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Take two electrical circuits containing coupled inductors. Let their constant self-inductances be $L_1,L_2>0$, mutual inductance $M$, and positive electrical resistances $R_1,R_2$. Choose current directions so that the off-diagonal entries of the inductance matrix are $+M$. Assume a linear, quasistatic model, with no capacitive effects or core hysteresis. The magnetic flux linkages are ${\lambda }_1=L_1{i}_1+M{i}_2$ and ${\lambda }_2=M{i}_1+L_2{i}_2$. Combining their time derivatives with resistive electric potential differences givesThe secondary is a closed, undriven electrical circuit. Require $M^2<L_1{L}_2$, making $\mathbf{L}$ a positive-definite matrix. The chosen current convention fixes the signs; reversing the secondary's reference direction changes the sign assigned to $M$ and $i_2$ without changing the physical circuit.

There is an essential switching assumption: the primary loop remains closed when $V$ becomes zero. An ideal voltage source then imposes zero electric potential difference, allowing the current to decay. Literally opening an inductive circuit requires additional physics to determine the interruption, including possible capacitance and an electric arc. This closed-return model explains transient induction while keeping that distinction from Faraday's historical making and breaking of contact explicit.