The circuit under consideration has two inductively coupled loops, one with a DC

Question

The circuit under consideration has two inductively coupled loops, one with a DC battery, inductor, and resistor in series. The other loop has two inductors, one inductively coupled to the first, the other not, and a resistor, all in series.

I would like to know the current in the top loop (the one without the battery) as a function of the given quantities after the battery is connected.

I tried adding the three inductors and using that value to find the bottom loop's current, then substituting that into the top loop to find d(phi)/dt and then current. This gives me the standard RL current expression for the bottom loop and an exponential decay model for the current in the top loop. This is counter-intuitive for me; I believed that the current in the top loop should start at zero, peak, and decay rather than start high and decay.

Thanks in advance.

Explanation / Answer

Your intuition is correct: the top loop inductor prevents its current from changing instantaneously, so it starts at zero. Qualitatively, the top loop current:

starts at zero (Here the inductors block current flow in both loops.)
rises to a max as the top loop uncoupled inductor allows current to flow through it while its source voltage (the coupled inductor voltage) simultaneously falls as the bottom loop current rises.
decays to zero as the coupled inductor voltage falls to zero.

An equivalent circuit model does not contain all 3 inductors in series, but rather two parallel branches (corresponding to the primary and secondary of the coupled inductor), with the "secondary" branch (corresponding to the top loop) containing an inductance which blocks initial current flow.

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