Circuit Theory/Example70

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original circuit

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[edit]

Node analysis

File:Example70mupad.png
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[edit]

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[edit]

Thevenin voltage requires opening the load

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[edit]

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[edit]

Example70impedance.png

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[edit]

Example70theveninEquivalent.png

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[edit]

Z_{L} = z_{th}^* = 0.537 - j1.5465

Find average power transfer with Load that maximizes[edit]

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[edit]

Example70simulation2.png

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[edit]

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

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

Current[edit]

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[edit]

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.