I'm trying to understand the behavior of inductors and transformers and I was trying out some circuits. I'm having trouble understanding what happens when we use a current source and a transformer. For the circuit below, the current source is constant in time: $i_1(t) = I_1$.

The KVL equations for this system are as follows: $$L_1 \frac{di_1(t)}{dt} + M \frac{di_2(t)}{dt} + R_1 i_1(t) = 0$$ $$L_2 \frac{di_2(t)}{dt} + M \frac{di_1(t)}{dt} + R_2 i_2(t) = 0$$ which reduces to: $$M \frac{di_2(t)}{dt} + R_1 I_1 = 0$$ $$L_2 \frac{di_2(t)}{dt} + R_2 i_2(t) = 0$$ Now, both equations give different solutions for $i_2(t)$ and there is no solution for the entire system. When using the initial value $i_2(0)=0$, the first equations gives $i_2(t) = \dfrac{-R_1 I_1}{M}t$ and the second equation gives $i_2(t)=0$. This second solution makes more sense because a constant current through the left inductor doesn't cause an emf in the right inductor. But why do I get two conflicting equations and what am I doing wrong?

Thanks in advance.

  • $\begingroup$ A constant current source with an inductor? Is it switched on at t = 0? Also, you ignored the voltage from the current source. It cannot be zero if there is resistance n the circuit. $\endgroup$ – R.W. Bird Oct 24 '19 at 16:56
  • $\begingroup$ Yes, it is switched on at $t=0$. And I didn't know about the voltage from the current source. How do I include it in my equations? What's the relationship with the current it provides? $\endgroup$ – Sudera Oct 24 '19 at 16:59
  • $\begingroup$ @R.W.Bird I think I see. The first equation becomes $V_I(t) + M \frac{di_2(t)}{dt} + R_1 I_1 = 0$ and when I solve the system now for $i_2(t)$ and $V_I(t)$, I do get a single solution for $i_2(t)$, where $i_2 = ce^{-\frac{R_2}{L_2}t}$. Is this correct? Thanks for your help. $\endgroup$ – Sudera Oct 24 '19 at 17:07
  • $\begingroup$ @Sudera A current source without a large parallel resistor representing its internal resistance is considered an ideal current source, which is what you have here. It is just like a voltage source without a low series resistor representing its internal resistance is an ideal voltage source. In this circuit, since the current is constant and the voltage across the ideal inductor is zero, the voltage across the current source will always be $I_{1}R_1$, regardless of the value of $R_1$. This is the case after the switch is closed a long time (transients vanish). $\endgroup$ – Bob D Oct 24 '19 at 17:18
  • $\begingroup$ @Sudera But if you are now including switching transients it is a completely different problem than what you initially presented. Are you considering switching transients? $\endgroup$ – Bob D Oct 24 '19 at 17:19

There are a number of mistakes in your assumptions of the circuit:

  1. You are neglecting the voltage of the current source. The right way to write the first equation would be:

$$L_1 \frac{di_1(t)}{dt} - M \frac{di_2(t)}{dt} + R_1 i_1(t) = V_S$$ $\qquad$where $V_S$ is the voltage of the source (in your diagram, $V_+-V_-$). Note that the sign for the mutual inductance is also incorrect.

  1. The second equation also has the wrong sign for the mutual inductance term, the correct thing would be:

$$L_2 \frac{di_2(t)}{dt} - M \frac{di_1(t)}{dt} + R_2 i_2(t) = 0$$

  1. You are neglecting the time dependence of the first current, but there has to be a moment where you "switch it on". So the correct expression would be (if it is instantaneous, you could more realistically switch it on gradually):


$\qquad$ where $\theta(t)$ is Heaviside's step function (the derivative is a Dirac delta "function").

So the reason for the contradictions is twofold. Firstly, the equations have mistakes. Secondly, (and I think this is the most important reason) you are assuming that the primary circuit is in a steady state but you want to calculate the transient solution for the secondary circuit... this is wrong because the systems are linked together.

  • 1
    $\begingroup$ I see. Thank you for your comprehensive answer! $\endgroup$ – Sudera Oct 24 '19 at 17:56
  • $\begingroup$ You're welcome. In fact, the result of $i_2=0$ is what you get for the secondary circuit in steady state, that is why you got that result (although the complete solution gives you a decaying exponential). You can also see that the solution you got that varies linearly in time cannot be right, this is because you would get an arbitrarily large current (i.e., infinite energy). $\endgroup$ – user137661 Oct 24 '19 at 18:08
  • $\begingroup$ Also, note that the source voltage cannot ever be zero in your configuration. You have a disipative element ($R_1$) in that side that needs to be powered, this power has to come from the voltage source (the power of the voltage source is given by $P=V_S I_1$). $\endgroup$ – user137661 Oct 24 '19 at 18:09

Hint: If the current is constant in the primary what does that say about the voltage across the primary winding and therefore the voltage induced in the secondary winding?

Hope this helps.

  • $\begingroup$ A constant current means no change in magnetic flux, so no voltage across the primary winding and no voltage induced in the secondary winding. However, I'm wondering why the general method for solving circuits (simply solving the equations) doesn't work here and how you can know you can't use it. Thank you for your answer. $\endgroup$ – Sudera Oct 24 '19 at 16:56
  • $\begingroup$ @Sudera What you said in the first sentence is correct. But assuming you are not considering transients immediately following the closing of a switch (which you did not show or explain was involved), under steady state conditions there is no voltage across $L_1$. Since the voltage across $L_2$ is the ratio of the secondary turns to the primary turns times the voltage across $L_1$, the voltage and therefore current in the secondary will be zero. $\endgroup$ – Bob D Oct 24 '19 at 17:25

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