How Do You Calculate the Internal Resistance of a Thevenin Voltage Source?

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Homework Help Overview

The problem involves calculating the internal resistance of a Thevenin voltage source, given its open circuit voltage and the voltage across a load resistor when connected. The context is rooted in circuit analysis and Thevenin's theorem.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the significance of the internal resistance in a real voltage source and how it affects the total current in the circuit. There are attempts to calculate the internal resistance based on voltage drop and current, as well as suggestions to use the voltage divider equation.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. Some guidance has been provided regarding the need to consider internal resistance in calculations, but there is no explicit consensus on the correct method or final answer yet.

Contextual Notes

Participants are navigating the implications of using a real voltage source versus an ideal one, and there is a focus on the calculations involving voltage drops and current through the load and internal resistance.

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Homework Statement


A real voltage source is of the Thevenin Type. The open circuit voltage of the source is 12 volts. If a 100 ohm load is added to the circuit, the voltage at the same terminals is 11.2 volts. What is the internal resistance of the source.


The Attempt at a Solution




I found the total current in the circuit to be 12/100 = 120mA

If there is .8 volts missing, the internal resistance should be: .8/120mA?
 
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When they say "real voltage source", that means that you should not ignore the built-in output resistance of the source. So the total current will not be 12V/100 Ohms... you need to include the Rout in that equation as well...
 
I need to find the internal resistance before I can include it in the equation right? After doing .8/120mA=6.66 Ohms. The new total current could be 12V/(100 + 6.66)
 
I'm not sure if that's right, but try it this way -- use the voltage divider equation, since you know the voltage source is 12V, and the divided down voltage across Rout and 100 Ohms is 11.2V...
 

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