Max Power Transfer Homework: Solve for RL & Calculate Pmax

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of the load resistance (RL) that maximizes power transfer in a given circuit. Participants analyze the circuit using mesh analysis and Thevenin's theorem, addressing both the methodology and the calculations involved.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • The initial attempt at solving the problem includes calculations for mesh currents and voltages, with the participant expressing uncertainty about the voltage across certain resistors.
  • One participant points out an error in the calculation of mesh current in mesh 1, suggesting that the current should be 0.5A instead of 2A, which affects the Thevenin voltage calculation.
  • A suggestion is made to convert the voltage source and series resistor to a Norton equivalent to simplify the analysis, which could eliminate one mesh from the calculations.
  • The original poster seeks confirmation on whether the identified mistake is the only significant error in their solution.

Areas of Agreement / Disagreement

Participants generally agree on the identification of the error in the current calculation, but the discussion does not reach a consensus on the overall correctness of the solution or the best approach to take.

Contextual Notes

The discussion highlights potential limitations in the original calculations, including assumptions about voltage across resistors and the method of analysis used. Specific values and circuit configurations are not fully detailed, which may affect the conclusions drawn.

Who May Find This Useful

Students and individuals interested in circuit analysis, particularly those studying mesh analysis and Thevenin's theorem in electrical engineering contexts.

timeforplanb
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Homework Statement


Determine the value of RL that will draw the maximum power from
the rest of the circuit shown below. Calculate the maximum power. (This problem was included in our exam this week, and I don't really recall the exact values of the parameters, but I'm pretty sure that the circuit looks more or less like this.)

attachment.php?attachmentid=41273&stc=1&d=1322295774.jpg


Homework Equations


V=IR
P=I2R
KVL

The Attempt at a Solution



Computing for Vab:
mesh 1:
-10+5I1+15(I1-I3)=0
but I3=0, so I1=2A

mesh 2:
I2=3Vx
but Vx=-6I3, and I3=0, so I2=0

Since no voltage exists in the 3 and 6 ohm resistors, I shorted them and Vab will now be equal to the voltage across the 15 ohm resistor. (Will it? I'm not really sure about this)

Therefore,
V5ohm=Vab=15I1=15(2)
Vab=30V

Computing for Rab:

mesh 1:
20I1-153=0 --> eqn.1

mesh 2:
I2=3Vx
but Vx=-6I3
I2=3(-6I3)
I2+18I3=0 --> eqn.2

mesh 3:
-15I1-3I2+24I3=0 --> eqn. 3

I1=-0.011A
I2=0.270A
I3=-0.015A

I=-I3=-(-0.015)
I=0.015A

Rab=1/I=1/0.015
Rab=66.667ohms

Computing for the maximum power:

Pmax=I2Rab=(30/(66.667+66.667))2(66.667)
Pmax=3.375W

That is what I did. Sirs, did I solve it the right way? I'm especially uncertain about how I solved Vab.
 

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The general method looks okay, but you've got a small problem with mesh1. The voltage is 10V and it's flowing through 5+15 = 20Ω. That should be a current of 0.5A, not 2A. As a result your Thevenin voltage does not turn out correctly.

Also, if I might make a suggestion that could simplify things a bit, why not first convert the 10V supply and its 5Ω series resistor to a Norton equivalent current? The Norton resistance will be in parallel with the 15Ω resistor so you can combine them. Then convert back to Thevenin. You'll have eliminated one mesh.
 
ah. thanks for pointing those out, sir. so it means that the major mistake is the 2A that's supposed to be 0.5A? nothing more?
 
timeforplanb said:
ah. thanks for pointing those out, sir. so it means that the major mistake is the 2A that's supposed to be 0.5A? nothing more?

That's about it!
 

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