# Power factor calculation of an offset sine wave across a non-ideal capacitor

I got into a discussion with another student while we were studying and we couldn't come to a conclusion.

Assume you have a non-ideal capacitor with some ESR and leakage resistance. If you subject that non-ideal capacitor to an offset sine wave (AC voltage signal with a DC offset) and you wanted to calculated the real power, would you:

1. Multiply the product of the offset RMS values of the offset voltage and current waveforms by cos(phase shift), or
2. Multiply the product of the RMS values of JUST the AC waveforms (without the DC offset) by cos(phase shift) then add the DC power which you can calculate from knowing the ESR and leakage resistance?

In your example, the total loss dissipated in heat corresponds to the sum of individual losses in the resistors. You correctly calculate them with $$\P_R=I_{rms}^2R\$$ with $$\R\$$ the considered resistance and $$\I\$$ the current in the resistor. I simulated the below sketch with a 100-$$\\Omega\$$ resistance across the capacitor. The source delivers 10 V peak at a 1-kHz frequency with a 5-V offset.

You compute the average power - or the real power - by multiplying $$\i_{in}(t)\$$ and $$\i_{tot}(t)\$$ and average the result across a cycle. This always works regardless of the type of waveform or distortion. If you deal with sinusoidal signals, then you can involve $$\cos\phi\$$ applied to the ac component as you mentioned and deal with rms voltage and current.

The below simulation results show the instantaneous waveforms:

If I average the instantaneous power which is plotted in the low-side window, the simulator gives me 744 mW. The rms current flowing in resistor $$\R_1\$$ is 85.74 mA while in the 1-$$\\Omega\$$ resistance it is 96.4 mA: $$\P_{avg}=85.74m^2\times 100 + 96.4m^2\times 1 = 735.1\;\rm mW + 9.3\;\rm mW = 744.4\;\rm mW\$$ in line with the measurement (these numbers should be perfectly equal but you need to exactly position the measurement cursors to isolate a period).

If we now measure the rms ac components (without the dc offset) of the input current and voltage, we find $$\V_{in,rms}=7.07\rm\;V\$$ and $$\I_{tot,rms}=82.7\rm\;mA\$$. The phase shift is 31.6° which leads to an intermediate real power of $$\V_{ac,rms}I_{ac,rms}\rm cos(\phi)= 498\rm\; mW\$$. The dc current is the offset (5 V) divided by the total resistance ($$\101\rm\; \Omega\$$) and it amounts to 50 mA or 248 mW dissipated in the resistance. Sum 498 mW to 248 mW and you obtain 746 mW in line with what we found.

you wanted to calculated the real power

You would add the powers. You would calculate the DC power and you would calculate the AC power and add the two together.