# 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.

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

where

${\displaystyle Z\ }$ is the impedance chosen as the extra element to be removed
${\displaystyle Z_{in}^{\infty }}$ is the input impedance with Z removed (or made infinite)
${\displaystyle Z_{e}^{0}}$ is the impedance seen by the extra element Z with the source shorted (or made zero)
${\displaystyle Z_{e}^{\infty }}$ 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 ${\displaystyle Z_{in}}$  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

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

Removing this capacitor from the circuit we find

${\displaystyle Z_{in}^{\infty }=2\|1+1={\frac {5}{3}}}$

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

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

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

${\displaystyle Z_{e}^{\infty }=2\|1+1={\frac {5}{3}}}$

Therefore using the EET, we find

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

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