The unit of $L/R$ is (where $L$ = inductance and $R$ = resistance).

An inductance coil has a time constant of $4 \, sec$. If it is cut into two equal parts and connected in parallel,then the new time constant of the circuit is.....$sec$.

In the circuit shown,$X$ is joined to $Y$ for a long time,and then $X$ is joined to $Z$. The total heat produced in $R_2$ is:

An inductor $(L = 100 \, mH)$,a resistor $(R = 100 \, \Omega)$ and a battery $(E = 100 \, V)$ are initially connected in series as shown in the figure. After a long time,the battery is disconnected by short-circuiting the points $A$ and $B$. The current in the circuit $1 \, ms$ after the short circuit is

In the given circuit,the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of $8 \text{ A/s}$. At an instant when $R = 12 \Omega$,the value of the current in the circuit will be . . . . . . $A$.

In an $L-R$ circuit,for the case of increasing current,the magnitude of current can be calculated by using the formula: