In a R+1/(jwC) load (AC)voltage lags current with θ degrees. Reactive power Q is negative because sin(θ) is negative. My question is how can a capacitive load deliver power to the source wich is supposed to supply the power? Im sure I do not completely understand the subject, can someone explain my faults? Thank you for your help, Nick.
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\$\begingroup\$ Do you mean \$j\omega L\$ or \$\frac{1}{j\omega C}\$? \$\endgroup\$– Andy akaMar 21, 2019 at 12:52
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\$\begingroup\$ Yes I meant 1/(jwC) \$\endgroup\$– N. BergMar 21, 2019 at 13:39
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2 Answers
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It is a common misunderstanding that a voltage source must also be a power source. That's not true, an ideal voltage source merely constrains the voltage between two points in a circuit. Current may flow in either direction through the source, causing it to consume power or provide power. Think about a rechargeable battery, for example. If you have some kind of standby power supply, the battery will be charging (consuming power) part of the time and supply power part of the time.
answered Mar 21, 2019 at 12:46
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\$\begingroup\$ A capacitive circuit still needs to be charged. This would mean the average power consumption would be zero in an ideal circuit right? Is +- in reactive power just an indicator for phasor angle? \$\endgroup\$– N. BergMar 21, 2019 at 13:47
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\$\begingroup\$ Power must be conserved in a circuit, so the average power supplied must equal the average power consumed. I'm not sure what you mean by "+-" but think you are correct, it is just an indication of the phase angle. \$\endgroup\$ Mar 21, 2019 at 16:03
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I found a good explanation of what I am trying to ask, I will post the link for anyone else who is interested. What does it mean for reactive power to be delivered or consumed