How Do You Calculate Thevenin Equivalent Circuits and Current Flow?

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SUMMARY

The discussion focuses on calculating the Thevenin equivalent circuit for an unknown circuit with a variable resistor (RL). The maximum power dissipation of 10W occurs at RL = 25 ohms, leading to a Thevenin equivalent voltage of 22.4V and a resistor value of 25 ohms. When RL is set to 5 ohms, the current calculated is 0.74A. Participants clarify the correct application of power formulas, emphasizing that the power dissipated is associated with RL alone, not the total resistance.

PREREQUISITES
  • Understanding of Thevenin's Theorem
  • Knowledge of electrical power formulas (P=V^2/R and P=I^2*R)
  • Familiarity with circuit analysis techniques
  • Basic concepts of resistors and current flow
NEXT STEPS
  • Study Thevenin's Theorem applications in circuit analysis
  • Learn about maximum power transfer theorem in electrical circuits
  • Explore advanced circuit simulation tools like LTspice
  • Investigate practical applications of variable resistors in electronic devices
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in circuit analysis and design will benefit from this discussion.

Solidsam
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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?
 
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Solidsam said:
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.
 
lewando said:
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?
 
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?
 
a 15.81 voltage power supply connected to a 25ohm resistor?
 
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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