Circuit Theory/2Source Excitement/Thevenin and Norton

Thevenin Voltage

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

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

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

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

VAB 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:

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

Solving for VN:

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

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

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

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

${\displaystyle V_{th}=118.0-118.0*i}$

Norton Current

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

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

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

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

Both IC1 and IC2 can be found through current dividers, so IN is:

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