Double check apparent power definition

[tim9000]

Hey just to make sure,
can we say apparent power S = V^2 / Z

for a complex impedance Z?
And do we have to worry about conjugates at all?

Cheers

 

[anorlunda]

The total power S = VI. Where S, V and I are all complex. I prefer that to $$\frac{V^2}{Z}$$ because VI works for any circuit.

Is apparent power the same as total power? I'm not exactly sure what you mean by apparent power.

Edit: no conjugations were needed.

 

[Averagesupernova]

I always thought apparent power is the voltage across the complex impedance multiplied by the current through the complex impedance. So, V * I, or V^2 / Z, or I^2 * Z. I don't think I have heard of a definition of total power. But that just might be semantics.

 

[milesyoung]

First, you have the complex power:
$$
\mathbf{S} = \mathbf{V}\overline{\mathbf{I}}
$$
where ##\mathbf{V},\mathbf{I}## are RMS phasors. The overline means to take the complex conjugate.

The apparent power is defined as:
$$
|\mathbf{S}| = |\mathbf{V}\overline{\mathbf{I}}| = |\mathbf{V}||\overline{\mathbf{I}}| = |\mathbf{V}||\mathbf{I}| = \frac{|\mathbf{V}|^2}{|\mathbf{Z}|}
$$
which is consumed by the impedance ##\mathbf{Z}## with the voltage ##\mathbf{V}## across it.

 

[tim9000]

The total power S = VI. Where S, V and I are all complex. I prefer that to $$\frac{V^2}{Z}$$ because VI works for any circuit.

Is apparent power the same as total power? I'm not exactly sure what you mean by apparent power.

Edit: no conjugations were needed.

I always thought apparent power is the voltage across the complex impedance multiplied by the current through the complex impedance. So, V * I, or V^2 / Z, or I^2 * Z. I don't think I have heard of a definition of total power. But that just might be semantics.

Yeah Averagesupernova, I think that's right, I always think of it as being the hypotenuse of the reactive and real power.

 

[tim9000]

First, you have the complex power:
$$
\mathbf{S} = \mathbf{V}\overline{\mathbf{I}}
$$
where ##\mathbf{V},\mathbf{I}## are RMS phasors. The overline means to take the complex conjugate.

The apparent power is defined as:
$$
|\mathbf{S}| = |\mathbf{V}\overline{\mathbf{I}}| = |\mathbf{V}||\overline{\mathbf{I}}| = |\mathbf{V}||\mathbf{I}| = \frac{|\mathbf{V}|^2}{|\mathbf{Z}|}
$$
which is consumed by the impedance ##\mathbf{Z}## with the voltage ##\mathbf{V}## across it.

Hi Miles, thanks.
Ah ok, it is for magnitudes. But what about if I still wanted it all to be in terms of comples voltages, currents and powers?

Specifically because I'm trying to work out a Newton raphson here:

https://www.physicsforums.com/threa...on-matlab-not-converging.839961/#post-5270803

Using the aforementioned power equation. I'm trying to work out the complex voltages for 4 busses in Matlab. Where I have worked out the matricies for the complex power for each bus and the admittance matrix.Thank you

 

[milesyoung]

But what about if I still wanted it all to be in terms of comples voltages, currents and powers?

##\mathbf{V}## and ##\mathbf{Z}## are complex.

When you need the apparent power, then you could just, for instance, use the 'abs' function in MATLAB to find ##|\mathbf{V}|,|\mathbf{Z}|##.

 

[tim9000]

##\mathbf{V}## and ##\mathbf{Z}## are complex.

When you need the apparent power, then you could just, for instance, use the 'abs' function in MATLAB to find ##|\mathbf{V}|,|\mathbf{Z}|##.

Hey Miles,
Yeah I actually tried that yesterday, but now I'm thinking I can't use apparent power. So is it incorrect to say that:

Complex power = complex voltage * complex admittance?
like P + Q = V^2 / (R + X)

I think one thing that might be wrong is my jacobian.
From what I remember NR uses the form y = f(x)
where y is a constant (in this case the powers) and x is the voltages ('volts' matrix), so my function is
(volts)^2)*Y_mat

so the jabobian is:
(2*volts*Y_mat)

? Or have I not differentiated that properly?