Autoinduzione in regime sinusoidale
Self-inductance in sinusoidal steady state
Summary: We discuss self-inductance by studying how an ideal inductor (a solenoid with no resistance and therefore no energy losses) behaves when it carries a sinusoidal alternating current.
We look at the directions of the self-induced EMFs and at how the reactive power varies.
1 - Flux linkage, or self-flux
Self-inductance occurs when an inductor (a winding, a coil, a solenoid in particular, but to a greater or lesser extent any electric circuit) carries a current that varies in time.
The inductor then produces a magnetic flux \(\small \Phi (t)\) that can be found with Hopkinson’s law, if the reluctance \(\small \mathcal{R} \) of the magnetic circuit is known. Assuming the inductor has N turns carrying the current i(t), the so-called generated flux is:
(1) \(\qquad \qquad \qquad \Phi (t) = \dfrac{N}{\mathcal{R}}\, i(t)\)
This flux links the N turns of the inductor itself, giving the flux linkage \( \Phi_c (t)\), also called self-flux:
(2) \(\qquad \qquad \qquad \Phi_c (t) = N\, \Phi (t) = N\, \dfrac{N}{\mathcal{R}}\, i(t) = \dfrac{N^2}{\mathcal{R}}\, i(t)\)
which we can write as:
(3)\(\qquad \qquad \qquad \Phi_c (t) = L\, i(t)\qquad \qquad L =\dfrac{N^2}{\mathcal{R}} \)
where L is the self-inductance, measured in henries (H).
As for the dimensions of L:
\(\qquad [L] = \dfrac{[flux]}{[current]} = \dfrac{[voltage \times time]}{[current]} = \dfrac{V\,s}{A} = \Omega\,s = H\)
2 - Self-inductance and magnetic field energy
According to the Faraday–Lenz law, any time-varying flux linked with an electric circuit produces an induced electromotive force (EMF):
(4) \(\qquad \qquad \qquad e(t) = -\dfrac{d \Phi(t)}{dt}\)
In our case, however, we must use the self-flux, since the change of flux involves all the turns of the same inductor that produces the flux:
(5) \(\qquad \qquad \qquad e(t) = -\dfrac{d \Phi_c(t)}{dt}\)
and taking L as constant (which is not the case for ferromagnetic circuits), from (3) we get:
(6) \(\qquad \qquad \qquad e(t) = -L\dfrac{d i(t)}{dt}\)
or, using finite differences:
(5') \(\qquad \qquad \qquad e(t) = -\dfrac{\Delta \Phi_c(t)}{\Delta t}\)
(6') \(\qquad \qquad \qquad e(t) = -L\dfrac{\Delta i(t)}{\Delta t}\)
The self-induced EMF (6) or (6') can be seen as a consequence of the storage and release of magnetic energy.
The magnetic field is in fact a form of energy. Building up the field of the inductor requires work. The current \( \small i(t)\) can do work on the inductor only if it is opposed by a potential difference, or counter-electromotive force (counter-EMF)(1).
Since the work \( \small \Delta L\) is done over a time interval \( \small \Delta t\), a power \( \small P = \Delta L /\Delta t\) is involved. In electric systems, as we know, power is the product of the current and the potential difference that makes the current flow:
(7) \(\qquad \qquad \qquad p(t) =v(t)\, i(t)\)
In an ideal inductor the power (7) is used entirely to build up the magnetic field. During this phase, instant by instant, \(\small v(t)\) must oppose the counter-EMF \(\small e(t)\):
(8) \(\qquad \qquad \qquad v(t) = - e(t) = L\dfrac{d i(t)}{dt}\)
(8') \(\qquad \qquad \qquad v(t) = -e(t) = L\dfrac{\Delta i(t)}{\Delta t}\)
so:
(9) \(\qquad \qquad \qquad p(t) = L\,\dfrac{d i(t)}{dt}\,i(t)\)
(9') \(\qquad \qquad \qquad p(t) = L\,\dfrac{\Delta i(t)}{\Delta t}\,i(t)\)
which we can write as the energy differential dW:
(10) \(\qquad \qquad \qquad dW = p(t)\,dt = L\,i(t)\,d i(t)\)
or as a finite difference:
(10') \(\qquad \qquad \qquad \Delta W = p(t)\,\Delta t = L\,i(t)\,\Delta i(t)\)
integrating (10):
(11) \(\qquad \qquad \qquad W(t) = \int p(t)\,dt = L\int i(t)\,d i(t)= \dfrac{1}{2}\,L\,i^2(t)\)
or, working with finite intervals, adding up the terms (10'):
(11') \(\qquad \qquad \qquad W(t) = \sum p(t)\,\Delta t = L \sum i(t)\,\Delta i(t)\)
It can be shown that this again gives:
(11'') \(\qquad \qquad \qquad W(t) = L \sum i(t)\,\Delta i(t) = \dfrac{1}{2}\,L\,i^2(t)\)
3 - Self-inductance in sinusoidal steady state
Now suppose the inductor carries a sinusoidal alternating current:
(12) \(\qquad \qquad \qquad i(t) = I_M\,sin(\omega t)\)
produced by a current source that imposes on the circuit a current of constant amplitude, regardless of the “load”, in this case the value of the inductance \(L\).
Looking at the board on the left, notice that during the 1st quarter period the EMF \(e(t)\) opposes the current. In this phase the current source supplies the energy needed to build up the magnetic field, and so it must overcome the EMF \(e(t)\); otherwise it could not do the necessary work. Notice also that the EMF \(e(t)\) points in the direction that would drive a hypothetical current \(i_{ind}\) (2) producing a magnetic flux that opposes the flux produced by \(i(t)\), in agreement with Lenz’s law:
“electromagnetic induction always acts so as to oppose the cause that produced it”.
Instead of the current source, now suppose the system is supplied by a voltage source v(t), which provides a sinusoidal alternating potential difference:
(12') \(\qquad \qquad \qquad v(t) = V_M\,sin(\omega t)\)
Here the current is driven by the source, which also supplies the power and the energy needed. The inductor behaves as a load, that is, it absorbs power, as shown by the fact that the instantaneous positive pole is at the terminal where the current enters. The induced potential difference \(e(t)\) is, to all intents and purposes, a counter-electromotive force.
Note the important fact that the voltage \(v(t)\) imposed by the source is exactly equal and opposite to the self-induced EMF \(e(t)\).
In the 2nd quarter period, with the current source, \(e(t)\) has the same direction as the current \(i(t)\), which is decreasing. This means that the strength of the magnetic field, and therefore the amount of stored energy, is decreasing too. The released energy cannot simply vanish: it is returned to the source (3). The voltage \(v(t)\) imposed by the voltage source opposes the current, which in energy terms means that the source is absorbing electric power, behaving rather as a load. The inductor takes on the role of generator and \(e(t)\) becomes an electromotive force. As in every generator, the instantaneous positive polarity is at the terminal where the current leaves.
With directions reversed, the 3rd quarter period repeats the situation of the first quarter. The voltage source \(v(t)\) drives the current (it has the same direction) and supplies the power for the magnetic field, which now points the opposite way. The counter-EMF \(e(t)\) opposes it and the inductor acts as a load.
The 4th quarter period again repeats, with directions reversed, the situation of the 2nd quarter. The inductor takes on the role of generator, the voltage source absorbs power because \(v(t)\) opposes the current, and the energy flows from the magnetic field of the inductor back to the source.
In conclusion, by looking at the directions of the EMFs and currents and at how the power varies (the sign of its sinusoid), we can say that the power, which we will call reactive, flows toward the load in the quarter periods where it is positive and returns to the source where it is negative.
Activity: using the simulation
4 - Self-inductance: a little mathematics
If the current \(\small i(t)\) is sinusoidal, \(\small i(t) = I_M\,sin(\omega t)\), then \(\small e(t)\) is found by differentiating with respect to time, as stated by (6):
(13) \(\ \ e(t) = -L\, \dfrac{d\,i(t)}{dt} = -L\,I_M\,\dfrac{d\, sin(\omega t)}{dt} =- \omega\,L\,I_M\,cos(\omega t) = \omega\,L\,I_M\,[-cos(\omega t)] \)
While \(\small i(t)\) varies as \(\small sin(\omega t)\), \(\small e(t)\) varies as:
\(\qquad \qquad \qquad \qquad -cos(\omega t) = sin( \omega t -\dfrac{\pi}{2}) \)
which means that \(\small e(t)\) lags \(\small i(t)\) by a quarter period: as we say, \( \small e(t)\) is in lagging quadrature with \(\small i(t)\).
The voltage imposed by the source is always opposite to the induced EMF:
(14) \(\qquad \qquad \qquad v(t)= -e(t) = \omega\,L\,I_M\,cos(\omega t)\)
that is:
(15) \(\qquad \qquad \qquad v(t)= \omega\,L\,I_M\,sin( \omega t +\dfrac{\pi}{2})\)
the imposed voltage \(\small v(t)\) is in leading quadrature with the current \(\small i(t)\).
5 - Reactive power
Recalling that \(\small p(t) = v(t)\,i(t)\), from formulas (12) and (15):
(16) \(\qquad \quad p(t)= \omega\,L\,I_M\,cos(\omega t)\,I_M\,sin(\omega t) = \omega\,L\,I^2_M\,cos(\omega t)\,sin(\omega t) \)
from which we see that the power depends on \(\small cos(\omega t)\,sin(\omega t) = \dfrac{1}{2} sin(2 \,\omega t)\)
and therefore:
(17) \(\qquad \qquad \qquad \qquad p(t)= \dfrac{1}{2} \omega\,L\,I^2_M\,sin(2\,\omega t) \)
varies sinusoidally, at twice the frequency of voltage and current and with amplitude \( \small \dfrac{1}{2}\,\omega\,L\, I^2_M\). In the 1st quarter period the power is positive, which we read as a flow of energy from the source to the load.
In the 2nd quarter period the power is negative: the energy stored in the inductor returns to the source.
In the 3rd quarter period the power is positive again: the current carries the energy needed to build a magnetic field pointing the opposite way from before. Finally, in the 4th quarter period the field gradually dies away, as in the second quarter. The energy is returned to the source and the power is negative.
This power, exchanged back and forth between source and inductor, is called reactive power. This power, with its energy, is essential for the system to work (it builds the magnetic field), but it stays inside the system. In other words, it cannot be used to make the system do work.
Learn more:
Power in sinusoidal AC circuits (in Italian)
6 - Directions of self-induced EMFs and the dot convention
The question of which direction to give self-induced EMFs often causes confusion and difficulty, because different and not always justified conventions are used.
The self-induced EMF is often given the same direction as the current, which wrongly leads to reading the source EMF and the self-induced EMF as acting in the same sense on the current in the circuit.
From the previous sections it should be clear that the inductor behaves alternately as a generator and as a load, and that the self-induced EMF is always opposite to the EMF imposed by the voltage source.
The most logical choice of directions is therefore the one used on the board. It is the one that respects the sequence of events and the related energy exchanges.
We have also used the dot convention to set a positive direction for the current and for the quantities associated with it (for example the magnetic flux). A current entering the inductor at the terminal marked with the dot is taken as positive, and so is an imposed EMF that drives the current in that direction (note that \(\small e(t)\) is not an imposed EMF).
Rev. 01/06/2021 · English translation 10/2026
Books by the author and further reading
(1) This happens in every physical system that can store energy in potential form. For example, to give potential energy to a mass \(\small m\) by lifting it to a height \(\small h\) above the ground, work \(\small \vec{F} \cdot \vec{h}\) must be done. The force \(\small \vec{F} = m \vec{g}\) must overcome the weight \( \small \vec{P}= - m \vec{g}\), which opposes it.
(2) It is worth repeating that \( \small i_{ind}\) is not actually present in the circuit. Its only purpose is to show the effect of opposing the changes of \(\small i(t)\), and therefore of the flux \( \small \Phi (t) \).
(3) An electric source absorbs power, and therefore energy, when the voltage across its terminals opposes the current it delivers. While this is fairly easy to see for a voltage source, it is less so for a current source.
The potential difference (voltage) across the terminals of a current source depends on the load connected to it and equals the potential difference across the load itself.
In our case the voltage across the current source is exactly the self-induced \(\small e(t)\) on the inductor. To make sense of this, note that the graph shows \(\small e(t)\) “as seen by the inductor”, that is, with the directions (positive or negative) defined for the inductor. So an EMF that is “negative” for the inductor (arrow pointing up) is “positive” from the point of view of the source.
