# 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

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

## 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

$i_n = i_1 - i_2 = \frac{V_{s1}}{5} - \frac{V_{s2}}{j\sqrt{3}}$
$i_n = -0.4667 + j0.3220$

## 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

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

## 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:

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

### 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.