Calculate the current,a long time after the closing of the key $K$ in a circuit containing a battery of $10 \ V$,a resistor of $2 \ \Omega$,and an inductor of $5 \ H$. (in $A$)

Inductance of a coil with $10^4$ turns is $10 \text{ mH}$ and it is connected to a $DC$ source of $10 \text{ V}$ with internal resistance of $10 \ \Omega$. The energy density in the inductor when the current reaches $(1/e)$ of its maximum value is $\alpha \pi \times (1/e^2) \text{ J/m}^3$. The value of $\alpha$ is . . . . . . . $(\mu_0 = 4 \pi \times 10^{-7} \text{ Tm/A})$.

$A$ coil of inductance $1\,H$ and resistance $100\,\Omega$ is connected to a battery of $6\,V$. Determine approximately:
$(a)$ The time elapsed before the current acquires half of its steady-state value.
$(b)$ The energy stored in the magnetic field associated with the coil at an instant $15\,ms$ after the circuit is switched on. (Given $\ln 2 = 0.693$,$e^{-3/2} = 0.25$)

In an $L-R$ decay circuit,the initial current at $t = 0$ is $I$. The total charge that has flown through the resistor until the energy in the inductor has reduced to one-fourth of its initial value is:

In the circuit shown,the switch $S$ is initially open. The switch $S$ is closed at $t = 0$. The difference between the maximum and minimum current that can flow in the circuit is.....$A$.

An inductor coil carries a steady-state current of $2.0 \ A$ when connected across an ideal battery of $emf$ $4.0 \ V$. If its inductance is $1.0 \ H$,find the time constant of the circuit in $sec$.