In a circuit that consumes some reactive power, the current waveform indicates current flow even when the supply voltage is at the zero crossing. This seems paradoxical since there cannot be current flow at zero volts, barring some kind of superconductivity effect.

What is the best way to understand this aspect of reactive power?

  • \$\begingroup\$ Faraday's Law... \$\endgroup\$ Commented Aug 15, 2018 at 18:12
  • 1
    \$\begingroup\$ Stored energy in inductive current lags phase of voltage is normal \$\endgroup\$ Commented Aug 15, 2018 at 18:19

2 Answers 2


The voltage across an inductor, V equals the indutance x the rate of change of current passing through the inductor. In short: -

$$V = L\dfrac{di}{dt}$$

So, if the rate of change of current through the inductor is zero then the voltage across its terminals is also zero BUT this doesn't mean that there isn't a value of current present and that it has reached a peak like this: -

enter image description here

Picture source.


Use a mechanical analogy with inertia and see what you come up with. Reactive loads have a lot of electrical inertia. And if you try to stop it, all of a sudden you have voltage (electrical force)!

Imagine a capacitor, for example, as a holding tank. It starts empty (no charge, zero volts), and you start putting charge into it. At the first instant, you have zero volts and non-zero current at the cap itself. As charge "piles up", you get a non-zero voltage at the cap. Likewise, you can discharge it through zero and continue going negative, and at some instant you again have zero voltage and non-zero current at the cap.

Same idea for an inductor, except that the energy is stored in a magnetic field, which you might think of as "stored current", or to continue the mechanical analogy, "inertia". If you really had a superconducting inductor*, then the current could continue forever just like charge/voltage would stay on a perfect capacitor forever.

*All real-world superconducting wires are like this because they only eliminate the resistance. The real-world parasititc inductance is still what it always was.

But in the real world, capacitors leak (imagine a resistor across it), and inductors are made from resistive wire. So some energy is lost through that resistance.

If you model a real-world part with a network of ideal parts, and then only look at the resistances, then by definition, they follow Ohm's law exactly: V=IR

And if you understand complex numbers, you'll find that the reactive ideal components also follow Ohm's law exactly...in the complex domain.

Instead of having a single number "x" to represent some quantity, a complex number is a 2-dimensional vector "a + bi" where i=sqrt(-1). It's a bit odd to think of math working like that, but it does. A bit of trigonometry from there gives you magnitude and phase angle. (a and b are the sides of a right triangle)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.