Thevenin equivalent circuit

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  • Thread starter Solidsam
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  • #1
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Homework Statement



A variable resistor (RL) is connected to an unknown circuit. It is found that the
maximum power that can be dissipated in the resistor (RL) is 10W, and that this
occurs when the value of the resistor (RL) is 25ohms.
Derive the Thevenin equivalent circuit for the unknown circuit, and hence find the
current which flows when RL = 5ohms.

Homework Equations



Maximum power dissaption occurs when RL=RS which was 25ohm

Which means P=V^2/R= V=sqrootof P*R = sqrootof 10W*(25+25)ohms=22.4V

Equivalent circuit is a 22.4 V power supply and a 25 ohm resistor

When RL is 5 I=V/(Rs+RL)=22.4/(25+5)=0.74Amps


are my calculations correct?
 

Answers and Replies

  • #2
lewando
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Which means P=V^2/R= V=sqrootof P*R = sqrootof 10W*(25+25)ohms=22.4V
P does equal V2/R, but the 10W of power is being dissipated by RL, not RL + RS.
 
  • #3
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P does equal V2/R, but the 10W of power is being dissipated by RL, not RL + RS.
so do i use P=(V^2/R) or P=I^2*R


if i use P=(V^2/R) i get V = sqrootof(25*10)=15.81V

if i use P=I^2*R i get I=sqrootof(10/25)=0.632A V=R*I=15.81V


Then I=V/(RL+RS)= 15.81/(25+5)=0.527A

are my calculations correct?
 
  • #4
lewando
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Your calculations are numerically correct but your result is wrong. You have gotten off the track with this line:
Then I=V/(RL+RS)= 15.81/(25+5)=0.527A
First off, what did you get for the Thevenin equivalent of the unknown circuit?
 
  • #5
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a 15.81 voltage power supply connected to a 25ohm resistor?
 
  • #6
lewando
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If you had a current going through RL=25ohms producing a voltage of 15.81V, what voltage would be produced across RS?
 
Last edited:

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