Three inductances are connected as shown below. Assuming no coupling,the resultant inductance will be $H$.

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

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 identical coils of inductance $L$ joined in series are placed very close to each other such that the winding direction of one coil is exactly opposite to that of the other. The net inductance is

Two inductors,each of inductance $L$,are connected in parallel. One more inductor of value $5 \text{ mH}$ is connected in series with this configuration,and the effective inductance is $15 \text{ mH}$. The value of $L$ is . . . . . . $\text{mH}$.

Two coils of self-inductance $L_{1}$ and $L_{2}$ are connected in series. If the mutual inductance of the coils is $M$,the equivalent self-inductance of the combination shown in the figure will be: