Power Dissipated - Kirchhoff's Rule

  • Thread starter Pete_01
  • Start date
  • #1
Pete_01
51
0

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
Science Advisor
Homework Helper
25,836
255
… 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
Pete_01
51
0
Hi Tim,

R0 is a constant resistor of value 100ohms.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,836
255
But where is it, and what's RL ? :confused:
 
  • #5
Pete_01
51
0
Oops, sorry. RL is a variable resistor that was changed in value from 10-1280ohms. I attached the circuit picture as well.

Thanks
 

Attachments

  • p.pdf
    51 KB · Views: 201
  • #6
tiny-tim
Science Advisor
Homework Helper
25,836
255
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
Pete_01
51
0
So it would be I = V/(R0+RL)?

Then your left with PRL = 102/(R0+RL)d(RL) Correct?
 
  • #8
tiny-tim
Science Advisor
Homework Helper
25,836
255
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
Pete_01
51
0
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
Science Advisor
Homework Helper
25,836
255
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
Pete_01
51
0
Ah yes, it is not. What is the best way to solve this?
 
  • #12
tiny-tim
Science Advisor
Homework Helper
25,836
255
erm :redface:

… one of Kirchhoff's rules ? :smile:
 

Suggested for: Power Dissipated - Kirchhoff's Rule

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
4K
Replies
1
Views
6K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
11
Views
14K
Replies
6
Views
2K
Replies
10
Views
4K
Replies
5
Views
2K
Replies
1
Views
9K
Top