gneill said:Check your Thevenin resistance. When you calculated 60 || 30 you seem to have ended up with 15, although you did it correctly elsewhere.
Check your ##V_{th1}## calculation. You have a voltage divider and shouldn't end up with the same voltage as the source driving it on its output.
For ##V_{th3}##, when you suppress the 1.2 A source you remove it, not short it. And suppressing the 18 V supply shorts it. Draw the remaining circuitry. You'll see that the 0.25 A supply does make a contribution to Vth (it's still in parallel with a 20 Ohm resistor).
Awesome. Thanks so muchgneill said:Right. When you suppress sources, currents are replaced by open circuits, voltages by short circuits.
A Thevenin Equivalent Circuit is a simplified representation of a complex electrical circuit. It consists of a single voltage source and a single resistor, which can be used to model the behavior of a more complex circuit.
To calculate a Thevenin Equivalent Circuit, you need to find the Thevenin voltage and the Thevenin resistance. Thevenin voltage is the open-circuit voltage at the output terminals of the circuit, while Thevenin resistance is the resistance seen at the output terminals when all voltage sources are removed and replaced with short circuits.
Thevenin Equivalent Circuits are used to simplify complex circuits and make them easier to analyze. They can also be used to find the maximum power transfer from a circuit to a load.
No, a Thevenin Equivalent Circuit can only be used to model linear circuits. It cannot accurately model circuits with non-linear elements such as diodes or transistors.
The output voltage of a Thevenin Equivalent Circuit will remain the same regardless of the load connected to it. However, the current flowing through the load will change, and the power delivered to the load will be maximum when the load resistance is equal to the Thevenin resistance.