Find the equivalent inductance of two inductors having inductances `L_1` and `L_2` connected in parallel with the help of appropriate DC circuit.

Explanation

Let two inductor `L_1` and `L_2` are connected in parallel across a battery whose emf can be varied continuously and having no internal resistance.

Let currents passing through `L_1` and `L_2` are `I_1` and `I_2` at time t and the rates of change of currents through them are `(dI_1)/(dt)` and `(dI_2)/(dt)` respectively.
`therefore` p.d. between their two ends are respectively `-L_1 (dI_1)/(dt)` and `-L_2(dI_2)/(dt)`
Current passing through the main circuit is I at time t then,
`epsilon =-L (dI)/(dt)=-L ((dI_1)/(dt)+(dI_2)/(dt))`...(1)
But, `L_1(dI_1)/(dt)=-epsilon` and `L_2(dI_2)/(dt)=-epsilon`
`therefore (dI_1)/(dt)=-epsilon/L_1` and `(dI_2)/(dt)=-epsilon/L_2`
Putting these values in eqn. (i)
`epsilon=-L (-epsilon/L_1-epsilon/L_2)`
`therefore 1/L=1/L_1+1/L_2`