${\displaystyle V_{s1}={\frac {\sqrt {2}}{6}}\cos(t+{\frac {\pi }{4}})}$

${\displaystyle V_{s2}=\cos(t+{\frac {\pi }{3}})}$

${\displaystyle \mathbb {V} _{s1}={\frac {1}{6}}(1+j)}$

${\displaystyle \mathbb {V} _{s2}={\frac {1}{2}}(1+j{\sqrt {3}})}$

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

Node Analysis

${\displaystyle {\frac {\mathbb {V} _{s1}-\mathbb {V} _{a}}{5}}-\mathbb {V} _{a}-{\frac {\mathbb {V} _{a}+\mathbb {V} _{s2}}{j{\sqrt {3}}}}=0}$

${\displaystyle \mathbb {V} _{a}=-0.42063+j0.065966}$

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

• 
  short out the load
• 
  split into two circuits so can use superposition
• 
  half the circuit is shorted out by shorting the voltage sources

${\displaystyle i_{n}=i_{1}-i_{2}={\frac {V_{s1}}{5}}-{\frac {V_{s2}}{j{\sqrt {3}}}}}$

${\displaystyle i_{n}=-0.4667+j0.3220}$

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:

• 
  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.