An inductor coil of inductance $L$ is divided into two parts and both parts are connected in parallel. The net inductance is

$A$ circuit contains two inductors of self-inductance $L_1$ and $L_2$ connected in series as shown in the figure. If $M$ is the mutual inductance between them,then the effective inductance of the circuit will be

Two inductors of $80 \ mH$ each are joined in parallel. The current passing through the combination is $2.1 \ A$. The energy stored in this combination of inductors is

Two inductors $A$ and $B$ when connected in parallel are equivalent to a single inductor of inductance $1.5 \ H$,and when connected in series are equivalent to a single inductor of inductance $8 \ H$. Find the difference in the inductances of $A$ and $B$. (in $H$)

Initially,the switch $S$ is open and the energy stored in the inductance $L$ connected in series with the battery is $E$. Now,the switch $S$ is closed. The energy stored in both the inductors after a long time is

Two coils with self-inductance $1 H$ and $3 H$ are connected in series. Their mutual inductance is $5 H$. The equivalent self-inductance of the combination is: (in $H$)