# Circuit Theory/Example70

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

## Node AnalysisEdit

## Mesh AnalysisEdit

Solving

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

## Thevenin voltageEdit

Make ground the negative side of , then:

Solving

## Norton CurrentEdit

## Thevenin/Norton ImpedanceEdit

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

check

yes! they match

## Evaluate Thevenin Equivalent CircuitEdit

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

yes! they match

## Find Load value for maximum power transferEdit

## Find average power transfer with Load that maximizesEdit

## SimulationEdit

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.

### PeriodEdit

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

### CurrentEdit

Current 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.

### VoltageEdit

The 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.