# Thevenin Equivalent Circuits

1. Dec 10, 2015

### DiamondV

1. The problem statement, all variables and given/known data
http://puu.sh/lQPvh/d0efa216ff.png [Broken]

2. Relevant equations

3. The attempt at a solution
http://puu.sh/lQPAR/2456db4783.jpg [Broken]
http://puu.sh/lQPCJ/f142c25f63.jpg [Broken]
http://puu.sh/lQPER/5eb4af107e.jpg [Broken]

Is this correct? I used principle of superposition to calculate the Thevenin voltage

Last edited by a moderator: May 7, 2017
2. Dec 10, 2015

### Staff: Mentor

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).

3. Dec 10, 2015

### DiamondV

Yeah. My thevenin resistance should be 40 and not 35.

Ah yes. I see my mistake there for Vth1. I added the resistances even though thevenin voltage was only across one of the resistors and not the sum of them so I should use voltage division for it since voltage is divided in series. Forgot about that.

So is this correct for Vth3 -
http://puu.sh/lR1gV/9cd3848fff.jpg [Broken]

That black oval, removing that completely?(and shorting the voltage) So the 0.25A would still be connected in parallel to 20ohms.

Last edited by a moderator: May 7, 2017
4. Dec 10, 2015

### Staff: Mentor

Right. When you suppress sources, currents are replaced by open circuits, voltages by short circuits.

5. Dec 10, 2015

### DiamondV

Awesome. Thanks so much