If a capacitor is connected with an inductor, then because $$Z=\frac{1}{j\omega C}+j\omega L,$$ the Z may be zero. Does that mean when I apply a voltage, the current will be infinite large?

What's more, in transmission line theory, the characteristic impedance could be $\sqrt{L/C}$ when $R=0 \ and~ G = 0$. Why capacitors and inductors could generate real impedance?


2 Answers 2


Does that mean when I apply a voltage, the current will be infinite large?

No, not even in the context of ideal circuit theory. It's a bit subtle since we're using phasor voltages and currents and that requires a couple of assumptions to hold in order to be valid.

When those assumptions don't hold, we have to see what the 'infinity' (division by zero) is trying to tell us.

It means that if a sinusoidal voltage of frequency $\omega = \frac{1}{\sqrt{LC}}$ is across the series LC combination, the amplitude of the (sinusoidal) current increases with time at a constant rate.

Which is to say that there is no AC steady state solution for the current. This is the correct interpretation of the 'infinity' when solving for the phasor current.

Recall that, in order to use the notion of impedance in this context, we assume the circuit

  • is driven by sinusoidal sources with the same frequency $\omega$
  • is in AC steady state which means the amplitudes of the voltage and current sinusoids are constant with time

But, in this case, the 2nd assumption doesn't hold. If one solves the differential equation for the series LC circuit driven by a sinusoidal voltage source, one finds that when the source frequency is $\omega = \frac{1}{\sqrt{LC}}$, the current is of the form

$$i(t) \propto t\cos(\frac{t}{\sqrt{LC}} + \phi)$$

so the amplitude of the current goes to infinity as $t \rightarrow \infty$. But note that at no time is the current actually infinite even in this ideal circuit theory context.

Also note that there is no phasor representation for a current of this form.

Now, for a physical LC circuit, the amplitude will increase until some physical limit is reached, e.g., the capacitor breaks down, or the source internal impedance limits the current.

Why capacitors and inductors could generate real impedance?

For a transmission line, the characteristic impedance is

$$Z_O = \sqrt{\frac{R + j\omega L}{G + j\omega C}} $$

In the case that $R = G = 0$, we have the (square root of the) ratio of two imaginary numbers which produces a real number.

  • $\begingroup$ Thanks! I think I realized something! Could you please have a look at the new edited question? Thanks! $\endgroup$ Commented Sep 21, 2014 at 14:53
  • $\begingroup$ @XiangruLian, I updated my answer though I'm not sure if this completely addresses your concern. That might be the topic of another question. $\endgroup$ Commented Sep 22, 2014 at 2:28

Essentially, the answer to your question is yes but your equation is not quite in the general form. Typically, impedance is $$Z=R + jX$$ with $R$ being the resistance, and $X$ being the reactance which is almost the equation you show, but without the imaginary component. Specifically, $$X = \omega L - \frac{1}{\omega C}$$. What this means is that a component with $Z=0$ would have zero values for both real and imaginary portions; $Z=0 \implies R=X=0$. In such a case, a voltage applied across such a component would lead to infinite current through it, due to Ohm's law.

Such devices don't actually exist, but one could approximate one with a large wire; in other words, a short-circuit.

  • $\begingroup$ The OPs equation is correct for a series connected L and C. $\endgroup$ Commented Sep 21, 2014 at 12:49
  • $\begingroup$ @AlfredCentauri: yes, it's correct for the case that $R=0$ (only). The equation I cite is the general form. $\endgroup$
    – Edward
    Commented Sep 21, 2014 at 12:52
  • $\begingroup$ the add symbol in your X expression should be minus symbol I think.. $\endgroup$ Commented Sep 21, 2014 at 12:58
  • $\begingroup$ @XiangruLian: Thanks, you're right. I've edited my answer. $\endgroup$
    – Edward
    Commented Sep 21, 2014 at 13:00

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.