0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Excuse me for my really bad use of the schematic feature I just didn't know how to switch to the phasor domain. Anyway, I'm asked to find the Power at the R2 Resistor knowing only that the capacitor absorbs -2.3 VARs of reactive power.

I've been stuck on this problem and just couldn't find a way to solve it so I just need something to focus at or maybe get some power properties I should know.

(And yes, we don't know how many volts the source is generating)

\$\endgroup\$

closed as off-topic by MCG, Elliot Alderson, Lior Bilia, Warren Hill, Finbarr Feb 7 at 11:56

  • This question does not appear to be about electronics design within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ The power source in the schematic is dc, there is no reactance here. I’m guessing you’re referring to an ac source instead. \$\endgroup\$ – rr1303 Feb 5 at 13:55
  • \$\begingroup\$ Yes I do mean that, this is my first time using a schematic \$\endgroup\$ – O. Sinno Feb 5 at 13:59
  • 4
    \$\begingroup\$ I'm voting to close this question as off-topic because this looks like a homework question without an attempt at a solution \$\endgroup\$ – MCG Feb 5 at 15:43
  • \$\begingroup\$ I see you're not the friendly type \$\endgroup\$ – O. Sinno Feb 5 at 18:10
  • \$\begingroup\$ I've updated the schematic to show V2 as AC. Do you know the frequency? Also if the values of L1 and C1 are Henry and Farad there should be no j in there. If they are impedances it should be ohms. \$\endgroup\$ – Warren Hill Feb 7 at 10:31
5
\$\begingroup\$

V2, L1, and R1 are red herrings - don't pay attention to them. You know the reactive power dissipated by the capacitor as well as its reactance. This yields the RMS voltage across the capacitor by Ohm's law. But the capacitor is in parallel with R2, so you immediately know the RMS voltage across the resistor. With the resistance and RMS voltage, you can easily determine the average power dissipation in the resistor.

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.