# Circuit Theory/Example70

!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

!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

!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

!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

<File:Example70norton1.png%7Cshort> out the load
<File:Example70norton2.png%7Csplit> into two circuits so can use
superposition <File:Example70norton3.png%7Chalf> the circuit is shorted
out by shorting the voltage sources

$$i_n = i_1 - i_2 = \frac{V_{s1}}{5} - \frac{V_{s2}}{j\sqrt{3}}$$

$$i_n = -0.4667 + j0.3220$$

## Thevenin/Norton Impedance

![](Example70impedance.png "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

![](Example70theveninEquivalent.png "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

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

## 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

![](Example70simulation2.png){width="800"}

The circuit was
simulated.
A way was found to enter equations into the voltage supply parameters
which improves accuracy:

<File:Example70-vs11.png%7Ctype> sqrt(3) instead of a numeric
approximation <File:Example70-vs2.png%7Cadded> 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

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.

