# Unknown Resistors

1. Jul 15, 2007

### scrgirl173

Given the network in the Figure, we wish to obtain a voltage of 4 V ≤ Vo ≤ 8 V across the full range of the pot. Determine the values of (a) R1 and (b) R2.

I'm quite confused as to what it means by "the full range of the pot." And I'm not quite sure how to approach this problem. Can anyone help me out?

Figure: V.S. = 10 V R3 = 1kOhm
___________________
l l
l R1
l l
l l
V.S. R3 <---- Vo
l l
l R2
l__________________l

2. Jul 16, 2007

### xez

If I'm reading this right (which I may not be):

From ohm's law:
V = I * R
R = V/I
I = V/R

The range of a pot is the maximum range it can be adjusted over.
Mechanically and electrically you have:

R3A C=Vo R3B
A ---/\/\/\/\/\/\/\/\/\/\/\--- B

So there's a fixed resistance of 1000 ohms (your pot's
specified maximum value of resistance) between 'A' and 'B'.
Mechanically there's a 'wiper arm' that literally wipes across
the resistor surface anywhere from 'A' to 'B' that is your
terminal 'C', so there's a variable resistance from
'A' to 'C' and 'C' to 'B'.

So as you move the pot wiper arm contact C moves between
(A) and (B) so that

at one maximum:
Contact C is at the place of Contact A and R3A=0, R3B=1000.

at the other maximum:
Contact C is at the place of Contact B and R3A=1000, R3B=0.

in the middle:
Contact C is in the middle of Contact A and Contact B and R3A=500, R3B=500.

where you're seeking to measure a voltage from 4V to 8V over the
possible range of the pot.

Since the pot R3 divides the circuit into two resistances with
a tap point in the middle you have electrically:

VS
R1
V1
R3A
VO
R3B
V2
R2
GROUND

points just for convenience of discussion, but in reality
R2(top) connects to R3B bottom,
Vo is connected between R3B(top) and R3A(bottom),
V1 is at the connection of R1(bottom) and R3A(top), etc.

Since it's a series network
I = current = constant through all series elements = VS/R(total)
so I = VS/(R1+R3A+R3B+R2).

The voltage difference between the terminals of any resistor is
by ohms law: V = I*R, so the voltage differences across each of your
resistors is:

V across R2 = I * R2
V across R3B = I * R3B
V across R3A = I * R3A
V across R1 = I * R1

so from bottom to top we have the actual circuit voltages:

Bottom of R2: 0V = Ground.
V2 at top of R2 = 0V + (I*R2)
VO at top of R3B = V2 + (I*R3B)
V1 at top of R3A = VO + (I*R3A)
VS at top of R1 = V1 + (I*R1) = your supply voltage.

So if you assume the pot is adjusted for one maximum extent,
R3A = 0, R3B = 1000, and
VO = I*(R2+R3B) = I*(R2+1000) which must be setup to become 8V since
this will represent your highest possible output voltage at VO.

So if you assume the pot is adjusted for the other maximum extent,
R3A = 1000, R3B = 0, and
VO = I*(R2+R3B) = I*(R2+0) which must be setup to become 4V since this
will represent your lowest possible output voltage at VO.

Fully explicitly we can combine the equations:
I = VS/(R1+R3A+R3B+R2)
VO = I*(R2+R3B)
VO = VS * (R2+R3B) / (R1+R3A+R3B+R2)

since it's a 1000 ohm pot, (R3A+R3B) = 1000, so:
VO = VS * (R2+R3B) / (R1+1000+R2)

at VO = 8V you must have R3B = 1000 and R3A=0 for maximum voltage VO, so
VO = 8V = VS * (R2+1000) / (R1+1000+R2)
VO = 8V = 10V * (R2+1000) / (R1+1000+R2)

at VO = 4V you must have R3B = 0 and R3A=1000 for maximum voltage VO, so
VO = 4V = VS * (R2+0) / (R1+1000+R2)
VO = 4V = 10V * (R2+0) / (R1+1000+R2)

You can use algebraic manipulations and simple evaluations
to solve for the actual values of the resistors (R1) and (R2) since
you've specified VS, two different values of VO, and two different
combinations of R3A;R3B which must generate your specified values of VO.

At VO=8V you're saying that
VS = 10V, R3A=0 (for maximum output VO),
VO=VS-(I*(R1+R3A)), i.e. you must have I*(R1+R3A) = 2V, a 2V
voltage drop across R1+R3A. Since we've already said that
R3A must be adjusted to be 0 to get the maximum voltage here,
you're just saying that:
(drop between VS and VO to get VO=8V) = 2V = I*(R1+R3A),
and since R3A is here adjusted to 0,
2V = I*(R1+0) = I*R1.
If 2 = I*R1, algebraically R1=2/I.

At VO=4V you're saying that there must be a 6V drop on top of VO,
and a 4V drop from VO to ground. In this minimum VO case R3B will
So VO = 4V = I*(R2+R3B), and since we've said that here R3B=0,
VO = 4V = I*(R2+0) = I*R2.
If 4 = I*R2, algebraically, R2=4/I.

Since the total series resistance never changes between VS and GROUND,
I will never change, so since we know from our above evaluations of
the limiting values of operation:
R2=4/I.
R1=2/I.

...we can say:
R2=2*R1 i.e. 2*(R1)=2*(2/I)=4/I=R2.

So we know the proportions of R1 and R2, and just have to pick
R1 to satisfy some other known constraint, such as this one
from above:

VO = 4V = 10V * (R2+0) / (R1+1000+R2)
4=10*(R2)/(R1+1000+R2)
4*(R1+1000+R2) = 10*R2
4*R1+4*R2+4000 = 10*R2
but R2=2*R1 from above, so,
4*R1+4*(2*R1)+4000 = 10*(2*R1)
4*R1+8*R1+4000 = 20*R1
(4*R1+8*R1+4000)/(20*R1) = 1
(12*R1+4000)/(20*R1)=1
(12*R1+4000)/(R1)=20
(12+(4000/R1))=20
(0+(4000/R1))=20-12
4000/R1=8
4000=8*R1
R1=4000/8 = 500.

From above,
R2=2*R1= so R2 = 1000.

Knowing that (R3A+R3B)=1000 since it's an 1000 ohm range pot,
I = VS/(R1+R3A+R3B+R2)
I = 10/(500+1000+1000) = 10/2500 = 0.004A.

Going back checking the 8V required case given R1=500 and R2=1000:
VO = 8V = 10V * (R2+1000) / (R1+1000+R2)
8 = 10 * (R2+1000) / (R1+1000+R2)
8 = 10 * (1000+1000) / (500+1000+1000)
8 = 10*2000/2500 = true, so that works to get 8V at one extreme pot setting.

Going back checking the 4V required case given R1=500 and R2=1000:
VO = 4V = 10V * (R2+0) / (R1+1000+R2)
4 = 10*(1000+0) / (500+1000+1000)
4 = 10000 / 2500 = true, so that works to get 4V at the other extreme pot setting.

So basically you just work with combinations of what you know must be true
given the specified circuit operations to try to define important
component relationships to each other and the known parameters, and
simplify, substitute algebraically, simplify, etc. until you have
solved for one unknown, then go back and substitute that known value
to get the values for other unknowns.