How Do You Calculate Thevenin Equivalent Circuits and Current Flow?

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Discussion Overview

The discussion revolves around calculating the Thevenin equivalent circuit for an unknown circuit connected to a variable resistor (RL). Participants explore the derivation of the Thevenin equivalent and the resulting current flow when RL is set to a specific value. The focus includes theoretical calculations and the application of relevant equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that the maximum power dissipation occurs when RL equals RS, which is 25 ohms, leading to a derived voltage of 22.4V for the Thevenin equivalent circuit.
  • Another participant challenges the initial calculation, emphasizing that the 10W of power is dissipated by RL alone, not by the sum of RL and RS.
  • There is a discussion on whether to use the formula P = V^2/R or P = I^2*R, with different participants arriving at different voltage and current values based on their chosen approach.
  • One participant calculates a voltage of 15.81V and a current of 0.527A based on their interpretation of the power dissipation, but another participant questions the correctness of this result.
  • Further inquiry is made about the voltage produced across RS when a current flows through RL.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct calculations or the interpretation of the power dissipation. Multiple competing views and calculations are presented, indicating uncertainty in the derived values.

Contextual Notes

There are unresolved assumptions regarding the definitions of the components in the circuit and the conditions under which the power calculations are made. The discussion reflects differing interpretations of the relationships between voltage, current, and resistance in the context of Thevenin's theorem.

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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