# Circuit Theory/Example70

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${\displaystyle V_{s1}={\frac {\sqrt {2}}{6}}\cos(t+{\frac {\pi }{4}})}$
${\displaystyle V_{s2}=\cos(t+{\frac {\pi }{3}})}$
${\displaystyle \mathbb {V} _{s1}={\frac {1}{6}}(1+j)}$
${\displaystyle \mathbb {V} _{s2}={\frac {1}{2}}(1+j{\sqrt {3}})}$

The series components can be lumped together .. which simplifies the circuit a bit.

## Node Analysis

File:Example70mupad.png
mupad and matlab code for all the work below
${\displaystyle {\frac {\mathbb {V} _{s1}-\mathbb {V} _{a}}{5}}-\mathbb {V} _{a}-{\frac {\mathbb {V} _{a}+\mathbb {V} _{s2}}{j{\sqrt {3}}}}=0}$
${\displaystyle \mathbb {V} _{a}=-0.42063+j0.065966}$

## Mesh Analysis

${\displaystyle i_{1}-i_{2}-V_{s1}+5*i_{1}=0}$
${\displaystyle i_{2}j{\sqrt {3}}-V_{s2}+i_{2}-i_{1}=0}$
${\displaystyle i_{3}=i_{1}-i_{2}}$

Solving

${\displaystyle i_{3}=-0.42063+j.065966}$

Which is the same as the voltage through the 1 ohm resistor.

## Thevenin voltage

Make ground the negative side of ${\displaystyle V_{s2}}$, then:

${\displaystyle V_{th}=V_{A}-V_{B}}$
${\displaystyle V_{th}=i*j{\sqrt {3}}-V_{s2}}$
${\displaystyle V_{th}={\frac {V_{s1}+V_{s2}}{5+j{\sqrt {3}}}}*j{\sqrt {3}}-V_{s2}}$

Solving

${\displaystyle V_{th}=-0.747977-j0.5492}$

## Norton Current

${\displaystyle i_{n}=i_{1}-i_{2}={\frac {V_{s1}}{5}}-{\frac {V_{s2}}{j{\sqrt {3}}}}}$
${\displaystyle i_{n}=-0.4667+j0.3220}$

## Thevenin/Norton Impedance

short voltage sources, open current sources, remove load and find impedance where the load was attached

${\displaystyle Z_{th}={\frac {1}{{\frac {1}{5}}+{\frac {1}{j{\sqrt {3}}}}}}=0.537+j1.5465}$

check

${\displaystyle Z_{th}={\frac {V_{th}}{I_{n}}}=0.537+j1.5465}$

yes! they match

## Evaluate Thevenin Equivalent Circuit

Going to find current through the resistor and compare with mesh current

${\displaystyle i={\frac {V_{th}}{Z_{th}+1}}=-0.42063+j0.06599}$

yes! they match

## Find Load value for maximum power transfer

${\displaystyle Z_{L}=z_{th}^{*}=0.537-j1.5465}$

## Find average power transfer with Load that maximizes

${\displaystyle Z_{th}+Z_{L}=0.537*2}$
${\displaystyle P_{avg}={\frac {Re(V_{th})^{2}}{Z_{th}+Z_{L}}}={\frac {\sqrt {0.747977^{2}+0.5492^{2}}}{2*0.537}}=0.8037watts}$

## Simulation

The circuit was simulated. A way was found to enter equations into the voltage supply parameters which improves accuracy:

To compare with the results above, need to translate the current and voltage through the resistor into the time domain.

### Period

Period looks right about 6 seconds ... should be:

${\displaystyle T={\frac {1}{f}}={\frac {2*\pi }{w}}=2*\pi =6.2832}$

### Current

Current through 1 ohm resistor, once moved into the time domain (from the above numbers) is:

${\displaystyle i(t)=0.4258*cos(t+171^{\circ })}$

From the mesh analysis, the current's through both sources were computed:

${\displaystyle i_{s1}=0.1192*cos(t+9.72^{\circ })}$

The magnitudes are accurate, they are almost π out of phase which is can be seen on the simulation.

### Voltage

The voltage is the same as the current through a 1 ohm resistor:

${\displaystyle v(t)=0.4258*cos(t+171^{\circ })}$

The voltage of the first (left) source is:

${\displaystyle V_{s1}={\frac {\sqrt {2}}{6}}cos(t+{\frac {pi}{4}})=0.2357*cos(t+45^{\circ })}$

The magnitudes match. The voltage through the source peaks before the current because the first source sees the inductor.

The voltage through the resistor should peak about 171 - 45 = 126° before the source ... which it appears to do.