Voltage Divider Problem

[nmacholl]

Homework Statement

You want to build a voltage divider for a product that uses as it's source a 12v battery and a non-zero internal resistance. You want to have an output voltage between 4 and 5 volts. The internal resistance varies from 5 ohms when fully charged to 100 ohms when the battery is nearly dead. Create a voltage divider that meets the requirements using the smallest possible resistors (doesn't have to be real resistors). You can neglect the tolerances of the resistors and the load resistance (assume it is infinite).
http://img811.imageshack.us/img811/8537/vdivdiagram.th.gif

Homework Equations

V=IR and Kirchoff's Rules

The Attempt at a Solution

r is internal resistance.
R₁ and R₂ are the resistors necessary for the voltage divider.
R_L is the load resistance (which is infinite)

12v - Ir - IR₁ - IR₂ = 0

If we assume infinite load resistance we can treat R_L as a gap.

12v = I * R_total (where R_total is the sum of the resistances in series)
R_total= r + R₁ + R₂

From Ohm's law:
I = 12v / (r + R₁ + R₂)

12v - Ir - IR₁ - IR₂ = 0
12v - I(r - R₁ - R₂) = 0

12v - (12v / (r + R₁ + R₂))
* (r - R₁ - R₂) = 0

Since the voltage across R₂ is equal to the voltage across R_L we need I * R₂ to be between 4 an 5 volts.

This is where I'm lost, specifically how do I use the internal resistance information in the problem to generate a solution that gives me 4-5volts across the load resistor?

 

[bob1182006]

You need to use the information about the internal resistance to calculate what the voltage across R2 will be.

So for r = 5 ohms what should the voltage across R2 be? The same thing for r = 100 ohms.

 

[nmacholl]

You need to use the information about the internal resistance to calculate what the voltage across R2 will be.

So for r = 5 ohms what should the voltage across R2 be? The same thing for r = 100 ohms.

Okay.
Is there any way to pull out another equation from the problem. If I am interpreting your suggestion correctly you want me to use r=5ohms then solve for the voltage across R₂ and get 5 volts, then r=100ohms and get 4v. But even then I have one equation with two unknowns: R₁ and R₂, am I missing an equation from somewhere?

 

[bob1182006]

Now you need to plug in the relationship between R1 and R2 back into the one of the equations for the voltage across R2 when r = 5 or 100 ohms. You will end up with the value for one resistor.

 

[nmacholl]

Now you need to plug in the relationship between R1 and R2 back into the one of the equations for the voltage across R2 when r = 5 or 100 ohms. You will end up with the value for one resistor.

I seem to be missing something. What relationship have I not used yet? I can't substitute the voltage equation back into itself and I've already used Ohm's law to replace "I" with 12v/R_total.

 

[bob1182006]

Some of the equations you have are:
R1+R2 = #
R2*I = V

Using the top one you can plug it into the bottom one for a specific value of r which will give you a specific value for V and that way you can solve for R1 or R2.

 

[nmacholl]

Some of the equations you have are:
R1+R2 = #
R2*I = V

Using the top one you can plug it into the bottom one for a specific value of r which will give you a specific value for V and that way you can solve for R1 or R2.

Forgive me but I don't seem to have equations like that.
My equation from Kirchoff's loop rule is: 12v - Ir - IR₁ - IR₂ = 0

This contains 3 unknowns: I, R₁, and R₂
Using ohm's law I can substitute; I = 12v / (r + R₁ + R₂)

It seems to me that the only equation I have that relates R₁ to R₂ is my loop rule equation and Ohm's law has already exhausted it's potential by eliminating the current "I".

:(

 

[bob1182006]

You have I = 12v / (r + R1 + R2), and at the load you have I*R2 = I = (R2*12v) / (r + R1 + R2).
There's two values for r and two values for I*R2 = VL and from those two equation you can get the relationship of R1+R2 = #, then you place this restriction back into either of the equations for I*R2.

 

[nmacholl]

You have I = 12v / (r + R1 + R2), and at the load you have I*R2 = I = (R2*12v) / (r + R1 + R2).
There's two values for r and two values for I*R2 = VL and from those two equation you can get the relationship of R1+R2 = #, then you place this restriction back into either of the equations for I*R2.

Following your suggestions I've gotten two sets of solutions by plugging in r={5ohms, 100ohms} and V_L={5v,4v}
To get:
R₁={-4,-99ohms}
R₂={5/7,-1/2ohms}

Now the resistors don't have to be "real" but I'm concerned because the resistors have different values for each load voltage and I don't think that's appropriate. Shouldn't the solution have fixed value resistors?

 

[bob1182006]

Yes, what was the relation that you obtained for R1+R2 = #?
Also when r = 5 ohms what voltage do you expect at the load?
How about when r=100 ohms, what voltage would be at the load?

I think If you have those backwards you'll get negative resistances.

