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 :

${𝕀}_t=\frac{{𝕍}_s}{X_{L1}+X_{L2}+R_1+X_{C2}+X_{C1}}$

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

${𝕍}_{AB}={𝕀}_t*(-4j)=0.0204-0.0344*i$

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:

${𝕀}_S-\frac{{𝕍}_N}{R_1+X_{L1}+X_{L2}}-\frac{{𝕍}_N}{X_{C1}+X_{C2}}=0$

Solving for V_N:

${𝕍}_N={𝕀}_S\frac{1}{\frac{1}{R_1+X_{L1}+X_{L2}}+\frac{1}{X_{C1}+X_{C2}}}$

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):

$V_{AB}=\frac{{𝕍}_N}{X_{c1}+X_{C2}}*X_{C2}-\frac{{𝕍}_N}{R_1+X_{L1}+X_{L2}}*X_{L2}=118.0-118.0*i$

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:

${𝕀}_N=\frac{{𝕍}_1}{R_1+X_{C1}+X_{L1}}$

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

${𝕀}_N={𝕀}_{C}1-{𝕀}_{C}2$

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

${𝕀}_N={𝕀}_s*(\frac{R1+X_{L1}}{R1+X_{L1}+X_{C1}}-\frac{X_{L2}}{X_{L2}+X_{C2}})$