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

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

$A$ $2500 \,\mu F$ capacitor is charged through a $1 \,k\Omega$ resistor by a $12 \,V$ d.c. source. The voltage across the capacitor after $5 \,s$ is ..... $V$.

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 $\ln 5=1.6, \ln 3=1.1$]

$A$ capacitor is discharging through a resistor $R$. Consider in time $t_{1}$,the energy stored in the capacitor reduces to half of its initial value and in time $t_{2}$,the charge stored reduces to one-eighth of its initial value. The ratio $t_{1} / t_{2}$ will be ................

$A$ charged capacitor is allowed to discharge through a resistor by closing the key at the instant $t = 0$. At the instant $t = (\ln 4) \, \mu s$,the reading of the ammeter falls to half the initial value. The resistance of the ammeter is equal to