I seek an intuitive understanding of a coupled inductor in a system. My professors, a generation ago, bemoaned that they'd had four semesters of magnetic circuits but had time to give us just one course including motors. This, I suspect, is what I missed. Please start with a general description before diving into the differential equations!

Take a core (such as a ferrite toroid) with two windings. Assume we will not saturate this core. Designate one winding primary and one secondary. The secondary has a non-reactive load. It may be linear (resistive) or non-linear (eg. resistor plus PN junctions).

Apply a voltage to the primary from a current-limited DC source. A voltage will appear across the secondary, and current will flow as the primary current increases.

What is happening with the fields from each winding? What energy does each hold?

Now the power source reaches its current limit. With no increase, how do the voltage and current in the secondary act? How does this affect conditions in the primary? As time goes to infinity, I expect the secondary current to go to zero and the primary's field to store energy equivalent to that stored in a simple inductor. Is this correct? How do the fields, currents, and voltages progress through time to reach this state?

If it will simplify the description, feel free to replace the voltage source with a variable current source which is ramped up at a slow, constant rate from zero to the final value.

Sorry it took so many words to describe. I hope the problem is clearly described. Thank you for aiding my education.

  • 1
    \$\begingroup\$ Welcome to EE.SE. Some words on this Q&A-site. You ask to get an encompassing explanation on a bunch of problems related to electromagnetism. It will be difficult to answer that with less than 10 pages of a textbook. You may somehow get useful answers, however this somehow breaks the concept of this site which is: A very detailed but concise problem which gets the best answers possible. Next time I recommend to break down such a problem in even smaller portions to be asked separately. You can then link the individual questions. This might even help you in understanding the conundrum better. \$\endgroup\$
    – Ariser
    Feb 26, 2021 at 8:54

1 Answer 1


Is this correct?

Yes, theoretically that is correct. As soon as the power supply starts to limit current, it's the end of the output voltage being constant. This is because induction stops i.e. the flux is no longer changing linearly with time because the current has abruptly halted at some value.

Given that voltage induction into the secondary is proportional to rate of change of primary magnetization current, the moment the primary current stops changing, the secondary voltage exponentially decays to zero based on the secondary inductance and the value of the resistive load. This phase is shedding the energy from the magnetic field. Here's a simulation: -

enter image description here

1 volt DC is applied at t = 0 and the following plot shows the secondary voltage (in red) and the primary current (in green): -

enter image description here

How does this affect conditions in the primary?

At the point when the current limit occurs, the primary current remains fixed and the primary voltage follows the secondary voltage (1:1 turns ratio) and decays to zero volts.

What is happening with the fields from each winding?

There are three fields: -

  1. magnetization field
  2. field caused by secondary current flow into the load
  3. field caused by secondary current flow into the load but from the primary

Fields 2 and 3 always cancel out and field 1 remains: -

enter image description here

Picture from here.

What energy does each hold?

The only energy is held in the magnetization field, \$½\cdot L\cdot i^2\$.

  • \$\begingroup\$ Exponential decay assumes a resistive load, right? If there were no resistance, the current would flow forever? What does this 'drawdown' do to the energy stored in the inductor, and does this process result in voltage changes on the primary? I'd assume 'yes', but my mental picture is simply not clear enough yet. And thanks for following this through. \$\endgroup\$ Feb 26, 2021 at 10:57
  • \$\begingroup\$ If the load is a short circuit then current flows forever and the core remains magnetized. With a secondary short, the secondary voltage is always zero and ditto the primary voltage (assuming 100% flux coupling) @user1468178 \$\endgroup\$
    – Andy aka
    Feb 26, 2021 at 11:00

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.