Summary: We discuss self-inductance by studying how an ideal inductor (a solenoid with no resistance and therefore no energy losses) behaves when it carries a sinusoidal alternating current.
We look at the directions of the self-induced EMFs and at how the reactive power varies.
1 - Flux linkage, or self-flux
Self-inductance occurs when an inductor (a winding, a coil, a solenoid in particular, but to a greater or lesser extent any electric circuit) carries a current that varies in time.The inductor then produces a magnetic flux \(\small \Phi (t)\) that can be found with Hopkinson’s law, if the reluctance \(\small \mathcal{R} \) of the magnetic circuit is known. Assuming the inductor has N turns carrying the current i(t), the so-called generated flux is:
(1) \(\qquad \qquad \qquad \Phi (t) = \dfrac{N}{\mathcal{R}}\, i(t)\)
This flux links the N turns of the inductor itself, giving the flux linkage \( \Phi_c (t)\), also called self-flux:
(2) \(\qquad \qquad \qquad \Phi_c (t) = N\, \Phi (t) = N\, \dfrac{N}{\mathcal{R}}\, i(t) = \dfrac{N^2}{\mathcal{R}}\, i(t)\)
which we can write as:
(3)\(\qquad \qquad \qquad \Phi_c (t) = L\, i(t)\qquad \qquad L =\dfrac{N^2}{\mathcal{R}} \)
where L is the self-inductance, measured in henries (H).
As for the dimensions of L:
\(\qquad [L] = \dfrac{[flux]}{[current]} = \dfrac{[voltage \times time]}{[current]} = \dfrac{V\,s}{A} = \Omega\,s = H\)
2 - Self-inductance and magnetic field energy
According to the Faraday–Lenz law, any time-varying flux linked with an electric circuit produces an induced electromotive force (EMF):(4) \(\qquad \qquad \qquad e(t) = -\dfrac{d \Phi(t)}{dt}\)
In our case, however, we must use the self-flux, since the change of flux involves all the turns of the same inductor that produces the flux:
(5) \(\qquad \qquad \qquad e(t) = -\dfrac{d \Phi_c(t)}{dt}\)
and taking L as constant (which is not the case for ferromagnetic circuits), from (3) we get:
(6) \(\qquad \qquad \qquad e(t) = -L\dfrac{d i(t)}{dt}\)
or, using finite differences:
(5') \(\qquad \qquad \qquad e(t) = -\dfrac{\Delta \Phi_c(t)}{\Delta t}\)
(6') \(\qquad \qquad \qquad e(t) = -L\dfrac{\Delta i(t)}{\Delta t}\)
The self-induced EMF (6) or (6') can be seen as a consequence of the storage and release of magnetic energy.
The magnetic field is in fact a form of energy. Building up the field of the inductor requires work. The current \( \small i(t)\) can do work on the inductor only if it is opposed by a potential difference, or counter-electromotive force (counter-EMF)(1).
Since the work \( \small \Delta L\) is done over a time interval \( \small \Delta t\), a power \( \small P = \Delta L /\Delta t\) is involved. In electric systems, as we know, power is the product of the current and the potential difference that makes the current flow:
(7) \(\qquad \qquad \qquad p(t) =v(t)\, i(t)\)
In an ideal inductor the power (7) is used entirely to build up the magnetic field. During this phase, instant by instant, \(\small v(t)\) must oppose the counter-EMF \(\small e(t)\):
(8) \(\qquad \qquad \qquad v(t) = - e(t) = L\dfrac{d i(t)}{dt}\)
(8') \(\qquad \qquad \qquad v(t) = -e(t) = L\dfrac{\Delta i(t)}{\Delta t}\)
so:
(9) \(\qquad \qquad \qquad p(t) = L\,\dfrac{d i(t)}{dt}\,i(t)\)
(9') \(\qquad \qquad \qquad p(t) = L\,\dfrac{\Delta i(t)}{\Delta t}\,i(t)\)
which we can write as the energy differential dW:
(10) \(\qquad \qquad \qquad dW = p(t)\,dt = L\,i(t)\,d i(t)\)
or as a finite difference:
(10') \(\qquad \qquad \qquad \Delta W = p(t)\,\Delta t = L\,i(t)\,\Delta i(t)\)
integrating (10):
(11) \(\qquad \qquad \qquad W(t) = \int p(t)\,dt = L\int i(t)\,d i(t)= \dfrac{1}{2}\,L\,i^2(t)\)
or, working with finite intervals, adding up the terms (10'):
(11') \(\qquad \qquad \qquad W(t) = \sum p(t)\,\Delta t = L \sum i(t)\,\Delta i(t)\)
It can be shown that this again gives:
(11'') \(\qquad \qquad \qquad W(t) = L \sum i(t)\,\Delta i(t) = \dfrac{1}{2}\,L\,i^2(t)\)
close this box3 - Self-inductance in sinusoidal steady state
Now suppose the inductor carries a sinusoidal alternating current:(12) \(\qquad \qquad \qquad i(t) = I_M\,sin(\omega t)\)
produced by a current source that imposes on the circuit a current of constant amplitude, regardless of the “load”, in this case the value of the inductance \(L\).
