Circuit Theory/Example70
The series components can be lumped together .. which simplifies the circuit a bit.
Node Analysis
editMesh Analysis
editSolving
Which is the same as the voltage through the 1 ohm resistor.
Thevenin voltage
editMake ground the negative side of , then:
Solving
Norton Current
edit
short out the load

split into two circuits so can use superposition

half the circuit is shorted out by shorting the voltage sources
Thevenin/Norton Impedance
editshort voltage sources, open current sources, remove load and find impedance where the load was attached
check
yes! they match
Evaluate Thevenin Equivalent Circuit
editGoing to find current through the resistor and compare with mesh current
yes! they match
Find Load value for maximum power transfer
editFind average power transfer with Load that maximizes
editSimulation
editThe 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
editPeriod looks right about 6 seconds ... should be:
Current
editCurrent through 1 ohm resistor, once moved into the time domain (from the above numbers) is:
From the mesh analysis, the current's through both sources were computed:
The magnitudes are accurate, they are almost π out of phase which is can be seen on the simulation.
Voltage
editThe voltage is the same as the current through a 1 ohm resistor:
The voltage of the first (left) source is:
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.