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 an $LR$ circuit,the time constant is the time in which the current grows from zero to the value (where ${I_0}$ is the steady-state current):

In the circuit shown, when the current $i$ is $3 \text{ A}$ and increasing at the rate of $1 \text{ A/s}$, the potential difference between $A$ and $B$ is $12 \text{ V}$. When the same current $3 \text{ A}$ is decreasing at the rate of $1 \text{ A/s}$, the potential difference between $A$ and $B$ is $6 \text{ V}$. The value of $R$ in the circuit is: (in $\text{ } \Omega$)

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

In a decaying $L-R$ circuit,the time after which energy stored in the inductor reduces to one-fourth of its initial value is

$A$ 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 $\ln 2 = 0.693$,$e^{-3/2} = 0.25$)