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I'm stuck at this problem, which doesn't seem to be so difficult, but, I guess there is some basic stuff that still confuses me. So, if we have a load represented with complex impedance \$Z=30+j40 \ \Omega\$ powered with a sinusoidal signal generator of voltage \$U=200 \ V\$, how can we find a true power on that load?

What I did, knowing that true power is: \$P=U\cdot I\cdot \cos \phi\$, where \$\phi\$ is phase difference between voltage and current through the load and equals \$\phi=\cos\big(\arctan\frac{X}{R}\big)\$ \$\big(X\$ - reactance, \$R\$- resistance \$\big)\$, since I don't know the current, I expressed it as: \$I=\frac{U}{Z}\$ and then calculated true power as: $$P=\frac{U^2}{Z}\cos\bigg(\arctan\frac{X}{R}\bigg)=742 \ W$$ but, I think I'm completely missing some parts regarding complex values of impedance and current.
Thank you for your time.

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    \$\begingroup\$ In your final equation, arctan should be cos(arctan. \$\endgroup\$ – Charles Cowie Jun 2 '16 at 21:15
  • \$\begingroup\$ Then calculating that out correctly should give you the correct answer. You made a calculation error. \$\endgroup\$ – Charles Cowie Jun 2 '16 at 21:18
  • \$\begingroup\$ Also, there is another way to do it by which you can solve it without a calculator. \$\endgroup\$ – Charles Cowie Jun 2 '16 at 21:21
  • \$\begingroup\$ @CharlesCowie What confuses me is putting an equal sign between \$I\$ and \$\frac{U}{Z}\$ (their effective values). Can I really do that? \$\endgroup\$ – A6SE Jun 2 '16 at 21:22
  • \$\begingroup\$ Yes you can do that. \$\endgroup\$ – Charles Cowie Jun 2 '16 at 21:23
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In the text below I will use \$\mathbf{Z}\$, \$\mathbf{U}\$, \$\mathbf{I}\$, \$\mathbf{S}\$ as complex numbers and \$Z\$, \$U\$, \$I\$, \$S\$ as their magnitudes.

Note that $$Z\cos\left(\arctan\frac{X}{R}\right) = \operatorname{Re}(\mathbf{Z})$$ so you can simplify the expression $$\frac{U^2}{Z}\cos\left(\arctan\frac{X}{R}\right) = \frac{U^2}{ZZ}Z\cos\left(\arctan\frac{X}{R}\right) = \frac{U^2}{Z^2}\operatorname{Re}(\mathbf{Z})$$

In fact it's much easier to avoid \$\cos(\varphi)\$ at all from the beginning, since the magnitude of the current is just $$I = \frac{U}{Z}$$ and, since the current is in phase with voltage at the active resistance, the active (real) power is just $$P = I^2 R = \frac{U^2}{Z^2}\operatorname{Re}(\mathbf{Z})$$ which is the same formula as above.


Also, there is a concept of complex power, usually denoted by \$\mathbf{S}\$. The complex power is defined as $$\mathbf{S} \equiv \mathbf{U}\mathbf{I}^*$$ where both \$U\$ and \$I\$ are complex numbers and \$\mathbf{I}^*\$ means the complex conjugate. The equivalent formulas are $$\mathbf{S} = \frac{U^2}{\mathbf{Z}^*} = I^2 \mathbf{Z}$$ Real power \$P = \operatorname{Re}(\mathbf{S})\$

So the alternative way to solve your problem is $$P = \operatorname{Re}\left(\frac{U^2}{\mathbf{Z}^*}\right)$$

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Solution without a calculator:

The impedance 30 + J40 can, by inspection, be represented by a 3-4-5 triangle, so Z = 50. I = U/Z = 200/50 = 4. Power = I squared R. 16 X 30 = 480 watts.

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