Energy conservation in induction

Question

Let's say we have 2 inductively coupled circuits with mutual inductance M, and Circuit 1 ist connected to a power source, changing its current by $\frac{dI_1}{dt}$, then the potential induced in Cuircit 2 would be $-M\frac{dI_1}{dt}$, resulting in a current $I_2=-\frac{M}{R_2}\frac{dI_1}{dt}$, thus consuming a power of $-M\frac{dI}{dt}I_2$. My question is: where does this power come from?

I would assume that there would be some resistance/counter-potential in Circuit 1 created by Circuit 2 that would then result in energy conservation. However, I have just learned about energy conservation in a single circuit with self-inductance, and from my understanding we did not take any induced resistance/counter-potential into consideration. Instead, when the current was increasing the induced potential simply reduced the energy output by opposing the direction of the current and thus did negative work. This reduced energy was then "stored" in the magnetic field, which would then be released when the current decreased again, because this time the induced potential is in the direction of the current, hence resulting in no net change of energy.

However, this idea of storing the energy in the magnetic field couldn't work with the 2 coupled circuits, since the induced potential is always in the direction of the current (assuming Circuit 2 has no other energy sources), so regardless of whether current is increasing or decreasing in Circuit 1, positive work is done in Circuit 2.

So how does this actually work? If there is such a thing as induced resistance/counter potential, why do we ingore it in the self inducting circuit, and if there is not such a thing, how is energy conserved for the 2 coupled circuits?

Answer 1

Let's say we have 2 inductively coupled circuits with mutual inductance M, and Circuit 1 ist connected to a power source, changing its current by $\frac{dI_1}{dt}$, then the potential induced in Cuircit 2 would be $-M\frac{dI_1}{dt}$,

"induced potential" is grossly incorrect term that may be a cause of your confusion. This formula gives induced electromotive force. Induced EMF has the same units as voltage, but it is not a difference of potentials. It is a net effect of induced, rotational electric field due to winding 1, in the winding 2. As a result of EMF, often charges redistribute so as to create a voltage, or difference of potentials, that partially counteracts effects of the EMF on the charge carriers; but except in ideal inductance, this is not complete, so in general, voltage and EMF have different magnitude and often different sign.

resulting in a current $I_2=-\frac{M}{R_2}\frac{dI_1}{dt}$

This would be true if we could ignore self-induced electromotive force. Let us assume that to be the case (this would happen with oscillating currents if the current $I_2$ had much lower amplitude than the current $I_1$).

thus consuming a power of $-M\frac{dI}{dt}I_2$

The energy "consumed" in circuit 2 per unit time, on time average, is actually given by

$$ R_2I_2^2 $$ or $$ R_2\left(M\frac{dI_1}{dt}\right)^2 $$ Some other energy is used to increase magnetic energy stored in the vicinity of the winding (assuming current $I_2$ increases).

where does this power come from?

Energy flows from the power source feeding the circuit 1, is funneled along the wires, through the common magnetic core, and part of it ends up in the wires of the circuit 2 (joule heating), part of it oscillates between being magnetic energy near the winding 2, winding 1, and the power source.