Total resistance in the circuit

[kaspis245]

Homework Statement

It is known that total resistance of the circuit is dependent on the position of the wiper in potentiometer $R$. Find the biggest and the smallest possible total resistances of the circuit. Potentiometer resistance is $R=9r$.

Homework Equations

Ohm's law.

The Attempt at a Solution

Here's a redrawn image:

where $R_1+R_2=R$.

This circuit reminds me of Wheatstone Bridge, so I approached the problem like this:
$V_1=\frac{2r+R_1}{3r+R_1}V$

$V_2=\frac{r+R_2}{4r+R_2}V$

I don't know what to do from here.

 

[Hesch]

You have no current through the circuit, and that is what you have to manage: Add a voltage-source between A and B ( I suggest 12V ). Assume "r" represents 1Ω.

So V_A = 12V, V_B = 0V.

Now you use Kirchhoffs voltage law ( KVL, 3 loops ) and calculate by algebra the current I_AB as R is divided into (x*9r) and (( 9-x)r):
Minimum current → maximum resistance, and vica versa.

You have 3 equations but only one unknown ( x ). Reduce the equations algebraically (right word?) and solve x as for I(x) = . . . . . .

 

[kaspis245]

http://imageshack.com/a/img673/9890/WLcnru.jpg
From Kirchhoff's first rule:
$I-I_3-I_1=0$
$I_2-I_1-I_5=0$
$I_3-I_4-I_5=0$

From Kirchhoff's second rule:
$I_3r+I_4(2r+9rx)=V$
$I_3r+I_5r-I_13r=V_1$
$I_2(r+(9-x)r)-I_4(2r+9rx)+I_5r=V_2$

 

[SammyS]

Homework Statement

It is known that total resistance of the circuit is dependent on the position of the wiper in potentiometer $R$. Find the biggest and the smallest possible total resistances of the circuit. Potentiometer resistance is $R=9r$.

Homework Equations

Ohm's law.

The Attempt at a Solution

Here's a redrawn image:

where $R_1+R_2=R$.

I don't know what to do from here.

Can you use a Y-Δ transform ? It works nicely with the three resistors on the left. (r, r, and 3r)

 

[kaspis245]

Can you use a Y-Δ transform ?

I've tried Y-Δ transform, but since I am not really familiar with it I don't want to get too much into it. I haven't studied it yet, so it would be better if I solve the problem without it.

By the way, I've found a simpler way to write the equations:

$\frac{V_1}{2r+9rx}+\frac{V_1-V_2}{r}+\frac{V_1-V}{r}=0$

$\frac{V_2}{2+(9-x)r}+\frac{V_2-V_1}{r}+\frac{V_2-V}{3r}=0$

$R_{total}=\frac{V}{\frac{V_1}{2r+9rx}+\frac{V_2}{r+(9-x)r}}$

 

[Hesch]

From Kirchhoff's first rule:
I−I3−I1=0
I2−I1−I5=0
I3−I4−I5=0

From Kirchhoff's second rule:
I3r+I4(2r+9rx)=V
I3r+I5r−I13r=V1
I2(r+(9−x)r)−I4(2r+9rx)+I5r=V2

What are you doing? 6 equations! Well, you are doing like everybody else: Very complicated. Here is how you do:

Assume V = 12V, V3 = 0V, r = 1Ω. Draw 3 current-loops:

La: clockwise in the upper left "window".
Lb: clockwise in the upper right window.
Lc: clockwise in the lower window.

Now use Kirchhoffs voltage law for the La-loop:

La: -5La + Lb + 3Lc = 0

Explanation: La is crossing 3r+r+r giving a voltage drop = -5La.
Lb is crossing r in the La-loop in opposite direction giving a voltage rise = Lb.
Lc is crossing 3r in the La-loop in opposite direction giving a voltage rise = 3Lc.

Make two more equation as for the Lb and Lc loops. You now have 3 equations with the unknown La, Lb, Lc and x. The current I through the circuit = Lc.
The resistance of the circuit = V / I. So now you can determine the resistance R(x).

 

[Hesch]

By the way, I've found a simpler way to write the equations:

That's right, but you have at least made one error in the equations ( the last fraction in the first equation ).

EDIT: No, sorry. It's correct ( just don't like your signs ).

 

[kaspis245]

I can't find a way how to derive anything useful from my equations. Is something missing?

