# Complex power, real power

Given:

$U_q = 220 V, f = 50 Hz, R_i=10 \Omega, R_a=40 \Omega, L_a=95.5 mH$

I'm asked to determine the real power transformed in $Z_a$. Here I break down the formulas that I use:

• $Z_{total}=Z_{La}+Z_{Ra}+Z_{Ri}$
• $I={U_q\over Z_{total}}$
• $P = Re[U_q . I^*]$

I'm really sure that my calculation is right using that. But it gives a really different answer than the answer key, can you see what's wrong with my formulas? Can I use $U_q$ directly as $U_{load}$?

I thought the current are in series so it's must be the same, and also the voltage is parallel so it's must be the same. So how can I determine the power only in $Z_a$?

EDIT:

Here is my calculation:

• $Z_{total}= j\omega L_a+R_a+R_i=(50+30j)\Omega$
• $I= {220\over 50+30j}={55-33j\over 17}A$
• $P=Re[220({55+33j\over 17})]=711 W$

And the correct answer is 284.7 W.

• "different answer than the answer key" - do you mean numerically or symbolically? You are right about the current being equal in all elements. Once you determine that current simply apply it to Ra by using P=I-squared R, and you have the real power consumed (transformed?) by the load Za. La does not contribute to real power dissipation/transformation in Za because it is reactive power - that is, it is stored in part of the 50 Hz cycle and released in the other part of the cycle. Commented Aug 16, 2016 at 16:20
– jonk
Commented Aug 16, 2016 at 16:21
• Well of course current is the same throughout the three series elements. But if you use Uq you are calculating power taken by the three elements, including that in Ri. Though you are asked power only in Za i.e. Ra +La, no Ri included. Commented Aug 16, 2016 at 16:40
• @FiddyOhm numerically, and I just copy the text of the question. Should it be 'consumed'? I've tried $P=I^2Z_a$ but when I take just the real value as real power, it gives -234.4 W. And yes, it's the same method as you said "La does not contribute to real power dissipation/transfor‌​mation in Za." Commented Aug 16, 2016 at 17:05
• @carloc then how can I calculate the current or voltage if it's not given? Commented Aug 16, 2016 at 17:06

I believe there are two issues preventing you from matching the answer key.

The first is that they really only want the power in Za, so you need to calculate U. Just take your current (I agree with that calculation) and multiply it by Za

U = I $\cdot$ (40 + j30)

Then calculate S = I $\cdot$ conj( U ). The real part of this should be 569 W. Now, that's still double what the book has...

So I would assume that the Uq given is peak voltage instead of the usual default of RMS. That reduces I_RMS by $\sqrt2$ and U_RMS by $\sqrt2$, reducing the product by a full factor of 2.

• Thanks a lot for your help! Now I got the same answer! Btw, I accidentally didn't put the half in the real power equation (the book says multiply it with half). And I remember when I solved few questions about power, sometimes when I multiplied by half, it gives the wrong answer. The right answer was just multiply the voltage and current. Can you explain a little bit further or example about this? Commented Aug 16, 2016 at 18:03
• sure! So the power consumed in an AC system varies in a sinusoidal fashion, and what we generally are concerned with is the average power. If you have a 1 V (peak) signal across a 1 ohm resistor, you'll get 1 A (peak) current. However, because these are sin waves, the average will be less than the peak. Of course the average for the V and I will be 0 in an AC system, so we take the mean (average) of the square of each signal and then take the square root of that (RMS). So the RMS current is 0.707 A and the RMS voltage is 0.707 V in the example above, and the average power is 0.5 W. Commented Aug 16, 2016 at 18:47
• similarly, another question might have started with giving you the RMS voltage 0.707 V and 1 ohm. So it's really important for the questioner to be specific about which one it is! Commented Aug 16, 2016 at 18:48
• Uh much simpler... once you've got current you don't have to calculate any voltage nor complex power. Active power only developes across resistor, so you only have to write $P=R_\text{a}|I|^2$ Commented Aug 16, 2016 at 21:02
• @MikeP problem solved :) thanks again for sharing the knowledge... Commented Aug 17, 2016 at 12:51

Well, we have for the input impedance:

$$\underline{\text{Z}}_{\space\text{in}}=10+\frac{1}{\text{j}\cdot2\cdot\pi\cdot50\cdot\frac{955}{10}\cdot10^{-3}}+40=50-\frac{20}{191\pi}\cdot\text{j}\tag1$$

The real power is given by:

$$\text{P}_{\space\text{in}}=\text{V}_\text{in rms}\cdot\text{I}_\text{in rms}\cdot\cos\left(\theta_{\space\text{in}}\right)\tag2$$

For $$\\theta_{\space\text{in}}\$$ we have:

$$_{\space\text{in}}=\left|\arg\left(\underline{\text{V}_{\space\text{in}}}\right)-\arg\left(\underline{\text{I}_{\space\text{in}}}\right)\right|=\left|\arg\left(\underline{\text{Z}_{\space\text{in}}}\right)\right|=\arctan\left(\frac{2}{955\pi}\right)\tag3$$

So:

$$\text{P}_{\space\text{in}}=220\cdot\frac{220}{\left|50-\frac{20}{191\pi}\cdot\text{j}\right|}\cdot\cos\left(\arctan\left(\frac{2}{955\pi}\right)\right)=$$ $$968\cdot\left(1-\frac{4}{4+912025\pi^2}\right)\approx967.999569841344\space\text{W}\tag4$$