# Circuit Theory/Extra Element

The Extra Element technique helps break down one complicated problem
into several simpler ones like superposition. However the goal is to
find impedance rather than current or voltage.

By removing the element that most complicates the circuit the overall
circuit impedance can be obtained.

$$Z_{in} = Z^{\infty}_{in} \left( \frac{1+\frac{Z^0_{e}}{Z}}{1+\frac{Z^{\infty}_{e}}{Z}} \right)$$

where

$$Z\$$ is the impedance chosen as the extra element to be removed

$$Z^{\infty}_{in}$$ is the input impedance with Z removed (or made
infinite)

$$Z^0_{e}$$ is the impedance seen by the extra element Z with the source
shorted (or made zero)

$$Z^{\infty}_{e}$$ is the impedance seen by the extra element Z with the
source opened (or made infinite)

Computing these three terms may seem like extra effort, but they are
often easier to compute than the overall input impedance.

### Example

!4 resistors, 1 capacitor
example !set up for
node analsysis !node
analysis in mupad ../code/ .. need to solve for
vs/current to compare
answers

Consider the problem of finding $Z_{in}$ for the circuit in Figure 1
using the EET (note all component values are unity for simplicity). If
the capacitor (gray shading) is denoted the extra element then

$$Z = \frac{1}{s}$$

Removing this capacitor from the circuit we find

<File:Example18b.png%7Ccapacitor> removed, looking at the input (source
input) <File:Example18c.png%7Ccapacitor> removed, reorganizing circuit
so can see input impedance $Z^{\infty}_{in}$

$$Z^{\infty}_{in} = 2\|1 +1 = \frac{5}{3}$$

Calculating the impedance seen by the capacitor with the input shorted
we find

<File:Example18d.png%7Cshorted> source, finding $Z^0_{e}$
<File:Example18h.png%7Cshorted> source redrawn

$$Z^0_{e} = 1\|(1+1\|1) = \frac{3}{5}$$

Calculating the impedance seen by the capacitor with the input open we
find

<File:Example18d.png%7Copen> source, finding $Z^{\infty}_{e}$
<File:Example18h.png%7Copen> source redrawn

$$Z^{\infty}_{e} = 2\|1+1 = \frac{5}{3}$$

Therefore using the EET, we find

$$Z_{in} = \frac{5}{3} \left(\frac{1+\frac{3}{5}s}{1+\frac{5}{3}s}\right)$$

Note that this problem was solved by calculating three simple driving
point impedances by inspection. 
