(B) $(i)$ At $t \rightarrow 0$,the capacitor acts as a short circuit (zero resistance wire). In this state,the total resistance of the circuit is $R_1

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$I = \frac{E}{R_1}, I = \frac{E}{R_1 + R_2}$

(B) $(i)$ At $t \rightarrow 0$,the capacitor acts as a short circuit (zero resistance wire). In this state,the total resistance of the circuit is $R_1$. Therefore,the current $I = \frac{E}{R_1}$.$(ii)$ At $t \rightarrow \infty$,the capacitor is fully charged and acts as an open circuit (no current flows through it). In this state,the resistors $R_1$ and $R_2$ are in series. The total resistance is $R_1 + R_2$. Therefore,the current $I = \frac{E}{R_1 + R_2}$.

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

Medium

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
$I = \frac{E}{R_1}, I = \frac{E}{R_1 - R_2}$
B
$I = \frac{E}{R_1}, I = \frac{E}{R_1 + R_2}$
C
$I = \frac{E}{R_2}, I = \frac{E}{R_1}$
D
$I = \frac{R_1 + R_2}{E + R_1}, I = \frac{E}{R_1 + R_2}$

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Class 12

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Electric Potential and Capacitance

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Charging and Discharging of Capacitance and RC circuit (DC)

$A$ capacitor of capacitance $C$ is discharging 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. Then the ratio $t_1 / t_2$ is:

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The stored charge in the capacitor in steady state of the following circuit is . . . . . . $\mu C$.

DifficultJEE MAIN 2026

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Under steady state condition,the potential difference across the capacitor in the circuit is . . . . . . $V$.

DifficultJEE MAIN 2026

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$A$ filter circuit used in a rectifier has a load resistance of $200 \Omega$ and a capacitance of $15 \mu\text{F}$. The value of the time constant is . . . . . . .

MediumGUJCET 2025

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At steady state,the charge on the capacitor,as shown in the circuit below,is . . . . . . $\mu \text{C}.$

MediumJEE MAIN 2025

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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 $t = 0$.

Similar Questions

$A$ capacitor of capacitance $C$ is discharging 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. Then the ratio $t_1 / t_2$ is:

The stored charge in the capacitor in steady state of the following circuit is . . . . . . $\mu C$.

Under steady state condition,the potential difference across the capacitor in the circuit is . . . . . . $V$.

$A$ filter circuit used in a rectifier has a load resistance of $200 \Omega$ and a capacitance of $15 \mu\text{F}$. The value of the time constant is . . . . . . .

At steady state,the charge on the capacitor,as shown in the circuit below,is . . . . . . $\mu \text{C}.$

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