An $L-R$ circuit has a cell of $e.m.f.$ $E$,which is switched on at time $t = 0$. The current in the circuit after a long time will be

$A$ 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 the circuit shown,the cell is ideal. The coil has an inductance of $4H$ and zero resistance. $F$ is a fuse of zero resistance and will blow when the current through it reaches $5A$. The switch is closed at $t = 0$. The fuse will blow after how much time?

In the circuit shown here, the point '$C$' is kept connected to point '$A$' till the current flowing through the circuit becomes constant. Afterward, suddenly, point '$C$' is disconnected from point '$A$' and connected to point '$B$' at time $t = 0$. The ratio of the voltage across the resistance $(V_R)$ and the inductor $(V_L)$ at $t = L/R$ will be equal to:

The resistance in the following circuit is increased at a particular instant. At this instant,the value of resistance is $10\,\Omega$. The current in the circuit will be

For an $L-R$ circuit,the time constant is equal to