Impedance can be complex!
Put your complex numbers and calculus to the test with this
impedance calculation.
Age
16 to 18
Challenge level
Problem
An AC generator of rms voltage E is connected in series with two reactive impedances,
$Z_1$ and $Z_2$, as shown below:
$\theta$ is the argument of $Z_1$ and $\phi$ is the argument of $Z_2$.
Image
Find the power dissipated in $Z_2$
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 $Z_2$.
Getting Started
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.
You may wish to read our articles on AC/DC circuits for the background physics and mathematics required.
Student Solutions
Part 1:
$P = IV$
I = $\frac{E}{Z_1 + Z_2} = \frac{E}{Z_1cos\theta + Z_2cos\phi + i(Z_1sin\theta + Z_2sin\phi)}$
|I| = $\frac{E}{\sqrt{(Z_1cos\theta + Z_2cos\phi)^2 + (Z_1sin\theta + Z_2sin\phi)^2}}$
The real power transferred to $Z_2$: P = I^2 R
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 )} $
Part 2:
To find the maximum power; we can differentiate the power expression with respect to $Z_2$ and set the derivative equal to zero.
$\frac{dP}{dZ_2} = \frac{d}{dZ_2}\frac{E^2 Z_2cos \phi}{Z_1^2 + Z_2^2 + 2Z_1Z_2cos( \theta - \phi )}$
Let U = $E^2 Z_2cos \phi$
Let V = $ Z_1^2 + Z_2^2 + 2Z_1Z_2cos( \theta - \phi ) $
$\frac{dP}{dZ_2} = \frac{VU' - UV'}{V^2}$
V' = $ 2Z_2 + 2Z_1 cos ( \theta - \phi)$
U' = $E^2 cos \phi$
Substituting thses values into the expression and setting it equal to zero we find that:
$Z_1 = Z_2$
For maximum power transfer between load and source we must therefore match the internal impedance of the load with the impedance of the source