(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 Electrostatic 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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Electrostatic Potential and Capacitance

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Charging and Discharging of Capacitance and RC circuit (DC)

An initially uncharged capacitor $C$ is being charged by a battery of emf $E$ through a resistance $R$. Up to the instant when the capacitor is charged to a potential $E/2$, the ratio of the work done by the battery to the heat dissipated by the resistor is given by

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In the circuit shown,the switch is shifted from position $1 \rightarrow 2$ at $t = 0$. The switch was initially in position $1$ for a long time. The graph between the charge on capacitor $C$ and time $t$ is:

Medium

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As shown in the figure,if a capacitor of capacitance $C$ is charged by connecting it with a resistance $R$ and a battery of $EMF$ $V$,then the total energy supplied by the battery is:

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$A$ capacitor is charged to a potential difference of $100\, V$ and then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After $1\, s$,the potential difference across the capacitor plates is $80\, V$. What is the fractional loss of the stored energy?

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An electric bulb, a capacitor, a battery, and a switch are all connected in series in a circuit. How does the intensity of light vary when the switch is turned on?

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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 $t = 0$.

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An initially uncharged capacitor $C$ is being charged by a battery of emf $E$ through a resistance $R$. Up to the instant when the capacitor is charged to a potential $E/2$, the ratio of the work done by the battery to the heat dissipated by the resistor is given by

In the circuit shown,the switch is shifted from position $1 \rightarrow 2$ at $t = 0$. The switch was initially in position $1$ for a long time. The graph between the charge on capacitor $C$ and time $t$ is:

As shown in the figure,if a capacitor of capacitance $C$ is charged by connecting it with a resistance $R$ and a battery of $EMF$ $V$,then the total energy supplied by the battery is:

An electric bulb, a capacitor, a battery, and a switch are all connected in series in a circuit. How does the intensity of light vary when the switch is turned on?

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