Kirchoff's conundrum

[Simon]

I have been given the attached questions to solve. I was assuming they would expect me to use Kuirchoffs laws however after a lot of attempts at trying to write equations I find I cannot write anything that will allow me to work this one out. However intuitively I already know the answer the potentiometer needs to be at 50% position because when it is 5K load and the 5K in the pot will produce 2.5 K leaving 5K in series which is a 2 to one ratio which is what is required to produce 3 V on the pot wiper.

The question doesn't seem to ask me to solve it in any particular way but I obviously don't want to fail the question. If I were asked to work out the percentage at another random voltage I would not be able to calculate it intuitively.

Can this thing be solved mathematically does it even involve Kirchoff or is it just writing of some simultaneous equations that will work out the resistance set up.

[Simon]

I'll give you a clue.
The pot is shown in the exact position where 3V will be output.

ha, ha, ha. Like I said I know the solution. I have to demonstrate it though, I guess....

[bobaruni]

I deleted my post after reading again that you knew that lol

[AndyC_772]

Split the pot into two separate resistors, say Rt and Rb, where Rt + Rb = 10k, and the 5k resistor is in parallel with Rb.

You can then apply Kirchhoff's current law to the node where the three resistors meet.

Given that you know the voltage across the 5k resistor, you also know the voltage across the top half of the pot (ie. Rt). This gives you the current through Rt, which equals the total current through Rb and the 5k resistor.

Since you already know the voltage across the 5k resistor, you can calculate the current through it without reference to the pot at all. Subtract this from the total flowing through Rt, and you get the current through Rb.

Now, you know two relationships between Rb and Rt, which should allow you to solve for them.

[Brumby]

Simple algebra will solve it for any desired voltage.

Just start by considering the pot as 2 resistors.

[Simon]

OK, I thought it would work like that, I have been trying to build into my equations the fact that I know the sum of both resistors is 10K but I still seem to tie myself in knots.

[Brumby]

Try using 'x' as one resistor and '10K -x' as the other.

[Simon]

Try using 'x' as one resistor and '10K -x' as the other.

Yes I've been trying that but not come up with anything yet that uses it properly.

[Brumby]

I'm out at the moment. .. and my shopping docket is too small.

I'll write it out when I get home ... unless someone beats me to it.

[Simon]

Hopefully i beat you to it

[Christopher]

What book is this from? Looks cool

[Simon]

What book is this from? Looks cool

it's from the PDF that is my assignment for an online based HNC course.

Oh and I've nearly solved it I tthink after a leap of faith that looked like the math would get harder but instead gets easier.

[Brumby]

Hopefully i beat you to it

There's your challenge.

If it helps, the way I was going to go about it was as follows:
1.  Derive an expression that gives the resistance of the two parallel elements.
2. Use this in a voltage divider calculation.
3. Shuffle the result around until you get 'x' on the LHS all by itself.  (Just follow the rules and be careful)

Since everything else has a given value , you have a number that you just use to find that position on the pot. (We ARE talking about a linear taper, I hope)

[coppice]

Try using 'x' as one resistor and '10K -x' as the other.

Yes I've been trying that but not come up with anything yet that uses it properly.

We want 1/3 of the input voltage, so we want the point where the upper resistor, x fulfils

x = 2*(10k-x)*5k/(10k - x + 5k).

Tidy this up:
x = (100M - 10k*x)/10k - x + 5k)

15k*x - x*x = 100M - 10k*x

25k*x - x*x - 100M = 0

Now extract the roots:
(x - 20k)(-x + 5k) = 0

So, x = 5k or 20k are the potential answers. 20k is a clearly impossible, as the companion resistance would be negative, so the 5k point on the 10k pot is the position you want.

[Simon]

yes thats one way although not how they want it i suspect

[Brumby]

Try using 'x' as one resistor and '10K -x' as the other.

Yes I've been trying that but not come up with anything yet that uses it properly.

We want 1/3 of the input voltage, so we want the point where the upper resistor, x fulfils

x = 2*(10k-x)*5k/(10k - x + 5k).

Tidy this up:
x = (100M - 10k*x)/10k - x + 5k)

15k*x - x*x = 100M - 10k*x

25k*x - x*x - 100M = 0

Now extract the factors:
(x - 20k)(-x + 5k) = 0

So, x = 5k or 20k are the potential answers. 20k is a clearly impossible, as the companion resistance would be negative, so the 5k point on the 10k pot is the position you want.

LOL

I did like that - but it gets very messy if you pick some other target voltage ... say 4.3V

[Zero999]

It doesn't require Kirchoff's law or any simultaneous equations.

I did this at work, when I was bored and left my notes there so can't post it immediately. I'd do it again but I'm going out in half an hour.

Derive the formula for Vout, the voltage across the 5k resistor. The 10k pot. is modeled as a potential divider. The 5k resistor is in parallel with the lower potential divider resistor R2.

Rearrange the formula to make R2 the subject.

[coppice]

Try using 'x' as one resistor and '10K -x' as the other.

Yes I've been trying that but not come up with anything yet that uses it properly.

