Circuit Theory/2Source Excitement/Thevenin and Norton
Thevenin Voltage
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removing the load R_{3}

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 :
V_{AB} due to the voltage source is going to be the drop across C2 and L2 which is (matlab code):
V_{AB} due to the current source is a little more difficult. Put ground in between C_{2} and L_{2}. Use node analysis at the junction of C_{1}, R_{2} and R_{1}:
Solving for V_{N}:
Now can find the current in the two parallel branches. Then V_{AB} is going to be the drop across C_{2} minus the drop across L_{2} (because the currents are going in opposite directions):
So the Thevenin voltage is going to be the addition of the two V_{AB}'s (matlab code):
Norton Current
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Load resistor R_{3} 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:
Current in the short due to the current source is a little more difficult. From the first drawing:
Both I_{C1} and I_{C2} can be found through current dividers, so I_{N} is: