# Circuit Theory/Example70

    $V_{s1}={\frac {\sqrt {2}}{6}}\cos(t+{\frac {\pi }{4}})$
    $V_{s2}=\cos(t+{\frac {\pi }{3}})$
    $\mathbb {V} _{s1}={\frac {1}{6}}(1+j)$
    $\mathbb {V} _{s2}={\frac {1}{2}}(1+j{\sqrt {3}})$
The series components can be lumped together .. which simplifies the circuit a bit.

## Node Analysis

    ${\frac {\mathbb {V} _{s1}-\mathbb {V} _{a}}{5}}-\mathbb {V} _{a}-{\frac {\mathbb {V} _{a}+\mathbb {V} _{s2}}{j{\sqrt {3}}}}=0$
    $\mathbb {V} _{a}=-0.42063+j0.065966$

## Mesh Analysis

    $i_{1}-i_{2}-V_{s1}+5*i_{1}=0$
    $i_{2}j{\sqrt {3}}-V_{s2}+i_{2}-i_{1}=0$
    $i_{3}=i_{1}-i_{2}$
Solving
    $i_{3}=-0.42063+j.065966$
Which is the same as the voltage through the 1 ohm resistor.

## Thevenin voltage

Make ground the negative side of $V_{s2}$, then:
    $V_{th}=V_{A}-V_{B}$
    $V_{th}=i*j{\sqrt {3}}-V_{s2}$
    $V_{th}={\frac {V_{s1}+V_{s2}}{5+j{\sqrt {3}}}}*j{\sqrt {3}}-V_{s2}$
Solving
    $V_{th}=-0.747977-j0.5492$

## Norton Current

* short out the load
* split into two circuits so can use superposition
* half the circuit is shorted out by shorting the voltage sources
    $i_{n}=i_{1}-i_{2}={\frac {V_{s1}}{5}}-{\frac {V_{s2}}{j{\sqrt {3}}}}$
    $i_{n}=-0.4667+j0.3220$

## Thevenin/Norton Impedance

short voltage sources, open current sources, remove load and find impedance where the load was attached
    $Z_{th}={\frac {1}{{\frac {1}{5}}+{\frac {1}{j{\sqrt {3}}}}}}=0.537+j1.5465$
check
    $Z_{th}={\frac {V_{th}}{I_{n}}}=0.537+j1.5465$
yes! they match

## Evaluate Thevenin Equivalent Circuit

Going to find current through the resistor and compare with mesh current
    $i={\frac {V_{th}}{Z_{th}+1}}=-0.42063+j0.06599$
yes! they match

## Find Load value for maximum power transfer

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

## Find average power transfer with Load that maximizes

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

## Simulation

The circuit was simulated. A way was found to enter equations into the voltage supply parameters which improves accuracy:
* type sqrt(3) instead of a numeric approximation
* added pi/2 to convert from cos to sin sources
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:
    $T={\frac {1}{f}}={\frac {2*\pi }{w}}=2*\pi =6.2832$

### Current

Current through 1 ohm resistor, once moved into the time domain (from the above numbers) is:
    $i(t)=0.4258*cos(t+171^{\circ })$
From the mesh analysis, the current's through both sources were computed:
    $i_{s1}=0.1192*cos(t+9.72^{\circ })$
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:
    $v(t)=0.4258*cos(t+171^{\circ })$
The voltage of the first (left) source is:
    $V_{s1}={\frac {\sqrt {2}}{6}}cos(t+{\frac {pi}{4}})=0.2357*cos(t+45^{\circ })$
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.