Looking at the board on the left, notice that during the 1st quarter period the EMF \(e(t)\) opposes the current. In this phase the current source supplies the energy needed to build up the magnetic field, and so it must overcome the EMF \(e(t)\); otherwise it could not do the necessary work. Notice also that the EMF \(e(t)\) points in the direction that would drive a hypothetical current \(i_{ind}\) (2) producing a magnetic flux that opposes the flux produced by \(i(t)\), in agreement with Lenz’s law:
“electromagnetic induction always acts so as to oppose the cause that produced it”.
Instead of the current source, now suppose the system is supplied by a voltage source v(t), which provides a sinusoidal alternating potential difference:
(12') \(\qquad \qquad \qquad v(t) = V_M\,sin(\omega t)\)
Here the current is driven by the source, which also supplies the power and the energy needed. The inductor behaves as a load, that is, it absorbs power, as shown by the fact that the instantaneous positive pole is at the terminal where the current enters. The induced potential difference \(e(t)\) is, to all intents and purposes, a counter-electromotive force.
Note the important fact that the voltage \(v(t)\) imposed by the source is exactly equal and opposite to the self-induced EMF \(e(t)\).
In the 2nd quarter period, with the current source, \(e(t)\) has the same direction as the current \(i(t)\), which is decreasing. This means that the strength of the magnetic field, and therefore the amount of stored energy, is decreasing too. The released energy cannot simply vanish: it is returned to the source (3). The voltage \(v(t)\) imposed by the voltage source opposes the current, which in energy terms means that the source is absorbing electric power, behaving rather as a load. The inductor takes on the role of generator and \(e(t)\) becomes an electromotive force. As in every generator, the instantaneous positive polarity is at the terminal where the current leaves.
With directions reversed, the 3rd quarter period repeats the situation of the first quarter. The voltage source \(v(t)\) drives the current (it has the same direction) and supplies the power for the magnetic field, which now points the opposite way. The counter-EMF \(e(t)\) opposes it and the inductor acts as a load.
The 4th quarter period again repeats, with directions reversed, the situation of the 2nd quarter. The inductor takes on the role of generator, the voltage source absorbs power because \(v(t)\) opposes the current, and the energy flows from the magnetic field of the inductor back to the source.
In conclusion, by looking at the directions of the EMFs and currents and at how the power varies (the sign of its sinusoid), we can say that the power, which we will call reactive, flows toward the load in the quarter periods where it is positive and returns to the source where it is negative.
Activity: using the simulation
4 - Self-inductance: a little mathematics
If the current \(\small i(t)\) is sinusoidal, \(\small i(t) = I_M\,sin(\omega t)\), then \(\small e(t)\) is found by differentiating with respect to time, as stated by (6):
(13) \(\ \ e(t) = -L\, \dfrac{d\,i(t)}{dt} = -L\,I_M\,\dfrac{d\, sin(\omega t)}{dt} =- \omega\,L\,I_M\,cos(\omega t) = \omega\,L\,I_M\,[-cos(\omega t)] \)
While \(\small i(t)\) varies as \(\small sin(\omega t)\), \(\small e(t)\) varies as:
\(\qquad \qquad \qquad \qquad -cos(\omega t) = sin( \omega t -\dfrac{\pi}{2}) \)
which means that \(\small e(t)\) lags \(\small i(t)\) by a quarter period: as we say, \( \small e(t)\) is in lagging quadrature with \(\small i(t)\).
