Does Mutual Inductance Affect Voltage in an Open Circuit?

[Master1022]

Homework Statement

You have two coils (eg. in a transformer). The primary coil is connected to a voltage source and the second coil is an open circuit? What is the voltage across the terminals of coil 2?

Relevant Equations

v = L di/dt

Hi, so my question is basically: Does the mutual inductance from coil 1 show up in the voltage of coil 2 even if coil 2 is an open circuit?

I would think that the answer is yes, as the Mutual inductance depends on the current passing through coil 1? However, the answer scheme seems to ignore the $M_{12} \frac{di_{1}}{dt}$ term and just think about the step-up/down aspect of the voltage.

I would appreciate any help in understanding why this is the case.

I apologise for not providing numbers or anything, but I am just interested in the concept here, rather than any numbers.

Thanks in advance.

 

[Charles Link]

I think what you are referring to in the answer provided is that for the power balance equation of the ideal transformer, $N_p I_p=N_s I_s$. If $I_s=0$, what that means is in the ideal case, $I_p=0$, and so you might incorrectly conclude that $\frac{dI_p}{dt}=0$. In the open circuit case for circuit 2, there is still a small (sinusoidal) current $I_{po}$ that makes the voltage via the mutual inductance for circuit 2. This voltage across the coils in circuit 2 stays essentially the same, and is unaffected even if large (sinusoidal) currents run through circuit 2. This was discussed in a previous thread. See posts 25-29 of https://www.physicsforums.com/threa...ce-in-transformers.941936/page-2#post-5960705 And I think I have the equation correct in this thread that $N_p I_p-N_s I_s=N_p I_{po}$ for a voltage driven transformer of the primary circuit. In looking at the level of (sinusoidal with time) magnetization in the iron core, which is responsible for the voltage, the mmf equations tell us that the important number there is $N_p I_p-N_s I_s$. For a voltge driven transformer, that must remain constant when the loading of the secondary changes, and it must equal $N_p I_{po}$, which is the value it takes on when $I_s=0$. $\\$ @jim hardy Perhaps you can also add something to this. And please correct me if I'm wrong, but I think I got it right.