The discussion centers on the power supplied by capacitors and inductors in AC circuits, specifically addressing the confusion surrounding reactive power (Q) and real power (P). The calculated reactive power is 179 VAR, but the textbook states that the real power is zero due to the nature of ideal capacitors, which do not supply net energy over a complete cycle. The average power formula must include the power factor, which is defined as the ratio of resistance (R) to impedance (Z). In ideal capacitive circuits, R is zero, leading to a power factor of unity.
PREREQUISITESElectrical engineering students, circuit designers, and professionals working with AC power systems will benefit from this discussion, particularly those focused on understanding reactive power and power factor in capacitive circuits.
Also, P is used for "real" power (which is 0 in this case), and Q is used for "reactive" power (which you calculated).engnrshyckh said:PQ=179VAR
How to find power factor in this case as P. F=COS@=R/ZTSny said:When AC voltage is applied to an ideal capacitor, the capacitor takes in energy for part of a cycle of voltage and the capacitor delivers energy back to the circuit for the remaining part of the cycle. The net energy delivered to the capacitor in one complete cycle is zero. So, over many cycles, there is no net energy supplied to the capacitor.
From the point of view of formulas, the average power supplied is not ##P_{\rm avg} = \large \frac{V^2}{Z}##. The formula should contain an additional factor called the "power factor". Is this something you have covered?
It should be zero if i am not wrong and cos0 becomes 1 so P. F of pure capacitive circuit become unityTSny said:For an ideal capacitor, what is the value of R?
Yes, R = 0.engnrshyckh said:It should be zero if i am not wrong
No, the formula says ##\cos \theta = R/Z##.and cos0 becomes 1