Maximizing Power Transfer: An Analysis of Resistance and Efficiency

  • #1
PhysicsTest
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Homework Statement
A battery of emf E and internal resistance r is hooked up to a variable "load"
resistance, R. If you want to deliver the maximum possible power to the load, what resistance
R should you choose?
Relevant Equations
Ohms Law V = IR
1691817049496.png

I equated dP/dR to 0 and find the solution for R. Is the solution correct?
 
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  • #2
PhysicsTest said:
Homework Statement: A battery of emf E and internal resistance r is hooked up to a variable "load"
resistance, R. If you want to deliver the maximum possible power to the load, what resistance
R should you choose?
Relevant Equations: Ohms Law V = IR

View attachment 330425
I equated dP/dR to 0 and find the solution for R. Is the solution correct?
It is clearly wrong since the final equation is dimensionally inconsistent.
 
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  • #3
PhysicsTest said:
I equated dP/dR to 0 and find the solution for R. Is the solution correct?
The method is correct but not the solution as @haruspex noted. Redo the algebra and, as you take the derivative, make sure that you get a dimensionally consistent equation after each step.
 
  • #4
PhysicsTest said:
I equated dP/dR to 0 and find the solution for R. Is the solution correct?
It's pretty clear that there is an error in taking the derivative,
 
  • #5
Yes the differentiation is wrong and i get R = r. But what i do not understand is that the power delivered to resistor is called loss, that how is it useful?
 
  • #6
PhysicsTest said:
Yes the differentiation is wrong and i get R = r. But what i do not understand is that the power delivered to resistor is called loss, that how is it useful?
The power dissipated in the internal resistance, r, is the loss; the power dissipated in R is the rate of doing useful work.
This (standard) result says that if you wish to maximise the rate of doing useful work then you have to accept suffering an equal amount of wasted work. In practice, you may settle for a lower rate of doing useful work in return for a higher efficiency.
 
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  • #7
haruspex said:
The power dissipated in the internal resistance, r, is the loss; the power dissipated in R is the rate of doing useful work.
This (standard) result says that if you wish to maximise the rate of doing useful work then you have to accept suffering an equal amount of wasted work. In practice, you may settle for a lower rate of doing useful work in return for a higher efficiency.
Yes, and in the real world, if you want to maximize the efficiency of delivering power to R, you will try to minimize r and reduce the source accordingly.

The reason this extremely common problem is given, in addition to practicing some analysis skills, is that this is a really important concept in more complex problems you will face later with high frequency circuits. I'll stop here, since it's beyond the scope of this thread, but this is the DC version of impedance matching, which is really, really, really important in high frequency circuit design.
 
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