circuit analysispassive-networksresonancevoltage
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.
In which condition the current through the inductor approach INFINITY and how can you explain that physically?
Yes, it is absolutely true - the current in the inductor can become infinite and the current in the capacitor can also become infinite however, because these two currents are anti-phase (one lags the sinewave source by 90deg and one leads the sinewave source by 90deg), the net current taken from an AC voltage source is zero other than what is consumed in the parallel resistor.
You are using a current source and the resultant voltage will be I*R. I'm also presuming you meant R to be a positive resistance value not as shown in your diagram although that won't alter things at all - if you meant R to be negative then the voltage developed across the resistor will be anti-phase to the current through it.
The actual current through the inductor is the voltage across it (as developed by the resistor and current source). Ditto the capacitor.
When you apply a voltage to a series resonant circuit of R, L and C, the reactances of L and C are equal but have opposite signs. This means their impedances totally cancel out and the only impedance that is left (at resonance) is R.
Regards the energy - this oscillates between L and C so at any one point the energies will not necessarily be equal but, if you averaged those energies out over time, you will find that they do become equal.
Because C and L reactances are equal, it means that the peak voltage attained on both is the same magnitude AND because they share the same current (series resonant) it duly follows that each will store the same instantaneous peak energy (but of course at different times in the AC cycle).
EDIT - reasons why the energies are the same: -
Here are what the waveforms look like for voltage and current in a capacitor: -
Here are what the waveforms look like for voltage and current in an inductor: -
Pictures taken from here
If you look at the voltage across the capacitor you will see that it lags the current by 90 degrees i.e. when the current is at a peak the voltage is zero. For the inductor, when the current is at a peak the voltage is also zero but, compared to the capacitor's voltage, it is 180 degrees different.
Because reactances are equal and current is equal the two voltages cancel each other out leaving only the resistor in circuit to take the full AC voltage of the driving source. It's like having two 9V batteries in series - one way they produce 18 volts and the other way they produce zero volts and you would not be able to distinguish two batteries wired series antiphase from a very low value ohmic resistor.
Also if you multiplied the capacitor's and inductor's respective current and voltage waveforms to produce a power waveform you'd find this: -
This means the peak energy stored in both L and C is the same.
Best Answer
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):
$$\left|\text{V}_\text{L}\right|=\left|\frac{\text{R}_2}{\text{R}_2+\text{j}\omega\text{L}}\cdot\frac{\underline{\text{V}}_{\space\text{in}}}{\underline{\text{Z}}_{\space\text{in}}}\cdot\text{j}\omega\text{L}\right|\space\Longrightarrow\space$$ $$\left|\text{V}_\text{L}\right|=\frac{20\pi\cdot\text{R}_2}{\sqrt{\text{R}_2^2+400\pi^2}}\cdot\frac{\sqrt{2}}{\sqrt{\left(\frac{1}{2}+\frac{400\pi^2\cdot\text{R}_2}{400\pi^2+\text{R}_2^2}\right)^2+\left(\frac{20\pi\cdot\text{R}_2^2}{400\pi^2+\text{R}_2^2}-\frac{1}{5066000\pi}\right)^2}}\tag2$$