# Circuit Theory/2Source Excitement/Thevenin and Norton

## Thevenin VoltageEdit

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 CurrentEdit

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: