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

In summary, Homework Equations: -The equivalent resistance is 20 ohms.-The open circuit voltage is 30V.-The short circuit current is 0.33A.-The Rth is 90 ohms.
  • #1
jdawg
367
2

Homework Statement


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

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
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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  • #2
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?
 
  • #3
Can you neglect the branch with the 30 ohm resistor? Could you then add the 10 and 15 ohm resistors in series?
 
  • #4
Nevermind, adding the resistors in series didn't work I still got V0=30
 
  • #5
Should V0=6.67V?
 
  • #6
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.
 
  • #7
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
 
  • #8
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).
 
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  • #9
jdawg said:
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...

jdawg said:
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.
 
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  • #10
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.
 

1. What is the Thevenin equivalent circuit?

The Thevenin equivalent circuit is a simplified representation of a complex circuit that contains a voltage source and a resistor. It can be used to analyze the behavior of a circuit without having to consider all of its individual components.

2. How do you find the Thevenin equivalent circuit?

To find the Thevenin equivalent circuit, you need to follow these steps:

  1. Disconnect all the sources in the circuit.
  2. Calculate the open circuit voltage (V0) by measuring the voltage across the open terminals.
  3. Calculate the equivalent resistance (RTH) by removing all the sources and finding the resistance between the open terminals.
  4. Reconnect all the sources and calculate the short circuit current (ISC) by finding the current that flows through the shorted terminals.
  5. The Thevenin equivalent circuit is then V0 in series with RTH.

3. What is the purpose of finding the Thevenin equivalent circuit?

The purpose of finding the Thevenin equivalent circuit is to simplify a complex circuit into a single voltage source and a resistor, making it easier to analyze and understand its behavior. This also allows for easier calculations and circuit design.

4. How is the Thevenin equivalent circuit used in real-world applications?

The Thevenin equivalent circuit is commonly used in circuit analysis and design, particularly in the field of electronics. It is also used in electrical power systems to model the behavior of large networks and in telecommunications to analyze signal transmission.

5. Are there any limitations to using the Thevenin equivalent circuit?

Yes, there are some limitations to using the Thevenin equivalent circuit. It only works for linear circuits, meaning that the relationship between current and voltage must be linear. It also assumes that the circuit is in a steady state and does not take into account any transient effects. Additionally, it is only accurate within a certain range of values and may not accurately represent the behavior of the circuit outside of that range.

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