We want 1/3 of the input voltage, so we want the point where the upper resistor, x fulfils

x = 2*(10k-x)*5k/(10k - x + 5k).

Tidy this up:
x = (100M - 10k*x)/10k - x + 5k)

15k*x - x*x = 100M - 10k*x

25k*x - x*x - 100M = 0

Now extract the factors:
(x - 20k)(-x + 5k) = 0

So, x = 5k or 20k are the potential answers. 20k is a clearly impossible, as the companion resistance would be negative, so the 5k point on the 10k pot is the position you want.

LOL

I did like that - but it gets very messy if you pick some other target voltage ... say 4.3V

If you change the voltage you only need to change the 1:2 ratio in my first line.

[Brumby]

That's fine until you get to the line:

Now extract the factors:

... but there is a simple way of stepping around that.

[AndyC_772]

You know the two halves of the pot add up to 10k, so Rb + Rt = 10k.

You also know the voltage across the 5k resistor, so you know the current through it = 3/5k = 600uA.

KCL at the node where the three resistors meet, gives:

6 / Rt = 600u + 3 / Rb

Since Rt = 10k - Rb, substitute in to get:

6 / (10k - Rb) = 600u + 3 / Rb

Multiply through by (10k - Rb):

6 = 6 - 600u Rb + 30k/Rb - 3

Cancel the sixes and multiply through by Rb:

0 = -600u Rb^2 - 3 Rb + 30k

Solve by quadratic formula to get Rb = -10k or Rb = 5k. Since the former is physically impossible, the latter is the only possible answer.

[Mechatrommer]

 

[coppice]

That's fine until you get to the line:

Now extract the factors:

... but there is a simple way of stepping around that.

There are numerous ways to solve the problem. I used only the series and parallel resistors rules, and simple algebra you learn early in high school.

[Simon]

Well I have my head around the quadratic equation method. Kirchoff still eludes me. I'll have another go as i think that is the method I am expected to use.

[IanB]

Well I have my head around the quadratic equation method. Kirchoff still eludes me. I'll have another go as i think that is the method I am expected to use.

The worked solution given by AndyC uses Kirchhoff's rules. The rules are fundamental, so making use of them is unavoidable.

[coppice]

Well I have my head around the quadratic equation method. Kirchoff still eludes me. I'll have another go as i think that is the method I am expected to use.

The worked solution given by AndyC uses Kirchhoff's rules. The rules are fundamental, so making use of them is unavoidable.

All solutions use Kirchhoff's rules, whether they are explicitly stated in the solution or not.

[Simon]

Well I have solved it by what I deemed to be a Kirchoff method however I still end up with a simultaneous equation. I expect many of the possible results are going to end in a simultaneous equation as a result is tied to the fact that there are 2 variables that are at the same time independent and linked together. So I think I will leave it at that.

[AndyC_772]

Of course you get simultaneous equations, there are two resistors that need solving: the top half of the pot, and the bottom half.

You know they add up to 10k, so that's one of the two necessary pieces of information that relates them. Kirchhoff's current law applied at the node where the three resistors meet is the other.

Two simultaneous equations, two unknowns. You might be able to 'hide' that fact by calling the two halves of the pot, say, X and 10k-X, but that doesn't change the underlying maths.

[bson]

The pot is two resistors, Rp1 and Rp2 such that Rp1+Rp2 = 10k
For a position P (0-1), Rp1=P*10k, Rp2=(1-P)*10k

Their voltages and currents Vp1,Vp2,Ip1,Ip2
The wiper resistor is Rw and its voltage,current Vw, Iw

KVL:
Vp1 + Vp2 = 9
Vp2 = Vw = 3
Vp1 = 9-3 = 6

KCL (direction downwards at wiper node):
Ip1 - Ip2 - Iw = 0
Ip1 = Ip2+Iw
Ip1 = 6/Rp1
Iw = 3/Rw
Ip2 = 3/Rp2

This looks quite overdetermined to me, but looks can be deceiving.  We want to solve for P.

Vp1 = Ip1 * Rp1 = (Ip2 + Iw) * P * 10k = 6
Ip2 = Vp2 / Rp2 = 3/(10k-Rp1) = 3/(10k-P*10k)
Iw = 3/5k

=>

((3/((1-P)*10k)+3/5k)*P*10k = 6
30kP/((1-P)*10k) + 30kP/5k = 6
3P/(1-P) + 6P = 6
0.5P/(1-P) + P - 1 = 0

Type this into wolframalpha... manually, since I can't get it to properly parse a urlencoded expression.

P=0.5; i.e. halfway (clearly P=2 is an uninteresting root)

Barring errors of course. 

[Brumby]

Got back too late last night.  Don't think I need to worry......

[IanB]

Well I have solved it by what I deemed to be a Kirchoff method however I still end up with a simultaneous equation. I expect many of the possible results are going to end in a simultaneous equation as a result is tied to the fact that there are 2 variables that are at the same time independent and linked together. So I think I will leave it at that.

