Mutual Induction — Question 6

Question 6

An ideal shorted secondary reduces the effective primary inductance

Two fixed lossless coils have positive M with $M^2<L_1L_2$. Coil 2 is shorted by a perfect conductor. Both currents start at zero. A prescribed current i1 is then established in coil 1.

1. Find secondary current by conserving its total linkage.

2. Derive the primary terminal inductance and total stored energy.

Question 6 - Solution

An ideal shorted secondary reduces the effective primary inductance

(a) Method and calculation

$$v_2=\frac d{dt}(L_2i_2+Mi_1)=0\Rightarrow L_2i_2+Mi_1=0.$$$$\boxed{i_2=-Mi_1/L_2}.$$

(b) Interpretation and further calculation

$$v_1=L_1\dot i_1+M\dot i_2=\left(L_1-M^2/L_2\right)\dot i_1.$$$$\boxed{L_{\mathrm{eff}}=L_1-M^2/L_2=L_1(1-k^2),\quad U=\tfrac12L_{\mathrm{eff}}i_1^2}.$$

Physical check

The opposing secondary current partly cancels primary linkage; the coupling bound keeps the effective inductance nonnegative.