The voltage imposed by the source is always opposite to the induced EMF:
(14) \(\qquad \qquad \qquad v(t)= -e(t) = \omega\,L\,I_M\,cos(\omega t)\)
that is:
(15) \(\qquad \qquad \qquad v(t)= \omega\,L\,I_M\,sin( \omega t +\dfrac{\pi}{2})\)
the imposed voltage \(\small v(t)\) is in leading quadrature with the current \(\small i(t)\).
5 - Reactive power
Recalling that \(\small p(t) = v(t)\,i(t)\), from formulas (12) and (15):
(16) \(\qquad \quad p(t)= \omega\,L\,I_M\,cos(\omega t)\,I_M\,sin(\omega t) = \omega\,L\,I^2_M\,cos(\omega t)\,sin(\omega t) \)
from which we see that the power depends on \(\small cos(\omega t)\,sin(\omega t) = \dfrac{1}{2} sin(2 \,\omega t)\)
and therefore:
(17) \(\qquad \qquad \qquad \qquad p(t)= \dfrac{1}{2} \omega\,L\,I^2_M\,sin(2\,\omega t) \)
varies sinusoidally, at twice the frequency of voltage and current and with amplitude \( \small \dfrac{1}{2}\,\omega\,L\, I^2_M\). In the 1st quarter period the power is positive, which we read as a flow of energy from the source to the load.
In the 2nd quarter period the power is negative: the energy stored in the inductor returns to the source.
In the 3rd quarter period the power is positive again: the current carries the energy needed to build a magnetic field pointing the opposite way from before. Finally, in the 4th quarter period the field gradually dies away, as in the second quarter. The energy is returned to the source and the power is negative.
This power, exchanged back and forth between source and inductor, is called reactive power. This power, with its energy, is essential for the system to work (it builds the magnetic field), but it stays inside the system. In other words, it cannot be used to make the system do work.
Learn more:
Power in sinusoidal AC circuits (in Italian) 6 - Directions of self-induced EMFs and the dot convention
The question of which direction to give self-induced EMFs often causes confusion and difficulty, because different and not always justified conventions are used.
The self-induced EMF is often given the same direction as the current, which wrongly leads to reading the source EMF and the self-induced EMF as acting in the same sense on the current in the circuit.
From the previous sections it should be clear that the inductor behaves alternately as a generator and as a load, and that the self-induced EMF is always opposite to the EMF imposed by the voltage source.
The most logical choice of directions is therefore the one used on the board. It is the one that respects the sequence of events and the related energy exchanges.
We have also used the dot convention to set a positive direction for the current and for the quantities associated with it (for example the magnetic flux). A current entering the inductor at the terminal marked with the dot is taken as positive, and so is an imposed EMF that drives the current in that direction (note that \(\small e(t)\) is not an imposed EMF).
Rev. 01/06/2021 · English translation 10/2026
Books by the author and further reading
(1) This happens in every physical system that can store energy in potential form. For example, to give potential energy to a mass \(\small m\) by lifting it to a height \(\small h\) above the ground, work \(\small \vec{F} \cdot \vec{h}\) must be done. The force \(\small \vec{F} = m \vec{g}\) must overcome the weight \( \small \vec{P}= - m \vec{g}\), which opposes it.
(2) It is worth repeating that \( \small i_{ind}\) is not actually present in the circuit. Its only purpose is to show the effect of opposing the changes of \(\small i(t)\), and therefore of the flux \( \small \Phi (t) \).
(3) An electric source absorbs power, and therefore energy, when the voltage across its terminals opposes the current it delivers. While this is fairly easy to see for a voltage source, it is less so for a current source.
The potential difference (voltage) across the terminals of a current source depends on the load connected to it and equals the potential difference across the load itself.
In our case the voltage across the current source is exactly the self-induced \(\small e(t)\) on the inductor. To make sense of this, note that the graph shows \(\small e(t)\) “as seen by the inductor”, that is, with the directions (positive or negative) defined for the inductor. So an EMF that is “negative” for the inductor (arrow pointing up) is “positive” from the point of view of the source.