What will be the current through the $200 \Omega$ resistor in the given circuit, a long time after the switch $K$ is closed?

The electric field between the two parallel plates of a capacitor of $1.5 \mu F$ capacitance drops to one third of its initial value in $6.6 \mu s$ when the plates are connected by a thin wire. The resistance of this wire is . . . . . . . $\Omega$. (Given,$\log_{e} 3 = 1.1$)

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

At time $t=0$,a battery of $10 \ V$ is connected across points $A$ and $B$ in the given circuit. If the capacitors have no charge initially,at what time (in seconds) does the voltage across them become $4 \ V$? [Take $\ln 5=1.6, \ln 3=1.1$]

The circuit shown has been connected for a long time. The voltage across the capacitor is ............. $V$.

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