# Thevenin Equivalent Circuits

• Engineering

## Homework Statement

http://puu.sh/lQPvh/d0efa216ff.png [Broken]

## 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[/B]

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gneill
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).

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).
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.

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

Right. When you suppress sources, currents are replaced by open circuits, voltages by short circuits.
Awesome. Thanks so much