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How is Pmax derived?

  • Thread starter xbuoix
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  • #1
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Homework Statement



Why is pmax=v^2/4r?

Homework Equations



I understand that p=v^2/r using the equations:
p=iv and v=ir


The Attempt at a Solution



My problem relates to when I have to find the pmax. Looking over my professors solutions for the homework, he uses pmax=vt^2/4rt.
 

Answers and Replies

  • #2
gneill
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Homework Statement



Why is pmax=v^2/4r?

Homework Equations



I understand that p=v^2/r using the equations:
p=iv and v=ir


The Attempt at a Solution



My problem relates to when I have to find the pmax. Looking over my professors solutions for the homework, he uses pmax=vt^2/4rt.
Hi xbuoix, Welcome to Physics Forums.

Can you post the original problem so that we can see the issue in context? It's hard to say anything meaningful about it without knowing where the 't' comes in or how it comes about.
 
  • #3
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Hi xbuoix, Welcome to Physics Forums.

Can you post the original problem so that we can see the issue in context? It's hard to say anything meaningful about it without knowing where the 't' comes in or how it comes about.
Thanks, I'm glad to be on this forum!

Sorry, vt and rt is the Thevenin equation for the voltage and resistance.

The question goes like this:
(There is a circuit.)
Draw the circuit on your solutions and show how you obtain a Thevenin equivalent for everything to the left of the load resistor. Then draw the Thevenin equivalent that you obtain connected to the load resistor. Using the results, what is the maximum power that can be delivered to Rl. What is the value of Wl would result in maximum power?

Finding the Thevenin equation, I found:
VT= 30 V
RT=3kohms
Imax=10mA
RLoad=RT
VMax=VT


At first, I thought that P=VT^2/RT. After I turned in my homework, I looked at the solution my professor posts online, and I see that he used P=VT^2/4RT. Where did the 1/4 come from?
 
  • #4
gneill
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20,795
2,773
Thanks, I'm glad to be on this forum!

Sorry, vt and rt is the Thevenin equation for the voltage and resistance.

The question goes like this:
(There is a circuit.)
Draw the circuit on your solutions and show how you obtain a Thevenin equivalent for everything to the left of the load resistor. Then draw the Thevenin equivalent that you obtain connected to the load resistor. Using the results, what is the maximum power that can be delivered to Rl. What is the value of Wl would result in maximum power?

Finding the Thevenin equation, I found:
VT= 30 V
RT=3kohms
Imax=10mA
RLoad=RT
VMax=VT


At first, I thought that P=VT^2/RT. After I turned in my homework, I looked at the solution my professor posts online, and I see that he used P=VT^2/4RT. Where did the 1/4 come from?
Okay, it's clearer now (By the way, you might try using the x2 and x2 icons in the edit panel to create superscripts and subscripts).

The 1/4 comes from the mathematical operations involved in determining the conditions for maximum power delivered to the load resistor. Find the value of RL that results in the maximum power delivered to the load, then use that value of RL to determine that power.
 
  • #5
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Okay, it's clearer now (By the way, you might try using the x2 and x2 icons in the edit panel to create superscripts and subscripts).

The 1/4 comes from the mathematical operations involved in determining the conditions for maximum power delivered to the load resistor. Find the value of RL that results in the maximum power delivered to the load, then use that value of RL to determine that power.
So this case, the RLoad=RT=3kΩ. I'm still confused on how it came about.
 
  • #6
gneill
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So this case, the RLoad=RT=3kΩ. I'm still confused on how it came about.
Have you studied the math behind finding the maximums and minimums of functions?
 
  • #7
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Have you studied the math behind finding the maximums and minimums of functions?
As in finding the double derivative of the function? Yes

Oh.. I get it now..I think.

The double integral of v2 would give 1/4? Shouldn't we have taken the derivative instead?

Edit: I meant 1/R. Why are we taking the maximum of the resistance and not the voltage?
 
  • #8
gneill
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20,795
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As in finding the double derivative of the function? Yes

Oh.. I get it now..I think.

The double integral of v2 would give 1/4? Shouldn't we have taken the derivative instead?

Edit: I meant 1/R. Why are we taking the maximum of the resistance and not the voltage?
You're looking for the value of RL which maximizes the power (in RL). Then use that RL to write the expression for the maximum power.
 

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