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 $A$)

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.....$sec$.

If $L$ is the inductance and $R$ is the resistance,then the $SI$ unit of $\frac{L}{R}$ is

$A$ coil of inductance $40 \, H$ is connected in series with a resistance of $8 \, \Omega$ and the combination is joined to the terminals of a $2 \, V$ battery. The time constant of the circuit is ...... $s$.

$A$ coil of inductance $1\, H$ and resistance $10\, \Omega$ is connected to a battery of $emf$ $50\, V$ and negligible internal resistance at time $t = 0.$ The ratio of the rate at which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at $t = 0.1\, s$ is:

The time constant of a circuit is $10 \, s$. When a resistance of $10 \, \Omega$ is connected in series to the circuit,the time constant becomes $2 \, s$. The self-inductance of the circuit is ....... $H$.