Find the time constant of the following circuit.

The current growth in two $L-R$ circuits $(b)$ and $(c)$ is shown in figure $(a)$. Let $L_1, L_2, R_1$ and $R_2$ be the corresponding values in the two circuits. Then:

The time constant of an inductance coil is $3 \text{ ms}$. When a $90 \Omega$ resistance is joined in series,the time constant becomes $0.5 \text{ ms}$. The inductance and the resistance of the coil are:

An inductor $L$,a resistance $R$,and two identical bulbs $B_1$ and $B_2$ are connected to a battery through a switch $S$ as shown in the figure. The resistance $R$ is the same as that of the coil that makes $L$. Which of the following statements gives the correct description of the happenings when the switch $S$ is closed?

The resistance and inductance of a series $RL$ circuit are $5 \, \Omega$ and $20 \, H$ respectively. At the instant of closing the switch,the current is increasing at the rate of $4 \, A/s$. The supply voltage is ... $V$.

$A$ coil of self-inductance $10\,mH$ and resistance $0.1\,\Omega$ is connected through a switch to a battery of internal resistance $0.9\,\Omega$. After the switch is closed,the time taken for the current to attain $80\%$ of its saturation value is: [take $\ln\,5 = 1.6$] (in $,s$)