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 $A$)

Two figures are shown as Fig. $A$ and Fig. $B$. The time constant of Fig. $A$ is $\tau_{A}$ and the time constant of Fig. $B$ is $\tau_{B}$. Then:

$A$ coil carrying a steady current is short-circuited. The current in it decreases $\alpha$ times in $t_0$. The time constant of the circuit is:-

The dimensions of the "time constant" $\frac{L}{R}$ during the growth and decay of current in an inductive circuit are the same as those of:

Assertion : The resistance offered by an inductor in a $d.c.$ circuit is always constant.
Reason : The resistance of an inductor in a steady state is non-zero.

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 $\text{ } \Omega$)