# Help needed with Kirchhoff's Law, any help would be greatly appreciated.

Discussion in 'Homework Help' started by ace2015, Jan 12, 2018.

1. ### ace2015 Thread Starter New Member

Oct 24, 2017
15
0
In the attached schematic use Kirchhoff's Law to find out the magnitude & direction of the current flowing through resistor R5.

2. ### WBahn Moderator

Mar 31, 2012
22,620
6,729
Okay. So show all of your workings. I'm not seeing ANY workings.

3. ### ace2015 Thread Starter New Member

Oct 24, 2017
15
0
Hi
Sorry; I have spent virtually all day on this problem. I have scanned a few sheets of my workings out so far. Please see attached.

I have watched many videos on YouTube on this subject.

It's just that resistor R5.

I worked it out with just the 4 resistors in parallel without the R5 resistor in the circuit and then tried various methods to solve to get the current flowing through R5

Best regards

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4. ### WBahn Moderator

Mar 31, 2012
22,620
6,729
The introduction of R5 is the whole point of the problem. Without it, the circuit is nothing but a very simple series/parallel combination. But with it, it is neither series nor parallel, and so the techniques you have learned thus far can't be used to reduce it to a single resistor from which you can solve for the total current and eventually work your way back out.

So that is why you are told to apply KVL and KCL to the circuit. They will let you solve for all of the voltages and all of the currents without having to combine any of the components or make any simplifications to the circuit. But this comes at a cost -- namely you will have to solve some simultaneous equations.

As I think you already know, your "Kirchhoffs 1" is just wrong. You calculated the voltages under the assumption that I5 (IL in the drawing) is equal to zero. But that can only be if the voltage on both sides of it are the same, which they aren't. But this still is enough to tell you two useful pieces of information. First, the current through R5 is NOT zero (it's always possible that it could have been) AND you know what direction it has to flow. You also get another useful piece of information, though you probably don't realize it right now. If you take the voltage across R5 that you would have if R5 were removed (or just very, very large) and divide that by R5, you get the MOST that the current in R5 could possible be. The actual current will be smaller. So if you end up with a final answer that is larger than this, you know you have done something wrong.

Your second diagram goes a very long way to identifying all of the variables you need.

Let's take stock of the basic situation. How many unknowns in the circuit do you have? Well, there are a total of six components and each has both a voltage across it and a current through it. That's twelve quantities. Of these, you only know one -- the voltage across the source. So you have eleven unknowns and hence you need eleven equations.

Sounds pretty daunting. But you get five right off the bat because Ohm's Law gives you the relationship between the voltage across each resistor and the current through that resistor. So now you need just six more. What about KCL? You have four nodes in the circuit and you can apply KCL at each of them. That's four more equations. Then there's KVL. Any loop though the circuit that doesn't visit the same node twice is a path over which you can apply KVL. Look carefully and you should see that there are seven of these. That's a total of sixteen.

But we only have eleven unknowns. Either the system is over constrained, or some of our equations are not linearly independent. Hint: It's the latter. So we need to pick a subset of these sixteen equations that gives us eleven linearly independent ones. It turns out there are a few ways we can do this systematically, and you will learn very powerful forms of them shortly known as Node Voltage Analysis and Mesh Current Analysis.

Let's lead into mesh analysis this way.

First, pick all of the smallest loops -- the "window panes" of the circuit, if you care to think of it that way. You have these identified as L1, L2 and L3. Write KVL around those three loops using Ohm's Law and the appropriate branch current to express the voltage across each resistor as needed. This gives you three equations (from the loops) and you used the five Ohm's Law equations, so now you just need three more.

Now use KCL at all but one of the four nodes to express the relationships between the various branch currents. You can ignore one of them because it is guaranteed to be a linear combination of all the others. Here's your other three equations.

Now you have six equations in terms of the six unknown currents. It's all algebra from here.

ace2015 likes this.
5. ### ace2015 Thread Starter New Member

Oct 24, 2017
15
0
WBahn

Thank you so much for taking the time for writing such a detailed explanation.

Seriously; THANKS.

I will let you know how I get on with the circuit problem.

You have given me a lot of solid info for me to work with.

Best regards

6. ### WBahn Moderator

Mar 31, 2012
22,620
6,729
Good luck --and be sure not to get discouraged by the math you have to slog through at this point. You have only started learning what tools are at your disposal and right now the only tools you have are pretty blunt hammers. When the only tool you have is a hammer, every problem looks suspiciously like a nail.

But soon you will get some very powerful tools that are far more surgical and then you will have a couple of ways in which you will be able to write down just three equations in three unknowns and do so by inspection.

