Find V0, ISC, & RTH: Thevenin Equivalent Homework Soln

[jdawg]

Homework Statement

Find the open circuit voltage V₀, short circuit current I_SC, and R_TH,

Homework Equations

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
I_SC=(VA)/30 I_SC=0.33A

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

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

 

[CWatters]

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?

 

[jdawg]

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

 

[jdawg]

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

 

[jdawg]

Should V₀=6.67V?

 

[gneill]

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.

 

[jdawg]

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 R_TH=20ohm. I was pretty confident about I_SC being correct, so I used this formula to find V_OC: R_TH=(V_OC)/(I_SC) and found V_OC=6.67

 

[gneill]

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).

 

[CWatters]

What I did was take out the independent source and did series and parallel combinations to get R_TH=20ohm. I was pretty confident about I_SC being correct, so I used this formula to find V_OC: R_TH=(V_OC)/(I_SC) and found V_OC=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.