(D) The charge on a capacitor during charging is given by $q(t) = CE(1 - e^{-t/RC})$.The time constant is $\tau = RC$.From the graph,circuit $2$ takes

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$\frac{R_1}{R_2} < \frac{C_2}{C_1}$

(D) The charge on a capacitor during charging is given by $q(t) = CE(1 - e^{-t/RC})$.The time constant is $\tau = RC$.From the graph,circuit $2$ takes more time to reach the maximum charge compared to circuit $1$,which means the time constant of circuit $2$ is greater than that of circuit $1$.Thus,$\tau_2 > \tau_1$,which implies $R_2C_2 > R_1C_1$.Rearranging this,we get $\frac{R_1}{R_2} Therefore,option $D$ is correct.

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

Medium

The charge across the capacitor in two different $RC$ circuits $1$ and $2$ are plotted as shown in the figure. Identify the correct statement$(s)$ related to the $R_1, R_2, C_1$ and $C_2$ of the two $RC$ circuits.

A
$R_1 > R_2$ if $E_1 = E_2$
B
$C_1 < C_2$ if $E_1 = E_2$
C
$R_1C_1 > R_2C_2$
D
$\frac{R_1}{R_2} < \frac{C_2}{C_1}$

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Electrostatic Potential and Capacitance

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Charging and Discharging of Capacitance and RC circuit (DC)

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The charge across the capacitor in two different $RC$ circuits $1$ and $2$ are plotted as shown in the figure. Identify the correct statement$(s)$ related to the $R_1, R_2, C_1$ and $C_2$ of the two $RC$ circuits.

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When a lamp is connected in series with a capacitor,then:

The charge stored by the capacitor $C$ in the given circuit in the steady state is . . . . . . $\mu C$.

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The capacitor of capacitance $C$ in the circuit shown is fully charged initially. The resistance is $R$. After the switch $S$ is closed,the time taken to reduce the stored energy in the capacitor to half its initial value is

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