# I Where does energy of the inductor go?

1. Jul 30, 2015

### goodphy

Hello.

Let's say lossless case where AC voltage source connects to inductor in series. As current increases, energy is stored in form of magnetic field within the inductor. When current decreases, stored energy is released. Question is where this energy goes? Since there is no load and system is lossless, the only place the energy should go is...source?

When the load is connected, the energy is released on both source and the load?

2. Jul 30, 2015

### Hesch

When the current decreases, the flux in the inductor decreases. Thus an emf is induced in the inductor in the same direction as the current:

Emf = dΨn/dt.

So when the current/magnetic energy is decreased, the voltage source ( say a generator ) is "pushed".

3. Jul 30, 2015

### Hesch

To be more exact: P(t) = I(t) * V(t).

Integrated for a period, the result will be zero, but not for a half period.

4. Jul 31, 2015

### Staff: Mentor

The best way to do this consistently is to use the passive sign convention consistently. In the passive sign convention a positive power (voltage * current) represents power going into the component from the rest of the circuit, whereas a negative power represents power going into the rest of the circuit from the component.

So, in the setting that you have described the voltage across the source and the voltage across the inductor would be the same, but the current through the source would be the opposite of the current through the inductor. So you will always have that $V \; I_{inductor}=-V \; I_{source}$. Whenever $V \; I_{inductor}>0$ then the source is supplying power to the inductor and whenever $V \; I_{inductor}<0$ then the inductor is supplying power to the source.

5. Jul 31, 2015

### anorlunda

Pity the OP. This thread is way off topic.

It sounds like you did not understand the following, which is the actual answer to your original question.

P=I*V is valid always. It applies for each instant of time, and it applies for the average over any period of time. Of particular interest is the average over a whole AC cycle, which we call AC analysis.

So, choose several points of time within a single AC cycle, and apply P=I*V to each time point. (I, V, and P are all signed scalar quantities) For an inductive load, you should find that P is positive for half the time and negative for half the time. If you average the P values over and entire AC cycle, the answer should be zero. In other words, the energy goes one direction for half the cycle and the other direction for half. All that back and forth adds up to zero; that is what we call "imaginary power".

If the load is pure resistance, do the same exercise. You should find that when V is +, I is + so that P=V*I is +. When V is - then I is -, so that P=V*I is also +. The direction of P is + for all times in the AC cycle. Obviously, when you add up all those + values, the sum is +. This is what we call "real power".

But if you stick to one time point at once, and forget averages, there is no real and imaginary, no zero sum over a whole cycle. There are just values of P(t)=V(t)*I(t) which are sometimes + and sometimes -. That applies always, DC or AC, or any non-sinusoidal signal, or non-repetitive signals. Much of the confusion comes only when trying to characterize behavior over extended intervals of time.

6. Aug 1, 2015

### Staff: Mentor

Instantaneous power = the product of instantaneous current and instantaneous voltage,
i.e., p(t) = i(t).v(t)

When the load is a pure inductance, for part of the sinusoid's cycle the instantaneous power is positive (i.e., above the axis) and this represents power from the source to the load, then for an equal amount of time the instantaneous power is negative, (i.e., below the axis), and this is represents power delivered from the load back to the source. The integral of p(t) is energy, and over one complete power cycle the average is found to be zero, meaning the pure inductance load is indeed lossless---it returns to the source all the energy delivered.

The energy question seems to have been adequately addressed.

Meandering thread cleaned up, and closed.