so $e^{-t_{1}/RC} = 2^{-1/2}$.Taking the natural logarithm,$\frac{t_{1}}{RC} = \frac{1}{2} \ln(2)$.In time $t_{2}$,the charge reduces to one-eighth,so $q(t_{2}) = \frac{Q_{0}}{8}$.Using the discharging equation,$\frac{Q_{0}}{8} = Q_{0} e^{-t_{2}/RC}$,so $e^{-t_{2}/RC} = 2^{-3}$.Taking the natural logarithm,$\frac{t_{2}}{RC} = 3 \ln(2)$.Now,the ratio $\frac{t_{1}}{t_{2}} = \frac{\frac{1}{2} \ln(2)}{3 \ln(2)} = \frac{1}{6}$.

$A$ 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 ................

Similar Questions

$A$ $4\,\mu F$ capacitor is charged to $400\,V$ and then its plates are joined through a resistance of $1\,k\Omega$. The heat produced in the resistance is ............... $J$.

In the given circuit diagram,when the current reaches a steady state,the charge on the capacitor of capacitance $C$ will be:

In a $CR$ circuit, the growth of charge on the capacitor is

$A$ $1 \mu F$ capacitor $C$ is connected to a battery of $10 V$ through a resistance of $1 \text{ M}\Omega$. The voltage across $C$ after $1 \text{ s}$ is approximately: (in $V$)

$A$ capacitor is connected to a $20\, V$ battery through a resistance of $10\, \Omega$. It is found that the potential difference across the capacitor rises to $2\, V$ in $1\, \mu s$. The capacitance of the capacitor is $....\, \mu F$. Given: $\ln(10/9) = 0.105$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module