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Maximum Power Dissipation at Internal Resistance Proof

  • Thread starter ATOMatt
  • Start date
  • #1
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Homework Statement


My physics teacher (i'm in year 12) set me a "challenge" to prove a graph in our textbook, which shows Power against Resistance, with a local maximum where R = internal resistance, r. He told me i'd need to use differentiation, P = I[tex]^{2}[/tex].R
and said that I was right in thinking that I had to do something to do with there being a stationary point at R = r, but in all honesty I am pretty stumped at where to start.

Homework Equations



P = I[tex]^{2}[/tex].R

P = [tex]\frac{V^{2}}{R}[/tex]


The Attempt at a Solution


Well all i've really said so far is that,

[tex]\frac{dP}{dR}[/tex] = I[tex]^{2}[/tex]

and that at R = r, [tex]\frac{dP}{dR}[/tex] = 0, so I = 0.

The problem is... I don't really understand what I am trying to prove, is it simple enough to just say that max power is dissipated when I = 0, and I = 0 when R = r, or am I skimming over some deeper truth?

Any help would be appreciated, but i'd prefer guidance instead of a full answer :)
 

Answers and Replies

  • #2
berkeman
Mentor
57,458
7,475

Homework Statement


My physics teacher (i'm in year 12) set me a "challenge" to prove a graph in our textbook, which shows Power against Resistance, with a local maximum where R = internal resistance, r. He told me i'd need to use differentiation, P = I[tex]^{2}[/tex].R
and said that I was right in thinking that I had to do something to do with there being a stationary point at R = r, but in all honesty I am pretty stumped at where to start.

Homework Equations



P = I[tex]^{2}[/tex].R

P = [tex]\frac{V^{2}}{R}[/tex]


The Attempt at a Solution


Well all i've really said so far is that,

[tex]\frac{dP}{dR}[/tex] = I[tex]^{2}[/tex]

and that at R = r, [tex]\frac{dP}{dR}[/tex] = 0, so I = 0.

The problem is... I don't really understand what I am trying to prove, is it simple enough to just say that max power is dissipated when I = 0, and I = 0 when R = r, or am I skimming over some deeper truth?

Any help would be appreciated, but i'd prefer guidance instead of a full answer :)
Write the equation for the voltage across the load resistor as a function of the load resistor value. Write the equation for the current through the load resistor as a funtion of the load resistor value. (Both assuming a constant source resistance).

Then the power dissipated by the load resistor as a function of its resistance is what (write the equation)? So you can see that if Rload is small, there is more current, but less voltage drop. And if Rload is large, there is more voltage drop, but less current. So if you plot the power dissipated by Rload as a funtion of Rload, you will get a maximum at some value for Rload.

To find that peak, you can use differentiation of the equation for Pload = f(Rload). What can you say about the slope (the first derivative) of that function at the peak?
 
  • #3
31
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Thanks for the fast reply;

Write the equation for the voltage across the load resistor as a function of the load resistor value. Write the equation for the current through the load resistor as a funtion of the load resistor value. (Both assuming a constant source resistance).
Then the power dissipated by the load resistor as a function of its resistance is what (write the equation)?
I'm not quite sure what you mean by writing the equation as a function of the other value, or at least, if I do, i've never heard it phrased in this way. Or is the voltage across the load resistor just V = P/R+r and I = sqrt(P/R).
 
  • #4
berkeman
Mentor
57,458
7,475
Thanks for the fast reply;



I'm not quite sure what you mean by writing the equation as a function of the other value, or at least, if I do, i've never heard it phrased in this way. Or is the voltage across the load resistor just V = P/R+r and I = sqrt(P/R).
Use the voltage divider equation to get V = f(Rload):

http://en.wikipedia.org/wiki/Voltage_divider

And just use the series resistance Rload + Rsource and the source voltage to get I = f(Rload).
 
  • #5
31
0
Ah right, I should be able to do this now. I'll have a go in a minute. Thanks a lot for your help and patience :)
 

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