Consider an RLC cirucit in AC as in figure.


simulate this circuit – Schematic created using CircuitLab

Suppose that I can change only the frequency of voltage. What frequency will maximize the dissipated power in the resistor?

I'm trying to use Thevenin's Theorem but I do not think that this is the right situation for it, since I'm not changing the resistor but the frequency. How should I approach this situation?


2 Answers 2


A general way to do this is to solve for the total complex impedance, then find the frequency that minimizes the impedance magnitude.

A more case-specific way is to notice that the resistor is in series with a series L-C. Consider the impedance of a series L-C. It is infinite at 0 frequency due to the capacitor. It is infinite at infinite frequency due to the inductor. It is actually 0 at the resonant frequency, and goes up either side of it.

Therefore use the standard formula for finding the resonant frequency of a L-C system. At that frequency, all the voltage applied by V1 will appear across R1, and R1 will dissipate V12/R1.

Actually cranking thru the numbers is your job.


What frequency will maximize the dissipated power in the resistor?

This is called the resonant frequency and occurs when the reactance of the inductor is numerically equal to the reactance of the capacitor in magnitude.

It's quite easy to show that this is: -

\$\dfrac{1}{2\pi\sqrt{LC}}\$ (and not dependent on the value of the resistor).

  • \$\begingroup\$ @Sørën do you have anything else to ask about either of these answers? Has one or both answers given you the information you need? \$\endgroup\$
    – Andy aka
    Jan 19, 2017 at 12:40
  • \$\begingroup\$ @Sørën Are you aware that the person asking a question is somewhat obliged, when they have received a good answer(s) to formally accept one of those answers. Call it a type of payment for good information. Both answers (mine and Olin's are perfectly good) so you have a decision to make. \$\endgroup\$
    – Andy aka
    Jan 23, 2017 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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