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

AI Thread Summary
To calculate the internal resistance of a Thevenin voltage source, the open circuit voltage is given as 12 volts, and the voltage across a 100-ohm load is 11.2 volts. The missing voltage of 0.8 volts indicates that internal resistance affects the output. The initial calculation for current is 120 mA, leading to an internal resistance of approximately 6.66 ohms using the formula 0.8 volts divided by 120 mA. The voltage divider equation can also be applied to confirm this, utilizing the known voltages and resistances in the circuit.
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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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