 

[nmacholl]

Yes, what was the relation that you obtained for R1+R2 = #?
Also when r = 5 ohms what voltage do you expect at the load?
How about when r=100 ohms, what voltage would be at the load?

I think If you have those backwards you'll get negative resistances.

I'm still not getting anywhere. I tried doing all the algebra in Mathematica and ended up with worthless results.

 

[bob1182006]

What new equations do you now have that aren't in your original post?

 

[nmacholl]

What new equations do you now have that aren't in your original post?

I made a mistake in my first post but I have:
12-I(r+R1+R2)=0
VL=I*R2

Then you wrote this:
VL = (R2*12v) / (r + R1 + R2) but I don't know where that came from.

Using mathematica I can solve what you wrote for R2 in terms of R1
For VL = 4 :: R2=5(r+R1)/7
For VL = 5 :: R2=(r+R)/2

 

[bob1182006]

If you solve your first equation for I and then plug it into the second equation you get the equation that I gave you.

In the equations for R2 in terms of R1 you didn't specify the value for r, when VL = 4 or 5 r should be either 5 or 100 ohms.

 

[nmacholl]

If you solve your first equation for I and then plug it into the second equation you get the equation that I gave you.

In the equations for R2 in terms of R1 you didn't specify the value for r, when VL = 4 or 5 r should be either 5 or 100 ohms.

Right well if
r=100 ohms when VL = 5v
R2=(500+5*R1)/7
and
r=5 ohms when VL = 4v
R2=(5+R1)/2

But I don't have an equation to use these in. We used my loop equation and ohms law to get these, don't I need another equation to plug these into?

 

[bob1182006]

You can reuse VL = (R2*12v) / (r + R1 + R2) to find the value for one of the resistances.

Also those equations seem to give negative values for R1. I know that if you obtain the two equations from the equation VL = (R2*12v) / (r + R1 + R2) then you will get the correct numbers.

 

[nmacholl]

You can reuse VL = (R2*12v) / (r + R1 + R2) to find the value for one of the resistances.

Also those equations seem to give negative values for R1. I know that if you obtain the two equations from the equation VL = (R2*12v) / (r + R1 + R2) then you will get the correct numbers.

VL=(12*R2)/(r+R1+R2)
<algebra happens>
R2=(VL*r+VL*R1)/(12-VL)

But I can't substitute this back into VL=(12*R2)/(r+R1+R2) because then I would be plugging the equation back into itself.

 

[bob1182006]

r=100 ohms when VL = 5v
R2=(500+5*R1)/7
and
r=5 ohms when VL = 4v
R2=(5+R1)/2

Just noticed these are wrong, when r = 100 ohms the current will be a minimum throughout the circuit so you should expect the voltage at the load to be also at a minimum, 4v not 5v.

To get the equations you want you should plug in the values for r and VL into the equation like so:
VL=(12*R2)/(r+R1+R2):
4 = (12*R2)/(100+R1+R2)
Then you get another equation with the other 2 values for r,VL
5 = (12*R2)/(5+R1+R2)

Both of them have a term of 12*R2 that you can cancel out and solve for R1+R2

 

[nmacholl]

Just noticed these are wrong, when r = 100 ohms the current will be a minimum throughout the circuit so you should expect the voltage at the load to be also at a minimum, 4v not 5v.

To get the equations you want you should plug in the values for r and VL into the equation like so:
VL=(12*R2)/(r+R1+R2):
4 = (12*R2)/(100+R1+R2)
Then you get another equation with the other 2 values for r,VL
5 = (12*R2)/(5+R1+R2)

Both of them have a term of 12*R2 that you can cancel out and solve for R1+R2

NVM

I solved for 12*R2 and pluged the first into the second to get R1+R2=375

 

[bob1182006]

That, or you can just solve for 12*R2 and equate them to get one equation.

(12*R2)=5*(5+R1+R2)=4*(100+R1+R2) and then solve for R1+R2

 

[nmacholl]

Okay now I got these solution sets:
VL = 5; r = 5
R1 = 216.67 ohms
R2 = 158.33 ohms

VL = 4; r = 100
R1 = 60.42 ohms
R2 = 314.58 ohms

I'm still uncomfortable with there being two values for each resistor.

 

[bob1182006]

So from looking at your solutions I assume you got R1+R2=375 right?
I don't get the second set of solutions though, I get:
5 = 12*(375-R1)/(5+375) = 11.84-0.0316*R1, from this I get R1 = 216.45, roundoff errors

then there's
4 = 12*(375-R1)/(100+375)

Where did you get the second solution from?

 

[nmacholl]

My equation for R2 is:
R2=(VL*r + VL(375))/12

I made a calculator mistake:
For VL= 4; r = 100
R2 = (4(100)+4(375))/12
R2 = 158.33
R1 = 375-R2 = 216.67

:) EEEEEEEEEEEeeeeeeeeeeeeeeeeeeeeeee!

 

[bob1182006]

Ah I see hehe.