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Three identical capacitors are given a charge $Q$ each and they are then allowed to discharge through resistances $R_1, R_2,$ and $R_3$. Their charges,as a function of time,are shown in the graph below. The smallest of the three resistances is

In the circuit shown in the figure,$R = \sqrt{\frac{L}{C}}$. Switch $S$ is closed at time $t = 0$. The current through $C$ and $L$ would be equal after a time $t$ equal to

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$A$ capacitor is connected to a cell of $emf$ $E$ having some internal resistance $r$. The potential difference across the

$A$ capacitor of capacitance $C = 10 \, \mu F$ is connected to a battery of emf $\varepsilon = 2 \, V$ through a resistor $R$. It is observed that it takes $t = 50 \, ms$ for the charge on the capacitor to reach $q = 12.6 \, \mu C$. Calculate the resistance $R$ of the circuit in $k \Omega$. (Take $1/e \approx 0.37$)

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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