Kirchoff's conundrum - Page 1
Forum section: electronics / beginners

author: Simon
author_rank: Unknown
timestamp: August 20, 2016, 08:14:56 am
content: 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.

author: Simon
author_rank: Unknown
timestamp: August 20, 2016, 08:25:21 am
content: Quote from: bobaruni on August 20, 2016, 08:24:25 amha, ha, ha. Like I said I know the solution. I have to demonstrate it though, I guess....

author: bobaruni
author_rank: Regular Contributor
timestamp: August 20, 2016, 08:27:06 am
content: I deleted my post after reading again that you knew that lol

author: AndyC_772
author_rank: Super Contributor
timestamp: August 20, 2016, 08:28:35 am
content: 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.

author: Brumby
author_rank: Unknown
timestamp: August 20, 2016, 08:32:32 am
content: Simple algebra will solve it for any desired voltage.Just start by considering the pot as 2 resistors.

author: Simon
author_rank: Unknown
timestamp: August 20, 2016, 08:32:51 am
content: 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.

author: Brumby
author_rank: Unknown
timestamp: August 20, 2016, 08:46:00 am
content: Try using 'x' as one resistor and '10K -x' as the other.

author: Simon
author_rank: Unknown
timestamp: August 20, 2016, 08:47:29 am
content: Quote from: Brumby on August 20, 2016, 08:46:00 amYes I've been trying that but not come up with anything yet that uses it properly.

author: Brumby
author_rank: Unknown
timestamp: August 20, 2016, 08:55:15 am
content: 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.

author: Simon
author_rank: Unknown
timestamp: August 20, 2016, 08:55:57 am
content: Hopefully i beat you to it

author: Christopher
author_rank: Frequent Contributor
timestamp: August 20, 2016, 09:06:52 am
content: What book is this from? Looks cool

author: Simon
author_rank: Unknown
timestamp: August 20, 2016, 09:10:16 am
content: Quote from: Christopher on August 20, 2016, 09:06:52 amit'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.

author: Brumby
author_rank: Unknown
timestamp: August 20, 2016, 09:14:05 am
content: Quote from: Simon on August 20, 2016, 08:55:57 amThere'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)

author: coppice
author_rank: Super Contributor
timestamp: August 20, 2016, 09:17:35 am
content: Quote from: Simon on August 20, 2016, 08:47:29 amWe want 1/3 of the input voltage, so we want the point where the upper resistor, x fulfilsx = 2*(10k-x)*5k/(10k - x + 5k).Tidy this up:x = (100M - 10k*x)/10k - x + 5k)15k*x - x*x = 100M - 10k*x25k*x - x*x - 100M = 0Now extract the roots:(x - 20k)(-x + 5k) = 0So, 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.

author: Simon
author_rank: Unknown
timestamp: August 20, 2016, 09:20:34 am
content: yes thats one way although not how they want it i suspect

author: Brumby
author_rank: Unknown
timestamp: August 20, 2016, 09:28:15 am
content: Quote from: coppice on August 20, 2016, 09:17:35 amLOLI did like that - but it gets very messy if you pick some other target voltage ... say 4.3V

author: Zero999
author_rank: Super Contributor
timestamp: August 20, 2016, 09:42:31 am
content: 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.

author: coppice
author_rank: Super Contributor
timestamp: August 20, 2016, 01:20:11 pm
content: Quote from: Brumby on August 20, 2016, 09:28:15 amIf you change the voltage you only need to change the 1:2 ratio in my first line.

author: Brumby
author_rank: Unknown
timestamp: August 20, 2016, 01:28:15 pm
content: That's fine until you get to the line:Quote from: coppice on August 20, 2016, 01:20:11 pm... but there is a simple way of stepping around that.

