# Help Understanding Thevenin Question

Discussion in 'Homework Help' started by James4553, Jun 11, 2008.

1. ### James4553 Thread Starter Active Member

Jun 7, 2008
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Hello everyone

I know how to solve Thevenim and Norton problems, but there is a problem on a worksheet that I just don't understand.

Q) A source has an open circuit source voltage of 12V and when loaded with a 100 OHM resistor, the voltage drops to 8V. Sketch the Thevenim Equivalent Circuit, giving the values of Vthev and Rthev.

From the problem, it seems that since there is only one resistor and one voltage source, the circuit is already as simplified as it can be.
Is there something I'm missing here?

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Since the addition of the 100 ohm load resistor causes the open circuit voltage to drop, then there must be a resistance in series with the 12V source. The challenge is to determine the value of the series resistance from the effect that the 100 ohm resistive load has on the circuit. Thevenin's provides you a means of determining that unknown resistance.

hgmjr

3. ### James4553 Thread Starter Active Member

Jun 7, 2008
35
0
Would a request for someone to sketch a diagram of this and upload it to this thread be asking too much?

4. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Most forum members will drop you hints, reading suggestions, and generally guide you towards your goal. The work is yours to do. They will help you succeed, but they want to see your effort in the process. Most forum members, if not all, discourage plagerism. They wouldn't want to assist you in becoming a plagerist.

5. ### veritas Active Member

Feb 7, 2008
167
0
They give you the thevenin voltage, Vth, as 12V.
Your objective is to find Rth, the thevenin resistance.
The 100-ohm resistor is not Rth, but a load on the thevenin equivalent circuit.

6. ### RiJoRI Well-Known Member

Aug 15, 2007
536
26
HINT: Voltage sources have internal resistances, which is why you need a load to test a battery properly.

--Rich

7. ### James4553 Thread Starter Active Member

Jun 7, 2008
35
0
Thank for all the hints here.
I've made a quick sketch of what I think the solution may be.
Please take a look and let me know.

Many thanks

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8. ### hgmjr Moderator

Jan 28, 2005
9,030
214
From your sketch, there are a few things about the concept of Thevenin equivalents that you are not clear on.

From your starting sketch, it is R1 that you are trying to determine. R1 should is therefore your unknown. R2 is the known value since you are told in the problem statement that the voltage across the 100 ohm load resistor is 8 volts.

HINT: if you have a known resistance and you know the voltage across the resistance, Ohm's Law permits you to compute the current flowing in the resistance.

Rather than devulge additional clues, let's see if you can take it the rest of the way from that hint.

hgmjr

9. ### James4553 Thread Starter Active Member

Jun 7, 2008
35
0
Please let me know if this is correct.
Thanks again.

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10. ### hgmjr Moderator

Jan 28, 2005
9,030
214
You have correctly calculated Rth. You still need to think about Vth. It is not 4V as you indicated.

Review your textbook material on Vth to see if you can see where you have gone wrong.

hgmjr

11. ### James4553 Thread Starter Active Member

Jun 7, 2008
35
0
Would it be 8V?

12. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Nope. Sorry.

Remember that the definition of the thevenin voltage is that voltage that is output from the circuit when there is no load. The output is 8 volts when it is loaded.

hgmjr

13. ### cumesoftware Senior Member

Apr 27, 2007
1,330
10
A clue:
You know that the voltage drop on the series resistor (Thevenin resistance) must be 4V.

I think you can calculate Rthev now. I already figured it out using my head.

14. ### James4553 Thread Starter Active Member

Jun 7, 2008
35
0
I have reviewed the information from the link provided, and at the summary it says:

Steps to follow for Thevenin's Theorem:
• (1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be.
If the original source voltage is 12V and the series resistor is 4V, are you telling me that the load resistor (and therefore Vthev) isn't 8V ?

15. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The Thevenin voltage is not 8V.

hgmjr

16. ### James4553 Thread Starter Active Member

Jun 7, 2008
35
0
I think I know where I went wrong.

Is the Vthev = 12V because the series resistor is not connected on one end and doesn't carry current, and therefore doesn't have a voltage drop?

17. ### hgmjr Moderator

Jan 28, 2005
9,030
214
You have figured it out on your own. Well done.

hgmjr

18. ### cumesoftware Senior Member

Apr 27, 2007
1,330
10
No, I'm telling you just the opposite. The drop across the load resistor (not Vthev) is 8V. What did you learned about how voltage divides across series resistors? You know that the Thevenin voltage, or the voltage generated the the ideal voltage source, is 12V. You know that the voltage drop across the load resistor is 8V. Thus the voltage drop across Rthev is 4V right?

Keep in mind that you are dealing with an hypothetic real voltage source, which is represented by an ideal voltage source and a series resistor. If there is no load, there will be no current across the series resistor Rthev and therefore there is no voltage drop. Therefore, in a no load situation you are measuring the Thevenin's voltage, with is the voltage of the ideal voltage source.

I'm practically doing your homework. You really have to learn this.