Find $V_A - V_B$ in steady state. (in $, V$)

When a certain circuit consisting of an emf $E$,an inductance $L$,and a resistance $R$ is closed,the current in it increases with time according to curve $1$. After one parameter ($L$ or $R$) is changed,the increase in the current follows curve $2$,when the circuit is closed a second time. Which parameter was changed?

The current in an $LR$ circuit builds up to $3/4$ of its steady state value in $4 \, s$. The time constant of this circuit is:

$A$ $50 \, V$ 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 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:

$A$ $20 \, H$ inductor coil is connected to a $10 \, \Omega$ resistance in series as shown in the figure. The time at which the rate of dissipation of energy (Joule's heat) across the resistance is equal to the rate at which magnetic energy is stored in the inductor is: