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Power Dissipated - Kirchhoff's Rule

  • Thread starter Pete_01
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  • #1
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Homework Statement


Using symbols, write PRL, the power dissipated in RL as a function of RL (your other symbols are R0 and the power supply voltage epsilon both of which are constant. 10V and 100ohms) and then take the derivative of this function with respect to RL for which PRL has an extreme value, in this case a maximum.


Homework Equations


V = IR


The Attempt at a Solution


So here's what I was thinking...
PRL = VI where I = V/R so replace that with Epsilon2/R0
Then plug in the constant values 10V and 100ohms to get PRL = 1W dRL

Not sure where to go from here though as the derivative gives zero. What am I missing? Did I do it all incorrectly? Thanks in advance!
 

Answers and Replies

  • #2
tiny-tim
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… Not sure where to go from here though as the derivative gives zero.
Hi Pete_01! :smile:

You're confusing R0 with RL.

I haven't worked out what R0 is (what is it? and what is the context?), but clearly P will depend on RL, either instead of or as well as R0. :wink:
 
  • #3
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Hi Tim,

R0 is a constant resistor of value 100ohms.
 
  • #4
tiny-tim
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But where is it, and what's RL ? :confused:
 
  • #5
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Oops, sorry. RL is a variable resistor that was changed in value from 10-1280ohms. I attached the circuit picture as well.

Thanks
 

Attachments

  • #6
tiny-tim
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Oops, sorry. RL is a variable resistor that was changed in value from 10-1280ohms. I attached the circuit picture as well.

Thanks
ohh! they're two resistors in series! :rolleyes:

ok, then your formula I = V/R0 should be changed to … ? :wink:
 
  • #7
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So it would be I = V/(R0+RL)?

Then your left with PRL = 102/(R0+RL)d(RL) Correct?
 
  • #8
tiny-tim
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So it would be I = V/(R0+RL)?
Yes. :smile:
Then your left with PRL = 102/(R0+RL)d(RL) Correct?
what's d ? :confused:

and what formula are you using for PRL ?
 
  • #9
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So I used P=VI where I is substituted in from above. 10 is the voltage.

PRL = VI = V*(V/R) = V2/(R0+RL)

It said to find RL where it is maximum I need to take the derivative of the function with respect to RL and set the derivative equal to 0, then solve for RL.
 
  • #10
tiny-tim
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So I used P=VI where I is substituted in from above. 10 is the voltage.
But 10 isn't the voltage across RL, is it? :wink:

(and that's not the quickest formula, anyway)
 
  • #11
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Ah yes, it is not. What is the best way to solve this?
 
  • #12
tiny-tim
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erm :redface:

… one of Kirchhoff's rules ? :smile:
 

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