Power Dissipated - Kirchhoff's Rule

[Pete_01]

Homework Statement

Using symbols, write P_RL, the power dissipated in R_L as a function of R_L (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 R_L for which P_RL 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...
P_RL = VI where I = V/R so replace that with Epsilon²/R₀
Then plug in the constant values 10V and 100ohms to get P_RL = 1W dR_L

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!

 

[tiny-tim]

… Not sure where to go from here though as the derivative gives zero.

Hi Pete_01!

You're confusing R₀ with R_L.

I haven't worked out what R₀ is (what is it? and what is the context?), but clearly P will depend on R_L, either instead of or as well as R₀.

 

[Pete_01]

Hi Tim,

R₀ is a constant resistor of value 100ohms.

 

[tiny-tim]

But where is it, and what's R_L ?

 

[Pete_01]

Oops, sorry. R_L is a variable resistor that was changed in value from 10-1280ohms. I attached the circuit picture as well.

Thanks

 

[tiny-tim]

Oops, sorry. R_L 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!

ok, then your formula I = V/R₀ should be changed to … ?

 

[Pete_01]

So it would be I = V/(R₀+R_L)?

Then your left with P_RL = 10²/(R₀+R_L)d(R_L) Correct?

 

[tiny-tim]

So it would be I = V/(R₀+R_L)?

Yes.

Then your left with P_RL = 10²/(R₀+R_L)d(R_L) Correct?

what's d ?

and what formula are you using for P_RL ?

 

[Pete_01]

So I used P=VI where I is substituted in from above. 10 is the voltage.

P_RL = VI = V*(V/R) = V²/(R₀+R_L)

It said to find R_L where it is maximum I need to take the derivative of the function with respect to R_L and set the derivative equal to 0, then solve for R_L.

 

[tiny-tim]

So I used P=VI where I is substituted in from above. 10 is the voltage.

But 10 isn't the voltage across R_L, is it?

(and that's not the quickest formula, anyway)

 

[Pete_01]

Ah yes, it is not. What is the best way to solve this?

 

[tiny-tim]

erm

… one of Kirchhoff's rules ? ​