Circuit Theory/2Source Excitement/Thevenin and Norton

Thevenin Voltage

removing the load R₃
redrawing up and down
opening the current source (using superposition)
shorting the voltage source (using superposition)

After opening the current source, have a series circuit. The overall current is going to be :

\mathbb {I} _{t}={\frac {\mathbb {V} _{s}}{X_{L1}+X_{L2}+R_{1}+X_{C2}+X_{C1}}}

V_AB due to the voltage source is going to be the drop across C2 and L2 which is (matlab code):

\mathbb {V} _{AB}=\mathbb {I} _{t}*(-4j)=0.0204-0.0344*i

V_AB due to the current source is a little more difficult. Put ground in between C₂ and L₂. Use node analysis at the junction of C₁, R₂ and R₁:

\mathbb {I} _{S}-{\frac {\mathbb {V} _{N}}{R_{1}+X_{L1}+X_{L2}}}-{\frac {\mathbb {V} _{N}}{X_{C1}+X_{C2}}}=0

Solving for V_N:

\mathbb {V} _{N}=\mathbb {I} _{S}{\frac {1}{{\frac {1}{R_{1}+X_{L1}+X_{L2}}}+{\frac {1}{X_{C1}+X_{C2}}}}}

Now can find the current in the two parallel branches. Then V_AB is going to be the drop across C₂ minus the drop across L₂ (because the currents are going in opposite directions):

V_{AB}={\frac {\mathbb {V} _{N}}{X_{c1}+X_{C2}}}*X_{C2}-{\frac {\mathbb {V} _{N}}{R_{1}+X_{L1}+X_{L2}}}*X_{L2}=118.0-118.0*i

So the Thevenin voltage is going to be the addition of the two V_AB's (matlab code):

V_{th}=118.0-118.0*i

Norton Current

Load resistor R₃ shorted, norton current identified
Current source opened (calculate using superposition)
Voltage source shorted (calculate using superposition)

Current in the short due to the voltage source is the total current:

\mathbb {I} _{N}={\frac {\mathbb {V} _{1}}{R_{1}+X_{C1}+X_{L1}}}

Current in the short due to the current source is a little more difficult. From the first drawing:

\mathbb {I} _{N}=\mathbb {I} _{C}1-\mathbb {I} _{C}2

Both I_C1 and I_C2 can be found through current dividers, so I_N is:

\mathbb {I} _{N}=\mathbb {I} _{s}*({\frac {R1+X_{L1}}{R1+X_{L1}+X_{C1}}}-{\frac {X_{L2}}{X_{L2}+X_{C2}}})

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