I know the formula of resonance frequency for both series and parallel circuits. But when we have a circuit that is shown in the image below where the RESISTOR and INDUCTOR are in parallel and their equivalent impedance is in series with the Capacitor, what formula should I use for resonant frequency?

I am well aware of the fact that resonance frequency is at the point where reactance is zero. But when I try solving for the equivalent impedance, I am not able to separate Resistance R from Z_C and Z_I to make an equation when Z_C = Z_I to form the equation for Resonant Frequency.

enter image description here

  • \$\begingroup\$ So long as \$R\gt \frac12\sqrt{\frac{L}{C}}\$, then there is a dampened frequency, \$\omega_d=\omega_p\sqrt{1-\frac14\cdot\frac{L}{R}\,\cdot\,\frac1{R\,C}}\$, which is less than the pole frequency, \$\omega_p=\frac1{\sqrt{L\,\cdot\, C}}\$. \$\endgroup\$ Jun 3, 2023 at 0:11

1 Answer 1


As you've already said, the resonant frequency is the frequency at which the reactance of the circuit is zero. This means that you have to derive a formula for the impedance of the entire circuit and then find the frequency at which its imaginary part is zero.

The overall impedance of the circuit is Z = Z_C + (Z_L || R). You can simplify that formula and then solve for Im(Z)=0.

Mathematically, this is no different than series / parallel combinations of resistors.

  • \$\begingroup\$ Yes but I was just finding it odd that Resistance was remanining in the equation when I was trying to make Resonant frequency the subject by solving the overall impedance equation algebraically. I was finding it odd because in the general equation for resonance frequency of both series and parallel circuits, there was no involvement of resistance in it. I solved it, I will know in 3-4 days whether my answer is correct. What I got is (in CAS) = sqrt( (R^2) / (4 * (pi^2) * L * C * ( (R^2) - L))). Anyways, thanks for your input. \$\endgroup\$
    – RK Eshat
    Jun 2, 2023 at 18:19
  • \$\begingroup\$ @RKEshat It's expected that R is present in the final solution. See: en.wikipedia.org/wiki/RLC_circuit#Other_configurations Note that your solution can't be correct since the term "R² - L" is subtracting quantities with incompatible units. This is mathematically impossible; you might want to check your derivation again. \$\endgroup\$ Jun 2, 2023 at 18:47
  • \$\begingroup\$ @RKEshat I think the last "L" needs to be L^2, not L. Have you tried out your expression against a SPICE simulator? \$\endgroup\$
    – qrk
    Jun 2, 2023 at 19:52

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