Thevenin equivalent circuit

  • Engineering
  • Thread starter sisyphus0321
  • Start date
  • #1

Homework Statement


A car battery has an open-circuit terminal voltage of 12.6 volts. The voltage drops to 10.8 volts when the battery supplies 240 amps to the car's starter motor. What is the Thevenin eq for this battery?


Homework Equations


I think the root of my problem is the setup of the equation. I just don't understand how to setup the circuit prperly. Am I looking for it to have a t>0 or t<0 setup? There just doesn't seem to be one solution for this problem. Should I be looking for an answer in the voltage drop?

The Attempt at a Solution


Rth=Vth/Isc=10.8/240=.045Ohms
but
Rth=Vth/Isc=12.6/240=.0525Ohms
 
Last edited:

Answers and Replies

  • #2
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The way I understand it, thevenin circuit is the equivalent open circuit voltage and the resistance/impedence seen looking in from the terminals.

In this problem to work out your thevenin resistance, you are given the current that flows and the voltage across the terminals when the circuit is attached, thus can work out the thevenin resistance from V/I , ie 10.8V/240A = 0.045 ohms.

So your thevenin equivalent circuit will be the open circuit voltage and this thevenin resistance, thus the circuit will be a 12.6V voltage source in series with a 0.045ohm resistor.

Correct me if I'm wrong anyone
 
  • #3
MATLABdude
Science Advisor
1,655
4
The way I understand it, thevenin circuit is the equivalent open circuit voltage and the resistance/impedence seen looking in from the terminals.

In this problem to work out your thevenin resistance, you are given the current that flows and the voltage across the terminals when the circuit is attached, thus can work out the thevenin resistance from V/I , ie 10.8V/240A = 0.045 ohms.

So your thevenin equivalent circuit will be the open circuit voltage and this thevenin resistance, thus the circuit will be a 12.6V voltage source in series with a 0.045ohm resistor.

Correct me if I'm wrong anyone
This is incorrect. Note the first diagram on the Wikipedia page:
http://en.wikipedia.org/wiki/Thévenin's_theorem

That's the Thevenin equivalent circuit that we all know and love. There are two ways of determining a Thevenin Equivalent. You can be given the inner workings of a black box, and you can do all your circuit analysis and find an equivalent open-circuit voltage (Voc), short circuit current (Isc) and/or Thevenin Resistance. Or you receive a black box and figure out what the equivalent Voc and Rth are empirically. This example is firmly the latter.

When no load is attached, you see the open circuit voltage across terminals A and B, as there is no drop across Rth. When a load (say, a resistor) is attached, it goes between A terminal A and B, and you form a voltage divider with an ideal voltage supply (Voc), the internal resistor, and your load resistor. This is why the voltage at the output terminal drops when you hook in a load resistor that draws current.

Thevenin Equivalents are sometimes a handy way to reduce big blocks of a circuit. But its true power is allowing you to treat something as a black box and, just by hooking up a load or two to it, to determine the two or three important parameters that are needed to model a black box. Now, getting back to your question, having V_oc and the voltage with load connected, how would you determine R_th? (Yes, it's possible, and, if you look at the previous paragraph, pretty trivial).
 
  • #4
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sorry, thanks for clearing that up matlab.
 
  • #5
MATLABdude
Science Advisor
1,655
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sorry, thanks for clearing that up matlab.
No prob! Mr. Sisyphus, done rolling this rock up the hill?
 
  • #6
Nearing the Top MATLABdude...nearing the top. Thank you very much for the nudge back to the obvious. Sometimes it's very easy to miss the forest for the trees so-to-speak.
 

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