Here's the equations using your method:
Lb: -(12+8x)Lb + (10-x)Lc = 0
Lc: -(13-x)Lc = 0

How can I use them? There are too many unknowns to solve them.

 

[Hesch]

First of all when you draw cirkulation-paths with signed direction, you must use the correct sign, otherwise the direction of the path doesn't matter.

Remember that Lb is the path in the upper right window, so how does 12V come into it? here it is:

Lb: La - ( 4+8x ) Lb + ( 10 - x ) Lc = 0

Lc is the circulation path in the lower window and crosses the 12V batteri:

Lc: 3 La + (10 - x) Lb - ( 13 - x ) Lc = -12

There are three equations with 4 unknown. You can solve Lc(x) algebraic.

I(x) = Lc(x).

 

[kaspis245]

I fail to derive anything useful. There are just too many unknowns getting in the way. Are you sure there are only 3 equations? There must be more.

 

[Hesch]

There are three equations with 4 unknown. You can solve Lc(x) algebraic.

Say you have one equation with two unknown:

4x + 8y = 10

You cannot calculate x and y, but you can express x(y):

x(y) = ( 10 - 8y ) / 4

Likewise you can express Lc(x) = . . . . . . . having 3 equations and 4 unknown.

 

[SammyS]

Error alert!

There has been an inconsistency in the use of x throughout this thread.

If x is a number ranging between 0 and 1, then the resistances for the two "pieces" of the potentiometer are (9r)(x) and (9r)(1-x) .

If x is a number ranging between 0 and 9, then the resistances for the two "pieces" of the potentiometer are (r)(x) and (r)(9-x) .

However, using (9r)(x) and (9-x)(r) together is incorrect.

 

[kaspis245]

(1) La: -5 La + Lb + 3 Lc = 0
(2) Lb: La - ( 4 + 8x ) Lb + ( 10 - x ) Lc = 0
(3) Lc: 3 La + (10 - x) Lb - ( 13 - x ) Lc = -12

I've derived Lb from the first equation, La from the second and put those in the third. This is what I got:

$Lc(x)=\frac{(4+8x)Lb+(50-5x)La+12}{53-5x}$

Did you get the same?

 

[SammyS]

(1) La: -5 La + Lb + 3 Lc = 0
(2) Lb: La - ( 4 + 8x ) Lb + ( 10 - x ) Lc = 0
(3) Lc: 3 La + (10 - x) Lb - ( 13 - x ) Lc = -12

I've derived Lb from the first equation, La from the second and put those in the third. This is what I got:

$Lc(x)=\frac{(4+8x)Lb+(50-5x)La+12}{53-5x}$

Did you get the same?

I don't know if it's correct or not.

but ... It's not the way to solve this set of equations. If you solve Loop a: for La, you should then plug that into the other two loop equations. You then have two equations with only currents, Lb and Lc.
Solve one of them for Lb & plug that into the other. Then solve for Lc.

Aside: In my view, it's unfortunate that Hesch labeled these currents as La, Lb, and Lc, rather than the more customary Ia, Ib, Ic except that this default font makes those look bad -- could use i_a, i_b, and i_c -- or change the font Ia, Ib, Ic -- or use LaTeX $\ I_a,\, I_b,\, I_c\$.

 

[SammyS]

Maybe it's time to consider the Y-Δ transformation .

It's really not that difficult, particularly for the values here.

The delta is the triangle on the left. Those are the resistors in Loop a . Point A is equivalent to point A in your overall circuit. Point B is at the node which you labeled with voltage V₁ on some of your diagrams. Point C is at the node which you labeled with voltage V₂ on some of your diagrams.

The idea here is to replace that given set of resistors connected in that way, with a set of resistors in a "Y" configuration, as seen on the right. We need this set of resistors in the Y-configuration to behave exactly as the original set in the Δ-configuration in regards to nodes A, B, and C.

What this boils down to is that the resistances between the three nodes (taken pair-wise) must be the same for both configurations.

Starting with the Δ-configuration, find the resistance between A & B, between B & C, and between A & C .

For example: between A&B: The single resistor, r, is in parallel with the series combination of r and 3r. R_AB = (4r²)/(5r) = (4/5)r .

R_BC gives the same result: R_BC = (4/5)r .

For R_AC we have 3r in parallel with the series combination of r and r, giving R_AB = (3r⋅2r)/(5r) = (6/5)r .

Now figure out what R_A, R_B, and R_C must be in the Y-configuration on the right. For each pair here, we only have two resistors in series.

