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

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

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.