If I am writing KVL for one loop of the circuit which contains a coil coupled with another coil from another loop, the sign is chosen considering whether both currents enter or leave the stars AND if the original current has the same sense or not as the one chosen for the loop?

EDIT: enter image description here

So, if I apply KVL to the left loop, in the shown direction, the voltage drop caused by the coupling should have the sign (-)*(-)?


1 Answer 1


By the passive sign convention, the reference direction for the current through a circuit element is into the positive labelled terminal of the circuit element:

enter image description here

Assuming the starred (dotted) terminal denote the positive labelled terminal of each inductor and assuming the passive sign convention

enter image description here

we have

$$v_1 = L_1 \frac{di_1}{dt} + M\frac{di_2}{dt}$$

$$v_2 = M \frac{di_1}{dt} + L_2\frac{di_2}{dt}$$


$$M = k\sqrt{L_1L_2} $$

Clearly, if you change the reference direction for $i_2$

enter image description here

there must be a sign change in the equations

$$v_1 = L_1 \frac{di_1}{dt} - M\frac{di_2}{dt}$$

$$v_2 = M \frac{di_1}{dt} - L_2\frac{di_2}{dt}$$

  • $\begingroup$ I edited my post. Is this the correct way to do it? Thank you. $\endgroup$
    – user42768
    Jun 17, 2014 at 13:08
  • $\begingroup$ I don't understand the second diagram. Energy would be flowing into both the primary and secondary, no? $\endgroup$
    – endolith
    Apr 9, 2015 at 14:08
  • $\begingroup$ @endolith, why do you think that? In the 2nd diagram, if both currents are positive, the flow of energy is into the primary and out of the secondary. $\endgroup$ Apr 9, 2015 at 14:38
  • $\begingroup$ I mean diagram 2 of 3, where the currents are both flowing inward. They aren't meant to indicate the flow of current? I2 has a negative value, so is actually flowing outward? $\endgroup$
    – endolith
    Apr 9, 2015 at 15:24
  • 1
    $\begingroup$ @endolith, the arrows indicate the current variable reference direction, i.e., the polarity of the ammeter, not the actual direction of the secondary current. $\endgroup$ Apr 9, 2015 at 15:30

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