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.

  • $\begingroup$ Is it possible to add a figure of what you're describing? That will be a lot less confusing. :) $\endgroup$ – Kitchi Feb 16 '13 at 17:43

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

  1. starts at zero (Here the inductors block current flow in both loops.)
  2. 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.
  3. 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.

| cite | improve this answer | |
  • $\begingroup$ Could you give me a figure or a description of how to draw the equivalent circuit? My background in this subject is limited to AP Physics C. Thanks for the help. $\endgroup$ – user21033 Feb 16 '13 at 21:31
  • $\begingroup$ @user21033: A coupled inductor can be modeled as inductors plus an ideal transformer (see, for example here: icosym-nt.cvut.cz/courseodl/node40.html , where you have to link back to see the coupled inductor in figure 55. I like the model in figure 57b, but the representation is not unique.) An ideal transformer is characterized by its turns ratio n (and nothing else): secondary voltage = n * primary voltage, and primary current = n * secondary current. Sorry, I don't know if this level is appropriate for you or not... $\endgroup$ – Art Brown Feb 16 '13 at 23:08
  • $\begingroup$ I understand how this works in AC, but this is a DC application. I'm interested in the initial current that is set up after the battery is connected. My interpretation of your previous answer gives me a system of two differential equations, both with expected behavior. I believe that this set-up gives the correct answer, so I'm marking the question answered. Thank you for your help. $\endgroup$ – user21033 Feb 16 '13 at 23:43
  • $\begingroup$ @user21033: You're welcome. Technically, you're considering a transient situation: at DC, the voltages across all the inductances are 0 (so no DC current flows in the "top loop"). The coupled-inductor model is applicable in any case. Yes, there will be two coupled differential equations; your intuition is correct and useful for checking their solutions. $\endgroup$ – Art Brown Feb 16 '13 at 23:52
  • $\begingroup$ Could you point me in a direction to solve these diff eqs, or should I start a new question? I can't seem to change variables, so I can't put it in matrix form. They are: c1-c2*y-c3*x = dy/dt and c4-c5*x-c6*y = dx/dt Thanks. $\endgroup$ – user21033 Feb 17 '13 at 0:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.