author: AndyC_772
author_rank: Super Contributor
timestamp: August 20, 2016, 03:36:41 pm
content: 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 / RbSince Rt = 10k - Rb, substitute in to get:6 / (10k - Rb) = 600u + 3 / RbMultiply through by (10k - Rb):6 = 6 - 600u Rb + 30k/Rb - 3Cancel the sixes and multiply through by Rb:0 = -600u Rb^2 - 3 Rb + 30kSolve by quadratic formula to get Rb = -10k or Rb = 5k. Since the former is physically impossible, the latter is the only possible answer.

author: coppice
author_rank: Super Contributor
timestamp: August 20, 2016, 04:52:51 pm
content: Quote from: Brumby on August 20, 2016, 01:28:15 pmThere 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.

author: Simon
author_rank: Unknown
timestamp: August 20, 2016, 08:40:08 pm
content: 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.

author: IanB
author_rank: Super Contributor
timestamp: August 20, 2016, 08:52:20 pm
content: Quote from: Simon on August 20, 2016, 08:40:08 pmThe worked solution given by AndyC uses Kirchhoff's rules. The rules are fundamental, so making use of them is unavoidable.

author: coppice
author_rank: Super Contributor
timestamp: August 20, 2016, 09:04:14 pm
content: Quote from: IanB on August 20, 2016, 08:52:20 pmAll solutions use Kirchhoff's rules, whether they are explicitly stated in the solution or not.

author: Simon
author_rank: Unknown
timestamp: August 20, 2016, 09:15:54 pm
content: 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.

author: AndyC_772
author_rank: Super Contributor
timestamp: August 20, 2016, 09:32:31 pm
content: 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.

author: bson
author_rank: Unknown
timestamp: August 20, 2016, 09:58:56 pm
content: The pot is two resistors, Rp1 and Rp2 such that Rp1+Rp2 = 10kFor a position P (0-1), Rp1=P*10k, Rp2=(1-P)*10kTheir voltages and currents Vp1,Vp2,Ip1,Ip2The wiper resistor is Rw and its voltage,current Vw, IwKVL:Vp1 + Vp2 = 9Vp2 = Vw = 3Vp1 = 9-3 = 6KCL (direction downwards at wiper node):Ip1 - Ip2 - Iw = 0Ip1 = Ip2+IwIp1 = 6/Rp1Iw = 3/RwIp2 = 3/Rp2This looks quite overdetermined to me, but looks can be deceiving. We want to solve for P.Vp1 = Ip1 * Rp1 = (Ip2 + Iw) * P * 10k = 6Ip2 = Vp2 / Rp2 = 3/(10k-Rp1) = 3/(10k-P*10k)Iw = 3/5k=>((3/((1-P)*10k)+3/5k)*P*10k = 630kP/((1-P)*10k) + 30kP/5k = 63P/(1-P) + 6P = 60.5P/(1-P) + P - 1 = 0Type 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.

author: Brumby
author_rank: Unknown
timestamp: August 21, 2016, 01:42:09 am
content: Got back too late last night. Don't think I need to worry......

author: IanB
author_rank: Super Contributor
timestamp: August 22, 2016, 05:07:16 am
content: Quote from: Simon on August 20, 2016, 09:15:54 pmIt 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\$.

author: Brumby
author_rank: Unknown
timestamp: August 22, 2016, 07:15:03 am
content: Quote from: IanB on August 22, 2016, 05:07:16 amVery 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.)

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: August 22, 2016, 10:00:15 am
content: Quote from: IanB on August 22, 2016, 05:07:16 amI totally agree with the solution posted by IanB, and just want to add some additional considerations :Quote from: Simon on August 20, 2016, 08:14:56 amThe problem with the intuitive resolution, is that it can be applied here only because the problem is simple and the valuesof the voltage and resistors are chosen in a way that all the calculus is simple.But if you take8.34 V ; 2.55 V ; 12kR ; 4.7 kRinstead of12 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 willlead 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 resistorsand 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 themwill 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 equationsB) resolution of the equations.Most often, the difficult part is A), but mixing A) and B) just leads to additional confusion.