Mutual Induction — Question 7

Question 7

A finite primary ramp produces a delayed secondary current and a fixed total charge

A primary current is zero before t=0, rises as $i_1=\alpha t$ for $0<t<T$, and remains constant afterward. A fixed secondary has inductance L2, resistance R2, positive M, and initial current zero. Positive currents enter dotted ends.

1. Solve $L_2\dot i_2+R_2i_2=-M\dot i_1$ during and after the ramp.

2. Find total signed secondary charge from zero to infinite time, including the decay tail.

Question 7 - Solution

A finite primary ramp produces a delayed secondary current and a fixed total charge

(a) Method and calculation

$$\tau_2=L_2/R_2,\quad\boxed{i_2(t)=-\frac{M\alpha}{R_2}(1-e^{-t/\tau_2})\quad(0<t<T)}.$$$$\boxed{i_2(t)=i_2(T)e^{-(t-T)/\tau_2}\quad(t>T)}.$$

(b) Interpretation and further calculation

$$\int_0^\infty(L_2\dot i_2+R_2i_2)dt=-M\Delta i_1,\quad i_2(0)=i_2(\infty)=0.$$$$\boxed{Q_2=-M\alpha T/R_2=-M\Delta i_1/R_2}.$$

Physical check

Secondary inductance reshapes the current pulse but does not change its integrated charge when the endpoint currents vanish.