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I am beginning to learn AC circuits and am trying to analyze the following RLC circuit: enter image description here

I want to find the real-time voltage across the resistor. I believe I will do this by finding the current with I = V_p / Z then use that current to find the voltage on the resistor. I'm not sure how to calculate Z in a parallel circuit. Would I do Z = (1/(Z_R + Z_C) + 1/Z_L)^-1 as I would with resistors in parallel?

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    \$\begingroup\$ There's a voltage divider formed by R and C, so no need to bother with L \$\endgroup\$ – Chu Sep 30 '15 at 19:58
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    \$\begingroup\$ Paraphrasing @chu, you have a perfect voltage source that puts a 141.42V RMS sinewave across the series combination of R1 and C2 - the inductor is nothing to do with finding the voltage across R1. \$\endgroup\$ – Andy aka Sep 30 '15 at 21:08
  • \$\begingroup\$ Okay, I understand that. How do I then only deal with R1 then? I'm a little confused on AC circuits. \$\endgroup\$ – TheStrangeQuark Sep 30 '15 at 21:13
  • \$\begingroup\$ @Andyaka and does that mean there is no current going through L1? \$\endgroup\$ – TheStrangeQuark Sep 30 '15 at 21:14
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    \$\begingroup\$ The same voltage appears across L, and the series combination of R & C. You can apply Ohm's Law to both of these branches separately to work out the currents through them. But to work out the voltage across R you only need to treat R and C as a voltage divider - no need to find the current and then multiply by R. \$\endgroup\$ – Chu Sep 30 '15 at 21:49
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Yes, when in parallel combination, 1/(Znet) = 1/Z1 + 1/Z2 + 1/Z3 .....

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