Power supplied by AC component + DC component = total power?

[hsuy]

Hi,

I have a question about ac component and dc component of a source. For example, there is a current source connected directly to a resistor.
If the current source supplies a sine wave with no dc offset, then I understand that power supplied by dc component is zero, and so power absorbed by resistor is the same as power supplied by ac component and dc component. ( in this case, power supplied by ac component is the same as power absorbed by resistor)
But what if the source supplies a sine wave with a dc offset? From what I understood, the power supplied by ac component needed to be calculated by using the effective value (rms). In this way, when I do the calculation, the power supplied by ac component is still the same as power absorbed by resistor; however, since there is a dc offset, there should be an additional power supplied by the dc component. Is that correct?
I feel really confused about this, hope someone helps me!

Thanks

 

[anorlunda]

You can superimpose an AC signal with DC and sum the powers.

If you do it right, you can also calculate the RMS of a signal that has a DC offset. The RMS answer is not the same as one without the offset.

Just take any time varying signal, any shape, symmetrical or not, repeating or not, DC offset or not. Sample it in many small samples. Square each sample value, find the mean of the squares, take the square root of that. RMS means [square] Root of the Mean of the Squares.

 

[NascentOxygen]

This might help you. https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-sine-wave-with-a-dc-offset/

 

[Babadag]

It is very interesting, I did the same calculation for an example:

Let's say Vac[rms]=200 V and Vdc=100 V and the load impedance it is a pure resistance of R=10 ohms.

If we take Iac=200/10=20 A and Idc=100/10=10 A and we calculate the dissipated power separately

then Pac=20^2*10=4000 W and Pdc=10^2*10=1000 W total 5000 W.

Let's take now the total current 20+10=30 A. Ptotal=30^2*10=9000 W.

If total current it is I=[(sqrt(2)*Vac*sin(ꞷ*t)+Vdc)]/R then the dissipated energy in a half of a cycle will be:

E= R∫I^2*dt|t=0 to T/2| where T=1/frq

I^2=[2*Vac^2*sin(ꞷt)^2+2*sqrt(2)*sin(ꞷt)*Vac*Vdc+Vdc^2]/R^2

and the energy E=integral{[2*Vac^2*sin(ꞷt)^2+2*sqrt(2)*sin(ꞷt)*Vac*Vdc+Vdc^2]/R}

in the interval t=0 to T/2 and the result it is:

E=[Vac^2*(1/frq/2)+4*sqrt(2)*Vac*Vdc/ꞷ+Vdc^2/frq/2]/R [J(W.sec)]

Replacing the proposed values above will result:

E=71.68 J and divide by 1/60/2 P=8601.6 W

 

[Babadag]

NascentOxigen, this article is very interesting, indeed. However, in my opinion it is a mistake in one of equation. This:

it has to be:

 

[NascentOxygen]

However, in my opinion it is a mistake in one of equation.

Yes, they do seem to have the wrong sign. Fortunately, it doesn't change their result.

 

[Babadag]

I was wrong at the energy integral calculation in #4, because I considered only half a cycle and multiplied by 2. Because a.c. current change sign in the second half of the cycle, but d.c. current does not the total average is lower. So 5000 W it is correct answer. Sorry.