6677
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1
2011-02-01T00:00:01
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<p>An AC generator of rms voltage E is connected in series with two reactive impedances,<br></br>
$Z_1$ and $Z_2$, as shown below:<br></br>
<br></br>
$\theta$ is the argument of $Z_1$ and $\phi$ is the argument of $Z_2$.<br></br>
<mdo:image alt="" height="242" src="Problem4.jpg" width="382"></mdo:image><br></br>
Find the power dissipated in $Z_2$<br></br>
<br></br>
If the phase of $Z_2$ ($\phi$) is fixed whilst allowing the magnitude to vary, find the relationship between $Z_1$ and $Z_2$ when we achieve maximum power transfer to $Z_2$.<br></br>
<br></br>
<br></br>
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Part 1:<br></br>
<br></br>
$P = IV$<br></br>
<br></br>
I = $\frac{E}{Z_1 + Z_2} = \frac{E}{Z_1cos\theta + Z_2cos\phi +
i(Z_1sin\theta + Z_2sin\phi)}$<br></br>
<br></br>
|I| = $\frac{E}{\sqrt{(Z_1cos\theta + Z_2cos\phi)^2 + (Z_1sin\theta
+ Z_2sin\phi)^2}}$<br></br>
<br></br>
The real power transferred to $Z_2$: P = I^2 R<br></br>
<br></br>
P = $ I^2 Z_2cos \phi = \frac{E^2 Z_2cos \phi}{Z_1^2 + Z_2^2 +
2Z_1Z_2(cos \theta cos \phi + sin \theta \phi)}$= $ \frac{E^2
Z_2cos \phi}{Z_1^2 + Z_2^2 + 2Z_1Z_2cos( \theta - \phi )} $<br></br>
<br></br>
Part 2:<br></br>
<br></br>
To find the maximum power; we can differentiate the power
expression with respect to $Z_2$ and set the derivative equal to
zero.<br></br>
<br></br>
$\frac{dP}{dZ_2} = \frac{d}{dZ_2}\frac{E^2 Z_2cos \phi}{Z_1^2 +
Z_2^2 + 2Z_1Z_2cos( \theta - \phi )}$<br></br>
<br></br>
Let U = $E^2 Z_2cos \phi$<br></br>
<br></br>
Let V = $ Z_1^2 + Z_2^2 + 2Z_1Z_2cos( \theta - \phi ) $<br></br>
<br></br>
$\frac{dP}{dZ_2} = \frac{VU' - UV'}{V^2}$<br></br>
<br></br>
V' = $ 2Z_2 + 2Z_1 cos ( \theta - \phi)$<br></br>
<br></br>
U' = $E^2 cos \phi$<br></br>
<br></br>
Substituting thses values into the expression and setting it equal
to zero we find that:<br></br>
<br></br>
$Z_1 = Z_2$<br></br>
<br></br>
For maximum power transfer between load and source we must
therefore match the internal impedance of the load with the
impedance of the source<br></br>
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This problem makes use of the concepts of AC generators and RMS
(root-mean-square) voltage.You will also need to be familiar with
complex numbers in order to attempt this problem.<br></br>
<br></br>
You may wish to read our articles on <a href="http://nrich.maths.org/6661&part=">AC/DC circuits</a> for
the background physics and mathematics required.<br></br>
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Impedance can be complex!
Put your complex numbers and calculus to the test with this
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