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I want to analyse a circuit that I've drawn above. It is a series connection of a resistor and an LC circuit (parallel type) driven by a sine wave of variable frequency. No assumptions can be made about the values of the voltage, frequency or the values of the components.

I'm particularly interested in the voltage across the capacitor. I would also like to know how to calculate the resonant frequency of this circuit, as well as its bandwidth and Q factor.

I understand that normally with a simple LC parallel circuit, resonance occurs when the impedances of the capacitor and the inductor are equal, and therefore the impedance of the overall circuit is maximum. But does the resistor change things, and if so, how? Next to the drawing of the circuit is my very primitive attempt at a solution, but I did not get far. The voltage of the source clearly equals the voltage across the resistor and the LC, and the voltage across the cap and the inductor are clearly equal. Also the current the resistor equals the currents on the cap and the inductor. But I really cannot make progress beyond this.

I'm also guessing that by assumption of R being very large, I could say that the voltage source with the resistor approximates a current source, but this is not what I want to do. I would like to analyse a basic LC circuit connected to a signal generator, with the R being the internal resistance of the signal generator. In an earlier question of mine I asked why I'm not getting the resonance at the frequency calculated with the usual formula of resonance of parallel LC circuit, and I understood that the internal resistance of the signal generator affects the circuit.

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  • \$\begingroup\$ Welcome to EE.SE! This appears to be a homework question. As such, you need to show us your work so far, and explain which part of the question you're having trouble with. For future reference: Homework questions on EE.SE enjoy/suffer a special treatment. We don't provide complete answers, we only provide hints or Socratic questions, and only when you have demonstrated sufficient effort of your own. Otherwise, we would be doing you a disservice, and getting swamped by homework questions at the same time. See also here. \$\endgroup\$ – Dave Tweed Dec 18 '18 at 19:14
  • \$\begingroup\$ @DaveTweed This is not a homework question - just personal interest. And I thought that the equations rovided in my sketch would be adequate as "my own work". I did mention this and also the fact that I could not progress anymore. What else is expected of me? \$\endgroup\$ – S. Rotos Dec 18 '18 at 19:16
  • \$\begingroup\$ You're wanting to find voltage across the inductor. And you've hinted that the source could be viewed as a current source. Converting that Thevenin source to a Norton equivalent makes the solution much easier, because now you have a parallel RLC circuit. \$\endgroup\$ – glen_geek Dec 18 '18 at 19:17
  • \$\begingroup\$ @DaveTweed I also mentioned my goals in the original post, that I want to analyse LC circuits using a signal generator. \$\endgroup\$ – S. Rotos Dec 18 '18 at 19:17
  • \$\begingroup\$ HINT: A series resistor dampens the efficiency of the LC tank circuit. If R is a low value then LC dominates the impedance at frequency f. As some value R will resonate with LC. If R has a very high value then it dominates the equations. \$\endgroup\$ – Sparky256 Dec 18 '18 at 19:40
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If you understand how to compute impedance, Zo at resonance and know that L//C is high impedance as a Bandpass filter.

\$Z_c=\dfrac{1}{j\omega C},~ Z_L=j\omega L\$

The Voltage gain is always unity or 0dB for this resonant circuit, unless a load R is added.

\$Voltage~ Gain = \dfrac{Z_o(LC)}{Z_o(LC)+R}\$

However if the driver was the current source of a collector in a transistor whose impedance is high >100k, then with current excitation to L//C you get voltage gain relative to emitter resistance.

Falstad Simulator

But if you search my answers and others, you will find similar Q&A's.

I'll let you find out what the resonant frequency formula is when \$|Z_C|=|Z_L|\$

Remember that impedance ratios are the key to voltage dividers and Amplifier gain.

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