$A$ coil of inductance $8.4 \, mH$ and resistance $6 \, \Omega$ is connected to a $12 \, V$ battery. The current in the coil is $1.0 \, A$ at approximately the time

The time constant of an $LR$ circuit is $10 \, s$. When a resistance of $10 \, \Omega$ is connected in series with the existing circuit,the time constant becomes $2 \, s$. The self-inductance of the circuit is ..... $H$.

As shown in the figure,a battery of $emf \; \varepsilon$ is connected to an inductor $L$ and resistance $R$ in series. The switch $S$ is closed at $t=0$. The total charge that flows from the battery between $t=0$ and $t=t_{c}$ ($t_{c}$ is the time constant of the circuit) is

$A$ constant voltage of $25 \, V$ is applied to a series $L-R$ circuit at $t=0$ by closing a switch. What is the potential difference across the resistor and the inductor at time $t=0$?

$A$ $50 \, V$ potential difference is suddenly applied to a coil with $L = 5 \times 10^{-3} \, H$ and $R = 180 \, \Omega$. The rate of increase of current after $0.001 \, s$ is ....... $A/s$.

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?