An inductor of inductance $L = 400 \ mH$ and resistors of resistance $R_1 = 2 \ \Omega$ and $R_2 = 2 \ \Omega$ are connected to a battery of emf $E = 12 \ V$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t = 0$. The potential drop across $L$ as a function of time is:

Regarding the given circuit,the correct statement is:

Consider an electrical circuit containing a two-way switch $S$. Initially,$S$ is open and then $T_{1}$ is connected to $T_{2}$. As the current in $R = 6 \, \Omega$ attains a maximum value of steady-state level,$T_{1}$ is disconnected from $T_{2}$ and immediately connected to $T_{3}$. The potential drop across the $r = 3 \, \Omega$ resistor immediately after $T_{1}$ is connected to $T_{3}$ is $.... \, V$. (Round off to the nearest integer)

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 ..... $H$.

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

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: