Circuit Theory/Example70

 The latest reviewed version was checked on 29 March 2013. There are template/file changes awaiting review.

${\displaystyle V_{s1}={\frac {\sqrt {2}}{6}}\cos(t+{\frac {\pi }{4}})}$

${\displaystyle V_{s2}=\cos(t+{\frac {\pi }{3}})}$

${\displaystyle \mathbb {V} _{s1}={\frac {1}{6}}(1+j)}$

${\displaystyle \mathbb {V} _{s2}={\frac {1}{2}}(1+j{\sqrt {3}})}$

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

Node AnalysisEdit

File:Example70mupad.png

mupad and matlab code for all the work below

${\displaystyle {\frac {\mathbb {V} _{s1}-\mathbb {V} _{a}}{5}}-\mathbb {V} _{a}-{\frac {\mathbb {V} _{a}+\mathbb {V} _{s2}}{j{\sqrt {3}}}}=0}$

${\displaystyle \mathbb {V} _{a}=-0.42063+j0.065966}$

Mesh AnalysisEdit

${\displaystyle i_{1}-i_{2}-V_{s1}+5*i_{1}=0}$

${\displaystyle i_{2}j{\sqrt {3}}-V_{s2}+i_{2}-i_{1}=0}$

${\displaystyle i_{3}=i_{1}-i_{2}}$

Solving

${\displaystyle i_{3}=-0.42063+j.065966}$

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

Thevenin voltageEdit

Make ground the negative side of ${\displaystyle V_{s2}}$, then:

${\displaystyle V_{th}=V_{A}-V_{B}}$

${\displaystyle V_{th}=i*j{\sqrt {3}}-V_{s2}}$

${\displaystyle V_{th}={\frac {V_{s1}+V_{s2}}{5+j{\sqrt {3}}}}*j{\sqrt {3}}-V_{s2}}$

Solving

${\displaystyle V_{th}=-0.747977-j0.5492}$

Norton CurrentEdit

• 

  short out the load

 
• 

  split into two circuits so can use superposition

 
• 

  half the circuit is shorted out by shorting the voltage sources

${\displaystyle i_{n}=i_{1}-i_{2}={\frac {V_{s1}}{5}}-{\frac {V_{s2}}{j{\sqrt {3}}}}}$

${\displaystyle i_{n}=-0.4667+j0.3220}$

Thevenin/Norton ImpedanceEdit

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

${\displaystyle Z_{th}={\frac {1}{{\frac {1}{5}}+{\frac {1}{j{\sqrt {3}}}}}}=0.537+j1.5465}$

check

${\displaystyle Z_{th}={\frac {V_{th}}{I_{n}}}=0.537+j1.5465}$

yes! they match

Evaluate Thevenin Equivalent CircuitEdit

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

${\displaystyle i={\frac {V_{th}}{Z_{th}+1}}=-0.42063+j0.06599}$

yes! they match

Find Load value for maximum power transfer

Find average power transfer with Load that maximizesEdit

${\displaystyle Z_{th}+Z_{L}=0.537*2}$

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

SimulationEdit

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

• 

  type sqrt(3) instead of a numeric approximation

 
• 

  added 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.

PeriodEdit

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

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

CurrentEdit

Current through 1 ohm resistor, once moved into the time domain (from the above numbers) is:

${\displaystyle i(t)=0.4258*cos(t+171^{\circ })}$

From the mesh analysis, the current's through both sources were computed:

${\displaystyle i_{s1}=0.1192*cos(t+9.72^{\circ })}$

The magnitudes are accurate, they are almost π out of phase which is can be seen on the simulation.

VoltageEdit

The voltage is the same as the current through a 1 ohm resistor:

${\displaystyle v(t)=0.4258*cos(t+171^{\circ })}$

The voltage of the first (left) source is:

${\displaystyle V_{s1}={\frac {\sqrt {2}}{6}}cos(t+{\frac {pi}{4}})=0.2357*cos(t+45^{\circ })}$

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.

This article uses material from the Wikipedia article  Metasyntactic variable, which is released under the  Creative Commons Attribution-ShareAlike 3.0 Unported License.