Inductors and Inductance — Question 7

Question 7

Parallel inductors share changing current inversely in proportion to inductance

Two uncoupled ideal inductors, L1=2.00 mH and L2=6.00 mH, are connected in parallel with zero initial branch currents. The total current is increased to 8.00 A.

1. Derive the equivalent inductance and branch-current ratio.

2. Find the final branch currents.

Question 7 - Solution

Parallel inductors share changing current inversely in proportion to inductance

(a) Method and calculation

$$v=L_1\dot i_1=L_2\dot i_2,\quad\dot i=\dot i_1+\dot i_2=v(1/L_1+1/L_2).$$$$\boxed{L_{\mathrm{eq}}=\frac{L_1L_2}{L_1+L_2}=1.50\,\mathrm{mH}},\quad i_1/i_2=L_2/L_1=3.$$

(b) Interpretation and further calculation

$$i_1+i_2=8.00,\quad i_1=3i_2.$$$$\boxed{i_1=6.00\,A,\quad i_2=2.00\,A}.$$

Physical check

The current ratio assumes zero initial currents. A pre-existing circulating current is not described by the ratio alone.