Thevenin Equivalent

Homework Statement

Find the open circuit voltage V0, short circuit current ISC, and RTH,

The Attempt at a Solution

I found the equivalent resistance to be 20 ohms(just by combining resistors in parallel and series) and V0 to be 30V, which I think might be wrong.

I did a nodal analysis at VA :
(VA-30)/10 + (VA)/15 +(VA)/30 =0
VA=10V
ISC=(VA)/30 ISC=0.33A

Then I found RTH =30v/0.33A=90 ohms

Shouldn't my RTH = Req?
I'm not really sure what I did wrong.

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CWatters
Homework Helper
Gold Member
Inspecting the original circuit shows that the open circuit voltage Vo cannot be 30V due to the potential divider effect of the 10 and 15R.

When the output is open circuit what's the effect of the 30R?

Can you neglect the branch with the 30 ohm resistor? Could you then add the 10 and 15 ohm resistors in series?

Nevermind, adding the resistors in series didn't work I still got V0=30

Should V0=6.67V?

gneill
Mentor
jdawg, can you write out the steps of the procedure for finding the Thevenin equivalent of a circuit? It seems that you're trying things at random and hoping to hit upon a correct result, which is never a good approach.

You caught me! Honestly I don't really know what I'm doing. I've missed the past few lectures and I have to submit this problem in about an hour so I'm desperate to get it finished.
What I did was take out the independent source and did series and parallel combinations to get RTH=20ohm. I was pretty confident about ISC being correct, so I used this formula to find VOC: RTH=(VOC)/(ISC) and found VOC=6.67

gneill
Mentor
Your text should spell out the procedure.
In this case one method boils down to:

Thevenin Voltage:
1. Remove any load so the circuit is open.
2. Determine the voltage presented at the open terminals.

Thevenin Resistance:
1. Remove any load so the circuit is open.
2. Suppress all sources: Replace voltage sources with shorts, current sources with opens.
3. Determine the resistance of the circuit between the load connection points (the resistance from the load's point of view).

jdawg and CWatters
CWatters
Homework Helper
Gold Member
What I did was take out the independent source and did series and parallel combinations to get RTH=20ohm. I was pretty confident about ISC being correct, so I used this formula to find VOC: RTH=(VOC)/(ISC) and found VOC=6.67

Unfortunately your value for Is and Vo are both incorrect. Follow the procedure gneill posted in #8.

Regarding the open circuit voltage...

Can you neglect the branch with the 30 ohm resistor? Could you then add the 10 and 15 ohm resistors in series?

You can't totally neglect the branch with the 30R but the fact that one end is open circuit does allow you to simplify the circuit. What would you replace the 30R with?

Then simply adding the 10 and 15R isn't sufficient to work out Vo. Look up potential divider circuit.

jdawg
Oops... I completely messed up that assignment. Thanks for y'alls help, I'll have to come back to this problem after I watch the lectures.