In some texts it mentions that the average voltage across an ideal inductor is always zero.

How can we derive this conclusion by using: V = L (dI/dt)

Why is average voltage across an ideal inductor is always zero in steady state? Can you give an example when the source voltage is a PWM signal and a sinusoidal signal?

edit: They sometimes give buck converters as example. But how can they assume the current will be constant before making this assumption?

  • \$\begingroup\$ It is not zero, as your formula shows, unless the current is constant over time. \$\endgroup\$ – sweber Jun 6 '16 at 11:27
  • \$\begingroup\$ This is by definition. If someone sees a voltage drop on an inductor (provided the current is constant) and still calls it ideal, they'd have to write their own electronics book. \$\endgroup\$ – Dmitry Grigoryev Jun 6 '16 at 11:28
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    \$\begingroup\$ Integrate dI/dT over infinite time. There are two possible answers : (a) the average of dI/dT is zero, or (b) the average of dI/dT is not zero. In case (b) the integral of dI/dT over infinite time (i.e. the final current) is infinity, which is absurd. Hence only (a) is possible, hence the conclusion. \$\endgroup\$ – user_1818839 Jun 6 '16 at 11:44
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    \$\begingroup\$ Steady state means \$\frac{dI}{dt}=0\$ \$\endgroup\$ – Chu Jun 6 '16 at 12:26
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    \$\begingroup\$ @chu, yes but \$<\frac{dI}{dt}>=0\$ is different from \$\frac{dI}{dt}=0\$. \$\endgroup\$ – The Photon Jun 6 '16 at 16:34

The correct answers were already given. As a supplement, I will try to give a formal mathematical derivation.

To begin with, you can integrate both sides of the equation to get a formula for the current through an inductor at time \$t\$:

$$V(t) = L\frac{\mathrm{d}I(t)}{\mathrm{d}t} \implies \int_0^t V(\tau)\,\mathrm{d}\tau = LI(t) \implies I(t) = \frac{1}{L}\int_0^t V(\tau)\,\mathrm{d}\tau$$ here we assume that \$I(0) = 0\$, otherwise add \$I(0)\$ to the expression for \$I(t)\$.

"The average voltage across an ideal inductor is always zero" actually means the average voltage over a period is zero (otherwise it's meaningless to impose such condition). That is, here we assume that the voltage across an inductor is periodic.

Assume that the voltage across an inductor is a periodic function with period \$T\$. "Is periodic with period \$T\$" is just another way to say that \$V(t) = V(t + T)\$ for any \$t\$.

Let's calculate the current after \$n\$ periods, i.e. when \$t = nT\$ (\$n\$ is an integer):

$$ I(nT) = I(\underbrace{T + T + \dots + T}_{\text{n times}}) = \frac{1}{L}\int_0^{\underbrace{T + T + \dots + T}_{\text{n times}}} V(\tau)\,\mathrm{d}\tau$$

Here we can use the property of an integral \$\int_0^{x+y} f(t)\,\mathrm{d}t = \int_0^x f(t)\,\mathrm{d}t + \int_x^y f(t)\,\mathrm{d}t\$ to break the integral into a sum:

$$ \begin{split} I(nT) &= \frac{1}{L}\left( \underbrace{ \int_0^T V(\tau)\,\mathrm{d}\tau + \int_T^{2T} V(\tau)\,\mathrm{d}\tau + \dots + \int_{(n-1)T}^{nT} V(\tau)\,\mathrm{d}\tau }_{\text{n times}} \right)\\ &= n\cdot\frac{1}{L}\int_0^T V(\tau)\,\mathrm{d}\tau \quad \text{because }V(\tau)\text{ is periodic with period }T\\ &= n\cdot I(T) \end{split} $$

From this expression you can see that if the integral over a whole period $$I(T) = \frac{1}{L}\int_0^T V(\tau)\,\mathrm{d}\tau$$ is not zero, then after \$n\$ periods the current through an inductor will be n times larger: $$\boxed{I(nT) = n\cdot I(T)}$$ As \$n\$ goes to infinity, so does the the current.

Thus, the only way to keep current from going to infinity is the condition \$I(T) = 0\$, which is equivalent to $$\int_0^T V(\tau)\,\mathrm{d}\tau = 0$$ because \$\frac{1}{L}\$ is just a constant factor. Just to remind you, \$T\$ is the period of the voltage. The lower limit of integration is \$t = 0\$. "Zero time" can be an arbitrary chosen instant of time, because the process is periodic.


The average voltage across an ideal inductor is zero just as the average current into an ideal capacitor is always zero. If it were not zero average current for a capacitor it would charge to infinite volts. If it were not zero average volts for an inductor it would take infinite amps.


Think about what would happen when the average (DC component) of the voltage across a inductor was not zero. The current would build up linearly. For a ideal inductor, this would continue as long as the voltage was applied, and the current could become arbitrarily large.

Real inductors have some resistance, which can be thought of as being in series with the pure inductance. Eventually the current will reach a steady state where all the voltage is across this resistance. Even in real inductors, that resistance can be small, so currents resulting from steady applied voltage would be large.

  • \$\begingroup\$ Is this another way of saying: The inductor keeps the average current passing through itself constant whatever the voltage across it ? \$\endgroup\$ – user16307 Jun 6 '16 at 12:18
  • \$\begingroup\$ @user16307, no, it's a way of saying that circuits don't store infinite energy. \$\endgroup\$ – The Photon Jun 6 '16 at 16:06

beacuse the current through an ideal inductor is proportional to the time integral of the voltage it sees

and if the average voltage is not zero the time integral is infinite.

so when using ideal inductors to model real-world problems the current is finite and the average voltage (over all time) is zero.

in thought experiments this rule can be ignored, but you won't get results that are applicable to the real world.


The fact that the average pure inductor is zero can be proven from the equation.

$$ V = L \cdot \dfrac{d}{dt} I $$

We can do this by considering what current would be flowing otherwise.

$$ I = \int^{+\infty}_{-\infty} V \text{ d}t$$

from this no matter how small \$ V\$ is, other than 0V, then the current \$ I\$ would eventually become infinity large which clearly is impossible.

This does not mean you can't ever measure a non zero average voltage across an inductor but that's due to all real inductors having series resistance.

A similar argument explains why the average current in a capacitor must be zero.

This answer is essentially that given by Andy and others except for my added mathematical notation.


The DC resistance of an ideal inductor is zero. If a DC current is flowing through it, the resulting voltage across the inductor is zero. The average voltage is the DC component of the voltage.

  • \$\begingroup\$ Yes but Im asking why? \$\endgroup\$ – user16307 Jun 6 '16 at 15:52

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