# what does reactive power mean here? simulate this circuit – Schematic created using CircuitLab

assuming this circuit is run for 5 seconds period, the switch is closed for 1 second starting at t=3s

we can calculate average power as average = 1.8 W but apparent power S = Vrms * Irms = 4.023 VA and the difference will be the reactive power which is >0, right? what does this reactive power mean? in rl or rc circuit it's the powered stored in inductor or capacitor but switch cannot store power.

I am not seeing any reactive power there, that would require and inductor or capacitor.

During the 1 second the switch is closed current is 3V/1 ohm = 3A, and power is 9W giving a total energy expended of 9W * 1s = 9 joules, the rest of the time there is no power dissipated, so the average is indeed 1.8W integrated over 5 seconds, but I do not see where you get ~4VA from?

RMS current is (4 * 0^2 + 1 * 3^2)/5 = 1.34V, and because of the 1 ohm resistor RMS current must be 1.34A. 1.34V * 1.34A = 1.8W * 5s = 9 joules.

• it's the apparent power (S) and S=V rms * I rms=4.023 VA
– sihs
Nov 19 '16 at 18:57
• actually i am asking about the physical meaning of the reactive power here, which is the imaginary component from S= Q J + P
– sihs
Nov 19 '16 at 18:59
• Vrms * Irms does NOT equal 4.023VA! Vrms is 1.34V measured at the resistor and Irms is 1.34A, giving 1.8VA, and 1.8W because a resistor has a power factor of 1.0. If you consider things between the switch and the power supply, then Vrms is clearly 3V, but IRms is now 0.6A (root 3^2)/5 again giving 1.8W average, and 9J over the 5s integration time. Nov 19 '16 at 19:07
• Note my square roots in my original answer seem to be AWOL, but in that case it should be effectively root (9/5) as opposed to root (9)/5, the difference is in where you measure. In no case is reactive power in any way involved. Nov 19 '16 at 19:11
• good, but dealing with switch and resistor as a single device and measuring V across its terminal then Vrms = 3V and I rms=1.34A
– sihs
Nov 19 '16 at 19:17