# Find Voltage of inductor in LC series circuit with resistor in parallel

I have a LC resonance circuit with a resistor between the capacitor and inductor:

simulate this circuit – Schematic created using CircuitLab

I want to find the Voltage at L1, more specifically how it depends on R2. I cannot figure out how I to set up an equation or such to find the voltage at L1. If anyone knows, can you please help? :)

NOTE: This is a hobby project, not related to education.

• To understand this circuit, you need to learn Ohm's Law, Watt's Law, Kirchoff's laws, resistance, capacitance, inductance, impedance, vector math/phasors, series AC RLC circuits, parallel AC RLC circuits. All in all not a tremendous volume. This isn't math I'm concentrating on right now, so I'll leave an answer for someone else to provide, but in the meantime, write down that list of subjects, in order, and start browsing through. – K H Jan 7 at 4:38
• @KH Thank you, I will look further into those. Do you know if this system will lead to a set of differential equations or if everything can be done using "purely algebra"? Asking so I don't spend lots of time trying to solve differential equations when there's a simpler way of doing it :P – Beacon of Wierd Jan 7 at 19:19

Well, we have:

$$\underline{\text{Z}}_{\space\text{in}}=\text{R}_1+\frac{1}{\text{j}\omega\text{C}}+\frac{1}{\frac{1}{\text{R}_2}+\frac{1}{\text{j}\omega\text{L}}}\space\Longrightarrow\space$$ $$\underline{\text{Z}}_{\space\text{in}}=\frac{1}{2}+\frac{400\pi^2\cdot\text{R}_2}{400\pi^2+\text{R}_2^2}+\left\{\frac{20\pi\cdot\text{R}_2^2}{400\pi^2+\text{R}_2^2}-\frac{1}{5066000\pi}\right\}\cdot\text{j}\tag1$$

So, we got that (assuming that $$\1\space\text{V}\$$ is the RMS-voltage of the source):

$$\left|\text{V}_\text{L}\right|=\left|\frac{\text{R}_2}{\text{R}_2+\text{j}\omega\text{L}}\cdot\frac{\underline{\text{V}}_{\space\text{in}}}{\underline{\text{Z}}_{\space\text{in}}}\cdot\text{j}\omega\text{L}\right|\space\Longrightarrow\space$$ $$\left|\text{V}_\text{L}\right|=\frac{20\pi\cdot\text{R}_2}{\sqrt{\text{R}_2^2+400\pi^2}}\cdot\frac{\sqrt{2}}{\sqrt{\left(\frac{1}{2}+\frac{400\pi^2\cdot\text{R}_2}{400\pi^2+\text{R}_2^2}\right)^2+\left(\frac{20\pi\cdot\text{R}_2^2}{400\pi^2+\text{R}_2^2}-\frac{1}{5066000\pi}\right)^2}}\tag2$$

• Thank you so much :D Quick follow up question, if the source Voltage isn't the RMS-voltage, I can just divide the peak Voltage by sqrt(2), right? There's no strange effects because of the AC circuit (I am really not familiar with AC circuits). – Beacon of Wierd Jan 7 at 22:30
• @BeaconofWierd Yes that us true! And I'm glad that I could help you ;) – Jan Jan 7 at 22:36
• So, the equations worked perfectly and everything :D But I tried to use them to calculate the efficiency of the circuit, as in how much of the input power goes into the inductor. I used that Power = Vrms^2/|Z| and then I calculated the power for the whole circuit using voltage in and Zin, and I calculated the power at the inductor by using that |Z| for inductor = wL. However for "large" L (above something like 2mH) I get efficiencies above 1 :S Am I doing something wrong or is it just the instantaneous power or something like that? :S – Beacon of Wierd Jan 9 at 21:28
• @BeaconofWierd yes you're doing something wrong. You've to look at AC power. – Jan Jan 9 at 21:32

You had a question in the comments that I wanted to reply to, but can't as I don't have enough reputation points :(.

Imagine that the inductors and capacitors are resistors with resistances $$R_L$$ and $$R_C$$ respectively.

What would the voltage be across the "pseudo-inductor" be? Take that result, and replace $$R_L$$ and $$R_C$$ with the actual impedances of the components:

$$Z_{L_1} = j \omega L_1$$

$$Z_{C_1} = \frac{1}{j \omega C_1}$$ where $$\omega = 2 \pi f$$.

The input voltage used would be the amplitude of the source signal. The magnitude of that expression would be your answer.

Notice how that only involved algebra.

If you wanted a "time-domain" result, then it will involve differential equations.

In summary:

If you want the output in the time-domain --> will require differential equations with initial conditions.

If you want the output in the frequency-domain --> simple algebra

If you want to know the technical reason why this is the case, then please let me know!

• Thank you for your explanation :) I'm interested in calculating the power efficiency of the circuit (how much power is put into the inductor), but I am getting efficiencies above 1 in frequency-domain, is it not possible to calculate power/efficiency in the frequency-domain? :S – Beacon of Wierd Jan 9 at 21:31