(C) The potential difference across a discharging capacitor is given by $V(t) = V_0 e^{-t/RC}$.Given that at $t = 1 \, \text{s}$, $V(1) = V/3$, we hav

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(C) The potential difference across a discharging capacitor is given by $V(t) = V_0 e^{-t/RC}$.Given that at $t = 1 \, \text{s}$, $V(1) = V/3$, we have:$V/3 = V e^{-1/RC} \implies e^{-1/RC} = 1/3$.We need to find the potential difference at $t = 2 \, \text{s}$:$V(2) = V e^{-2/RC} = V (e^{-1/RC})^2$.Substituting the value of $e^{-1/RC}$:$V(2) = V (1/3)^2 = V/9$.

Class 12 Physics Electric Potential and Capacitance Charging and Discharging of Capacitance and RC circuit (DC)

Medium

The plates of a capacitor are charged to a potential difference of $V \, \text{volts}$ and then connected across a resistor. The potential difference across the capacitor decreases exponentially with respect to time. After one second, the potential difference between the plates is $V/3$; then after two seconds from the start, the potential difference between the plates is

A
$V/3$
B
$V/6$
C
$V/9$
D
$2V/3$

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Class 12 Questions

Class 12

Physics Questions

Chapter

Electric Potential and Capacitance

Subtopic

Charging and Discharging of Capacitance and RC circuit (DC)

Find the charge in steady state of the capacitor. (in $nC$)

MediumAIIMS 2019

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The plates of a capacitor are charged to a potential difference of $320 \, V$ and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After $1 \, s$ the potential difference between the plates of the capacitor is $240 \, V$,then after $2 \, s$ and $3 \, s$ the potential difference between the plates will be:

Difficult

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Find the current from the battery at $(i)$ $t \rightarrow 0$ (initial) and $(ii)$ $t \rightarrow \infty$ (after a long time) when the switch is closed at $t = 0$.

Medium

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$A$ capacitor of capacitance $C$ is being discharged through a resistor $R$. Let $t_1$ be the time taken for the energy stored in the capacitor to reduce to half of its initial value,and $t_2$ be the time taken for the charge to reduce to one-fourth of its initial value. Find the ratio $t_1/t_2$.

Difficult

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In the following circuit,the switch $S$ is closed at $t = 0.$ The charge on the capacitor $C_1$ as a function of time will be given by $\left( {{C_{eq}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right).$

DifficultJEE MAIN 2018

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Similar Questions

Find the charge in steady state of the capacitor. (in $nC$)

Find the current from the battery at $(i)$ $t \rightarrow 0$ (initial) and $(ii)$ $t \rightarrow \infty$ (after a long time) when the switch is closed at $t = 0$.

$A$ capacitor of capacitance $C$ is being discharged through a resistor $R$. Let $t_1$ be the time taken for the energy stored in the capacitor to reduce to half of its initial value,and $t_2$ be the time taken for the charge to reduce to one-fourth of its initial value. Find the ratio $t_1/t_2$.

In the following circuit,the switch $S$ is closed at $t = 0.$ The charge on the capacitor $C_1$ as a function of time will be given by $\left( {{C_{eq}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right).$

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