I have a parallel RLC circuit where the inductor is in series with the resistor. The value for the frequency is 200hz and the value for the current of the circuit is 5mA. The experimental answer is 1.241v but when I do the equations I’m given in my lectures and online(1/Z=SQRT((1/R)^2+(1/Xl - 1/Xc)^2) then times that by 5x10-3 I get a completely wrong answer. I just cannot figure out And as it’s for coursework I cannot just ignore it.any help would be greatly appreciated. parallel RLC circuit this is the equation that comes up in my notes from my lectures but it does not provide the correct results
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  • \$\begingroup\$ Can you draw the equivalent circuit that you get and you're trying to analyze with the voltage and amp. meter in the circuit? After that I can try to help you solve it. \$\endgroup\$ Nov 27 '21 at 20:59
  • \$\begingroup\$ It's good that you're saying it's for coursework (adding the homework would have helped), but it would also help to show what you did, because then people would be able to point out the mistake (as opposed to someone doing the homework for you). \$\endgroup\$ Nov 28 '21 at 7:49
  • \$\begingroup\$ @aconcernedcitizen as I have experiment data I have done the entire thing in excel. it is a two part coursework where you have to use the experimental date to create a graph to find the resonant frequency. then using you original values(for frequency and voltage/current) create theoretical answers and see how they compare. for the series RLC circuit this worked completely fine and I have two sets of near-identical answers(using the equation Root(R^2+(XL-Xc)^2). but for the parallel circuit as show above I simply cannot figure out the equation needed to find the correct total impedance. \$\endgroup\$ Nov 28 '21 at 12:55

I suspect that your notes have two standard formulas, one for a circuit where RLC are all in series and one where they are all in parallel. This circuit is neither of those cases, so you may need to work a little harder.

Notice that the series RL and the capacitor are in parallel. This means that the voltage across the RL and the capacitor must always be the same. So, you can find the impedance of the RL and the impedance of the C and treat them as parallel impedances.

Since I don't know what you have been taught about impedance, or what you are expected to learn independently, I will stop at that hint. I strongly suggest that you contact the instructor if it is still not clear to you.

  • \$\begingroup\$ You are correct. In our notes we only have notes for parallel and series circuit which is why I have been left stumped by this circuit. I have tired many combinations of impedance equations to find the impedance but cannot find one that gives the correct total impedance. I have tried the current divider rule which did not work. I tried so many things I cannot fit them all in the character limit. I asked for help from my tutor but he said as he was marking the coursework he couldn’t personally help me. I have searched all over for an explanation on This kind of circuit but have found nothing \$\endgroup\$ Nov 28 '21 at 13:55
  • \$\begingroup\$ I just need an equation on how to find the total impedance and I can do everything else myself it has also left my other teacher stumped and when I went to them for help they also tired too do current divider equations too no avail. \$\endgroup\$ Nov 28 '21 at 13:56
  • \$\begingroup\$ Do you understand how to represent impedance as a magnitude and an angle, or as a real part and an imaginary part? Do you understand how to combine such impedances in series or parallel? \$\endgroup\$ Nov 28 '21 at 17:28
  • \$\begingroup\$ i understand how to use imaginary numbers to represent impedance but i cannot turn that into a real number and put in into a graph to compare to my real results i did in the lab. \$\endgroup\$ Nov 28 '21 at 22:29
  • \$\begingroup\$ How did you measure or otherwise obtain your "real results"? You probably need to consider just the magnitude of the voltages and currents, unless you measured them simultaneously with an oscilloscope. \$\endgroup\$ Nov 28 '21 at 23:16

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