We must have:

R_A + R_B = (4/5)r

R_B + R_C = (4/5)r

R_A + R_C = (6/5)r​

What do you get for these?

 

[kaspis245]

I am not really sure what you are asking for. Is this what you want:
R_A = r
R_B = 3r
R_C = r

 

[SammyS]

I am not really sure what you are asking for. Is this what you want:
R_A = r
R_B = 3r
R_C = r

No.

Take the triangular (like Δ) configuration on the left. Find the equivalent resistance between points A and B. (Yes, I apologize for using B. It's not the same B as in your overall circuit.) I actually have found this resistance in post #15.

Do the same thing for the equivalent resistance between A and C, then also for the equivalent resistance between B and C.

 

[kaspis245]

I think I got this:

$R_A+R_B+R_C=\frac{r⋅3r+3r⋅r+r^2}{r+r+3r}=\frac{7}{5}r$

$R_A+R_B+R_C-(R_B+R_C)=R_A=\frac{7}{5}r-\frac{4}{5}r=\frac{3}{5}r$

$R_A+R_B+R_C-(R_A+R_C)=R_B=\frac{7}{5}r-\frac{6}{5}r=\frac{1}{5}r$

$R_A+R_B+R_C-(R_A+R_B)=R_C=\frac{7}{5}r-\frac{4}{5}r=\frac{3}{5}r$

 

[gneill]

Okay, now re-draw your circuit and plug in the new values in the appropriate locations:

Note that "r" can be treated as a scaling value since every resistance in the circuit is a multiple of it. So you can drop the "r" from the equations while you work out the minimums and maximums without losing generality. Just tack the r's back on at the end!

 

[kaspis245]

Well I started calculating total resistance of the circuit and got this expression:

$R=\frac{25.6+86.4x-81x^2}{12.8}r$

Is this correct? If so, how should I interpret it?

 

[gneill]

Well I started calculating total resistance of the circuit and got this expression:

$R=\frac{25.6+86.4x-81x^2}{12.8}r$

Is this correct? If so, how should I interpret it?

Looks okay. How do you mean "interpret it"? It's an expression for the total resistance. Clearly it's a function of x, where x lies between 0 and 1.

Why don't you pick a value for r (say, r = 1 Ohm), and plot the function R(x)? See if that gives you any ideas.

 

[kaspis245]

Ok, I see that $25.6+86.4x-81x^2$ is a parabola, so I need to find $y_{max}$ and $y_{min}$ in the interval $0≤x≤1$. Then the values are $y_{max}=48.64$ , $y_{min}=25.6$ and so $r_{max}=48.64Ω$, $r_{min}=25.6Ω$.

 

[gneill]

Don't forget that the values should be expressed as multiples of r.

 

[SammyS]

Ok, I see that $25.6+86.4x-81x^2$ is a parabola, so I need to find $y_{max}$ and $y_{min}$ in the interval $0≤x≤1$. Then the values are $y_{max}=48.64$ , $y_{min}=25.6$ and so $r_{max}=48.64Ω$, $r_{min}=25.6Ω$.

Don't forget to divide by 12.8, and as gneill points out, multiply by r.

The minimum value will occur at one of the end points. And you're correct about it occurring for x = 0. $\displaystyle \ R_{min} = \frac{25.6}{12.6}r=2r\$

However, the quadratic function, $\ 25.6+86.4x-81x^2\$ Opens downward and has its vertex near the middle of the interval [ 0, 1] . So you need to find the value of $\ R\$ at the vertex. There are several ways to do that .

For some easier numbers to work with:
$\displaystyle \ \frac{25.6+86.4x-81x^2}{12.8}\$

$\displaystyle \ =\frac{-405x^2+432x+128}{64}\ \ \$ (Multiplied the above by 5/5)
$\displaystyle \ =\frac{(-)5\cdot3^4x^2+3^3 2^4x+2^7}{2^6}\ \ \$ (prime factorizations of coefficients)
$\displaystyle \ =-5\left(\frac{9}{8}\right)^2x^2+6\left(\frac{9}{8}\right)x+2\$​

Now let u = (9/8)x to get $\displaystyle \ -5u^2+6u+2\$.

Find the maximum for this quadratic function or equivalently find its vertex.

 

[kaspis245]

$−5u^2+6u+2$
This quadratic function's maximum is 3.8, so I believe that $R_{max}=3.8r$.