Where does energy of the inductor go?

In summary: 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. 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
  • #1
goodphy
216
8
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?
 
Physics news on Phys.org
  • #2
goodphy said:
When current decreases, stored energy is released. Question is where this energy goes?
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
Khashishi said:
Calculate the power from/to the source using P = IV.
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
goodphy said:
I'm right now actually very confusing about power calculation for the source. In this case, voltage source is directly connected to the inductor. EMF of inductor is the same to voltage source and only difference is current flow direction. (In source, current flows out from + terminal while at the same time, current flows into + terminal of inductor.) Power calculation is P(t) = I(t)×V(t) and I and V are scalar value. Hmm...how can I put polarity of current into power calculation? I mean calculation should give me the picture that inductor gains energy while source loses and via versa.
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.
 
  • Like
Likes goodphy and Hesch
  • #5
Pity the OP. This thread is way off topic.

goodphy said:
I'm right now actually very confusing about power calculation for the source. In this case, voltage source is directly connected to the inductor. EMF of inductor is the same to voltage source and only difference is current flow direction. (In source, current flows out from + terminal while at the same time, current flows into + terminal of inductor.) Power calculation is P(t) = I(t)×V(t) and I and V are scalar value. Hmm...how can I put polarity of current into power calculation? I mean calculation should give me the picture that inductor gains energy while source loses and via versa.
It sounds like you did not understand the following, which is the actual answer to your original question.

Hesch said:
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.
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.
 
  • Like
Likes Hesch
  • #6
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.
 

1. Where does the energy of the inductor go when the current is turned off?

When the current is turned off, the energy stored in the inductor is released in the form of an electromagnetic field. This field can either dissipate as heat or be transferred to another component in the circuit.

2. Does the energy of the inductor ever fully disappear?

The energy stored in an inductor can never fully disappear. It can only be transferred to other components in the circuit or dissipated as heat. This is due to the law of conservation of energy, which states that energy cannot be created or destroyed, only transformed.

3. How does the energy of the inductor affect the circuit?

The energy stored in an inductor affects the circuit by creating a magnetic field. This field can cause current to continue flowing even after the power source is removed, as the inductor tries to maintain its current flow. This phenomenon is known as inductance and can have both positive and negative effects on circuit performance.

4. Can the energy of an inductor be harnessed and used for other purposes?

Yes, the energy stored in an inductor can be harnessed and used for other purposes. Inductors are commonly used in electronic devices to store energy and create oscillating currents, which can then be used to power other components in the circuit.

5. How does the resistance of an inductor affect its energy storage capabilities?

The resistance of an inductor can affect its energy storage capabilities by causing energy to dissipate as heat. This is known as the inductor's internal resistance and can decrease the amount of energy that can be stored in the inductor. In order to increase energy storage, inductors with lower internal resistance are typically used in circuits.

Similar threads

  • Electromagnetism
Replies
5
Views
899
Replies
9
Views
379
Replies
4
Views
1K
  • Electromagnetism
Replies
6
Views
991
Replies
50
Views
11K
  • Electromagnetism
Replies
9
Views
2K
Replies
24
Views
6K
Replies
13
Views
3K
  • Electromagnetism
Replies
6
Views
2K
Back
Top