No load voltage / resistance question

[orangeincup]

Homework Statement

The no load voltage in the voltage divider circuit is 8V. The smallest load resistor that is ever connected to the divider is 3.6 kΩ. When the divider is loaded v0 is not to drop below 7.5V.

Part 2)
Assume the voltage divider has been constructed from 1 W resistors. What is the smallest resistor that can be used as rL before one of the resistors in the divider is operating at its dissipation limit.

Homework Equations

r1*r1/(r1+r2)
v(r1/(r1+r2))=v0

The Attempt at a Solution

I started by saying
eq 1) Vo=40*R2/(R1+R2), therefore R1=4R2

eq 2) Re=R2*RL/(R2+RL)

eq 3) 40*Re/(Re+RL)=7.5V, which reduces to R1=4.33Re, therefore 4R2=4.33Re.

I'm not sure where to go from here. What do I do with the value of 3600 kΩ? I tried that below but didn't get very far with it:
eq 4)
Re=4.33Re*3600/(3600+R2)

 

[BvU]

Can you explain your first relevant equation?

I suppose the second is for the case RL is not connected.

eq 1) OK, eq 2) OK, eq 3) OK. Now substitute Re from eq 2) in eq 3): one equation (namely 4R2=4.33 R2*RL/(R2+RL) ) with one unknown. Because you know the RL that causes this worst case, namely ...

Gives you R2 !

For part 2 of your exercise: which of the two would get to this limit first ? What current ? What voltage drop ?

 

[orangeincup]

Can you explain your first relevant equation?

I suppose the second is for the case RL is not connected.

eq 1) OK, eq 2) OK, eq 3) OK. Now substitute Re from eq 2) in eq 3): one equation (namely 4R2=4.33 R2*RL/(R2+RL) ) with one unknown. Because you know the RL that causes this worst case, namely ...

Gives you R2 !

For part 2 of your exercise: which of the two would get to this limit first ? What current ? What voltage drop ?

Okay, I calculated R2=1200Ω and R1=300Ω by subbing in the equations.

For part 2:
eq1) vr1^2//1200=1 W, vR1=34.6 V

eq2) 40V-34.6V=5.4V for R2+RL

eq3) 40*Re/(1200+Re)=5.4V. Re=188.4Ω

eq4) 300*RL/(300+RL)=188.4Ω, therefore RL=487Ω

Not sure if this is right. Is there an easier way of calculating the RL, or do I have to calculate Re first like I did here?

 

[BvU]

1200Ω is spot on.
So is your part 2 calculation. I don't see any real shortcuts there: all paths are more or less the same. Want to see the one I was thinking of? Here goes:
eq1), eq 2: same as you
Vr1/R1 = 28.87 mA Vr2/R2 = 17.86 mA → Vr_L/R_L = 11.00 mA → R_L = 5.36 V/11.00 mA

It's not very generalized, sophisticated, high-brow, etc. But good enough here...

Comes down to using Ohm's law for the various current paths. And Kirchoff's law (which I never really absorbed -- I thought of it as 'the current has to go somewhere' ).

 

[HalfLostAndSearching]

Re=4.33Re*3600/(3600+4R2)
Re=4.33Re*3600/(3600+4*(4.33Re))
Re=4.33Re*3600/(3600+17.32Re)
0=4.33Re*3600-7.5*(3600+17.32Re)
0=15600*Re-2700*Re+7.5*17.32Re
0=15600*Re-2700*Re+129.9Re
0=129.9*Re-2700*Re
0=129.9*Re-2700*Re
-2570.1*Re=0
Re=0

It seems like you are on the right track with your equations, but you may have made a mistake in equation 4. Instead of setting Re equal to 0, try solving for R2 to find the smallest resistor that can be used as rL before one of the resistors in the divider is operating at its dissipation limit. This will give you a numerical value for R2 that you can then use in your calculations. Additionally, make sure to take into account the 1 W limit for the resistors when solving for R2.