For the circuit illustrated below, I am trying to compute the sinusoidal voltage across resistor \$R_{1}\$ when the parallel LC component is in resonance:

I setup my problem with the following information:

  1. My voltage source produces signal: \$V = 8\times sin(\omega t)\$

  2. My voltage across \$R_{1}\$ is going to be a voltage divider: \$V_{R} = V \times \frac{Z_{R}}{Z_{R} + (Z_{C} + Z_{L})}\$

The Circuit

enter image description here


At resonance, I know that my angular frequency will be: \$\omega_{0} = \frac{1}{\sqrt{LC}}\$.

This should remove the impedance component for the parallel LC part of my circuit. I show that now:

$$V_{R} = V \times \frac{R}{R + \biggl(\frac{1}{j \bigl (\omega C - \frac{1}{\omega L} \bigr )}\biggr)} = V \times \frac{R}{R - j\biggl(\frac{1}{\bigl(\omega C - \frac{1}{\omega L}\bigr)}\biggr)} = V \times \frac{R}{R - j(\frac{1}{0})}$$

This results in an undefined solution though. However, I also know:

  • If there is no impedance, then the resistor is the only component of the circuit providing resistance

  • Thus the voltage is simply determined by the resistor.

  • In which case won't the voltage across the resistor simply be the same as the voltage produced by the source?

What confuses me is trying to reconcile these two explanations I have been given. I am confused as to why I cannot obtain something like \$V_{R} = V\$, assuming that is indeed the correct measurement for the voltage across \$R_{1}\$.

  • \$\begingroup\$ @G36 Oh my mistake, yes the resistor is not in parallel. The schematic is correct my title is wrong. I will try to amend it now \$\endgroup\$
    – Micrified
    Commented Oct 9, 2019 at 17:30
  • \$\begingroup\$ At resonance, you have an open circuit, no current will flow through 50 ohms resistor. Only the current will circulated inside the LC tank circuit. \$\endgroup\$
    – G36
    Commented Oct 9, 2019 at 17:40
  • \$\begingroup\$ Look here wolframalpha.com/input/… \$\endgroup\$
    – G36
    Commented Oct 9, 2019 at 17:47
  • \$\begingroup\$ you can use an RLC IMpedance Nomograph for 10 % accuracy and instant intersection of lines from any knowns electronics.stackexchange.com/questions/369378/… \$\endgroup\$ Commented Oct 9, 2019 at 18:44

3 Answers 3


At resonance, the LC tank circuit behaves like an open circuit.

Therefore, there is no current flowing through it. And therefore there is no current through the resistor.

Since there is no current through the resistor, there is no voltage across it.

Therefore the full voltage of the voltage source appears across the LC tank.

To show specifically where your analysis went wrong, you said,

$$V_{R} = ... = V \times \frac{R}{R - j(\frac{1}{0})}$$

This results in an undefined solution though.

While this is mathematically undefined, you have a value going to infinity in the denominator of your right hand side expression. This means the total quantity goes to zero, so you have

$$V_R = 0.$$

From there you can reach the conclusion I presented above.

  • \$\begingroup\$ And the impedance goes to... \$\endgroup\$
    – Voltage Spike
    Commented Oct 9, 2019 at 22:04

I think a better way to solve this is to find what the actual load is:

\$Z_L = j\omega L\$

\$Z_C = \frac{1}{j\omega C}\$

\$Z_{load} = Z_R+\frac{Z_L Z_C}{Z_L+Z_C}\$

then look at only the parallel impedance of L and C

\$\frac{Z_L Z_C}{Z_L+Z_C}=\frac{j\omega L}{1-\omega^2 L C } \$


\$ \frac{j\omega C^{-1}}{(L C)^{-1} -\omega^2 } \$

When you plug in the numbers, the impedance of the LC portion get's very large around the resonant point

  • 1
    \$\begingroup\$ Maybe I missed a step...should the denominator of the final value be \$1 - \omega^2LC\$? \$\endgroup\$ Commented Oct 9, 2019 at 19:08
  • \$\begingroup\$ But now the denominator is zero at resonance, so we are right back where we started. \$\endgroup\$ Commented Oct 9, 2019 at 20:25
  • \$\begingroup\$ Yeah, I was translating it after I worked it out, but I had \$ \frac{-j\omega l}{\omega^2 L C -1}\$ and when translating forgot both. When I plugged in numbers I got a very large number on the bottom, but it is dependent on the frequency. \$\endgroup\$
    – Voltage Spike
    Commented Oct 9, 2019 at 20:26
  • \$\begingroup\$ If you think about it, the resonance point has to be frequency dependent, I've never seen an RLC filter that isn't. \$\endgroup\$
    – Voltage Spike
    Commented Oct 9, 2019 at 20:32

Well, we can use Mathematics to compute what happens.

We write, for the input voltage:

$$\text{V}_\text{in}\left(t\right)=\hat{\text{v}}_\text{in}\cdot\sin\left(\omega t+\varphi\right)=\hat{\text{v}}_\text{in}\cdot\cos\left(\omega t+\varphi+\frac{\pi}{2}\right)\tag1$$

So, the complex input voltage is given by:


Now, the complex input impedance is given by:


The complex input current is given by:


The time function for the input current is given by:

$$\space\text{I}_\text{in}\left(t\right)=\left|\frac{\underline{\text{V}}_{\space\text{in}}}{\underline{\text{Z}}_{\space\text{in}}}\right|\cdot\cos\left(\omega t+\arg\left(\frac{\underline{\text{V}}_{\space\text{in}}}{\underline{\text{Z}}_{\space\text{in}}}\right)\right)=$$ $$\frac{\left|\underline{\text{V}}_{\space\text{in}}\right|}{\left|\underline{\text{Z}}_{\space\text{in}}\right|}\cdot\cos\left(\omega t+\arg\left(\underline{\text{V}}_{\space\text{in}}\right)-\arg\left(\underline{\text{Z}}_{\space\text{in}}\right)\right)\tag5$$

The complex voltage across the parallel part is given by:

$$\underline{\text{V}}_{\space\text{p}}=\underline{\text{Z}}_{\space\text{p}}\cdot\underline{\text{I}}_{\space\text{p}}=\underline{\text{Z}}_{\space\text{p}}\cdot\underline{\text{I}}_{\space\text{in}}=\frac{\text{L}}{\text{C}}\cdot\frac{1}{\frac{1}{\omega\text{C}}-\omega\text{L}}\cdot\text{j}\cdot\frac{\hat{\text{v}}_\text{in}\cdot\exp\left(\left(\varphi+\frac{\pi}{2}\right)\text{j}\right)}{\text{R}+\frac{\text{L}}{\text{C}}\cdot\frac{1}{\frac{1}{\omega\text{C}}-\omega\text{L}}\cdot\text{j}}=$$ $$\frac{\text{L}}{\text{C}}\cdot\frac{\hat{\text{v}}_\text{in}\cdot\exp\left(\left(\varphi+\frac{\pi}{2}\right)\text{j}\right)}{\text{R}\left(\frac{1}{\omega\text{C}}-\omega\text{L}\right)+\frac{\text{L}}{\text{C}}\cdot\text{j}}\cdot\text{j}\tag6$$

At resonance we know that:







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