It seems to me that you always get a quadratic equation. But there is no need for simultaneous equations since there is only one unknown variable--the slider position on the pot.

Here is what seems to me to be a simple solution using Kirchhoff's current law:

Let \$R\$ be the unknown resistance of the lower section of the potentiometer in \$\text{k}\Omega\$. Then the resistance of the upper section is \$10-R\$.

A current balance (KCL) around the potentiometer wiper junction gives:
$$\frac{6}{10-R}=\frac{3}{5}+\frac{3}{R}$$
We can cancel the 3's:
$$\frac{2}{10-R}=\frac{1}{5}+\frac{1}{R}$$
And then multiply it out:
$$10R=(10-R)(5+R)$$
Simplifying and rearranging:
$$\begin{align*}
& & 10R &= 50-5R+10R-R^2 \\
&\Rightarrow & R^2+5R-50 &= 0 \\
&\Rightarrow & (R-5)(R+10) &= 0 \\
&\Rightarrow & R &= 5 \text{ or } {-10} \end{align*}$$
Choosing the positive root we find that the potentiometer setting has \$R=5\text{ k}\Omega\$.

[Brumby]

Here is what seems to me to be a simple solution using Kirchhoff's current law:

Let \$R\$ be the unknown resistance of the lower section of the potentiometer in \$\text{k}\Omega\$. Then the resistance of the upper section is \$10-R\$.

A current balance (KCL) around the potentiometer wiper junction gives:
$$\frac{6}{10-R}=\frac{3}{5}+\frac{3}{R}$$
We can cancel the 3's:
$$\frac{2}{10-R}=\frac{1}{5}+\frac{1}{R}$$
And then multiply it out:
$$10R=(10-R)(5+R)$$
Simplifying and rearranging:
$$\begin{align*}
& & 10R &= 50-5R+10R-R^2 \\
&\Rightarrow & R^2+5R-50 &= 0 \\
&\Rightarrow & (R-5)(R+10) &= 0 \\
&\Rightarrow & R &= 5 \text{ or } {-10} \end{align*}$$
Choosing the positive root we find that the potentiometer setting has \$R=5\text{ k}\Omega\$.

Very nicely done.  A solution that is simple to work and follows all requirements - inferred or not.

(OK, I admit it ... mine would have been messier.)

[JacquesBBB]

Well I have solved it by what I deemed to be a Kirchoff method however I still end up with a simultaneous equation. I expect many of the possible results are going to end in a simultaneous equation as a result is tied to the fact that there are 2 variables that are at the same time independent and linked together. So I think I will leave it at that.

Let \$R\$ be the unknown resistance of the lower section of the potentiometer in \$\text{k}\Omega\$. Then the resistance of the upper section is \$10-R\$.

A current balance (KCL) around the potentiometer wiper junction gives:
$$\frac{6}{10-R}=\frac{3}{5}+\frac{3}{R}$$

I totally agree with the solution posted by  IanB,  and just want to add some additional  considerations :

I have been given the attached questions to solve. I was assuming they would expect me to use Kuirchoffs laws however after a lot of attempts at trying to write equations I find I cannot write anything that will allow me to work this one out. However intuitively I already know the answer the potentiometer needs to be at 50% position because when it is 5K load and the 5K in the pot will produce 2.5 K leaving 5K in series which is a 2 to one ratio which is what is required to produce 3 V on the pot wiper.

The problem with the intuitive resolution, is that  it can be applied  here only because the problem is simple and  the values
of the voltage and resistors are chosen in a way that all the calculus is simple.
But if you  take
8.34 V ;  2.55 V ;  12kR ; 4.7 kR
instead of
12 V ; 3V ; 10 kR ; 5 kR;
you will see that there will no longer be an intuitive solution, and only a proper setting of the equations will
lead to the result.

Now faced to any problem, you need to ask yourself about the number of unknowns in the problem.
An electrical circuit  with  a network of resistors and a voltage source is entirely determined once the  values of the resistors
and voltage source are known : all node voltage can be computed as well as all  currents in the  branches of the network.

Here in the original setting, you have two unknowns \$R\$, the lower resistance of the pot and \$R'\$ the upper part.

You thus need  two additional  relations to solve the circuit.

The first one is given by the fact that the two resistors are linked as the two sides of the pot  :
$$ R + R' = 10 $$
(by the way, in agreement with IanB, I advise to drop the units when you set up your equations, as leaving them
will just obfuscate the computations).
So \$R'=10-R \$ and you are left with a single unknown.

The second relation will be provided by the additional information that is provided with the  3V voltage at one node.
The best way to set up the equation is to avoid  the introduction of any additional variable. So
$$\frac{6}{10-R}=\frac{3}{5}+\frac{3}{R}$$

is a good relation.

Once this is written, you know that your problem is solved. You have one equation, with only one unknown.
The remaining part is just calculus.

Face to any problem of the kind, even very complicated, you have to distinguish the two phases :
A) proper physical modelisation and setting of the equations
B) resolution of the equations.

Most often, the difficult part is A), but mixing A) and B) just leads to additional confusion.