Having to go through this pain now will make you really appreciate the beauty and power of the techniques to come.

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7. ### ace2015 Thread Starter New Member

Oct 24, 2017
15
0
WBahn

I hope you are well.

I have spent a lot of time and effort on this problem and I have followed you guidance as best as I could.

Could you check my work and tell me if it is good or bad.

I will do a clean and neat write up once I am 100% sure it is correct. This is only a draft but as far as I know it is correct.

A LOT of WORK to say the least.

Best regards

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8. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,429
1,256
Unfortunately your solution (math) is WRONG. After you got your solution you should always check whatever the solution you get has any sense.

For exampel the KVL for L1 loop

5V = I2*22.5Ω + I4*22.5Ω = 55mA*22.5Ω + 55mA*22.5Ω = 2.475V

As you can see 5V is not equal to 2.475V.

But what is more important your equations are fine. You simply mess up the math somewhere.

Because when I plug your equations into the WolframAlpha I get the correct results.
The last equation X1 = X2+X3 is not needed.

http://www.wolframalpha.com/input/?i=-22.5*X_2+-+22.5*X_4+++5+=+0+,+-30*X_3+++50*X_6+++22.5*X_2+=+0+,+-50*X_6+-+25*X_5+++22.5*X_4+=+0,+X_6+=+(X_2+-X_4),+X_5+=+(X_3+++X_6),+X_1+=+X_2+X_3

Where
I1 = X1
I2 = X2
I3 = X3
I4 = X4
I5 = X5
IL = X6

And now we can check the KVL for L1
5V = I2*22.5Ω + I4*22.5Ω = 112.63mA*22.5Ω +109.6mA*22.5Ω = 5V

9. ### WBahn Moderator

Mar 31, 2012
22,620
6,729
Sorry it took so long to get back to you. I lost touch with this thread.

One thing that you need to start doing is checking your answers. Fortunately, in most engineering problems, the validity of the answer can usually be verified from the answer itself pretty easily.

You initial equations are correct. It would be a very good practice for you to get in the habit of properly tracking units, as many mistakes that you will make will mess up the units and can be caught very quickly IF the units are there to get messed up.

But it's very good that you didn't try to do a bunch of math at the same time that you set up your equations. Those equations are were all the EE is -- everything after that is just math. Which means that if you DID mess up those equations, doing all the math right will just be wasted effort. So you want your initial equations to be as bullet proof as possible and as easy to check the correctness of as possible.

Next, when you have the option, I'd recommend making the effort to harmonize your variables. Make I1 the current through R1 and I4 the current through R4. Use Is or I0 for the current in the supply. This will make it easier to keep everything straight, easier to spot check your results as you go along, and easier to track down and correct any errors you make.

Another very good habit to get into is to put bounds on your answer so that if you make a mistake in the math it is very likely going to mess up the result enough so that it violates your bounds. Plus, sometimes you can put the bounds in tight enough that you know that any answer within those bounds will be good enough and you can call it a day (though usually not on homework, but sometimes on multiple choice exams).

So what would the source current be under the following two conditions: (1) the bridge resistor is so large that it becomes an open, and (2) the bridge resistor is so small that it becomes a short?

If it's an open, then the source sees a resistance of (22.5 Ω + 22.5 Ω) || (30 Ω + 25 Ω) = 24.75 Ω and the source current would be 202.0 mA.

If it's a short, then the source sees a resistance of (22.5 Ω || 30 Ω) + (22.5 Ω || 25 Ω) = 24.70 Ω and a source current would be 202.4 mA

This tells us a couple of things. First, we have such a tight bound on the source current that it is unlikely that we can make a mistake and not catch it. Second, it's pretty clear that the presence of that bridge resistor has very little impact on the circuit.

We can now put an upper bound on the current in the bridge resistor by figuring out how much current is flowing in the short. That's a bit more work, but not a lot. The voltage at the bridge would be 2.397 V. That means the current in R1 would be 115.7 mA and the current in R3 would be 106.5 mA, making the bridge current about 9.2 mA. So any answer more than that is wrong.

We can put another limit on the bridge current by dividing the open circuit voltage across the bridge (which is 417 mV) by the 50 Ω resistor which give 8.3 mA. The actual current must be less than this, so we have a new, tighter upper bound on the current of 8.3 mA.

So now let's move on to page two over your work.

Whoa!!!

Where did all these values for all these currents come from?

Present your work in a logical sequence so that the reader (for instance, the person giving you a grade) can follow your work without having to bounce around trying to figure out what tortured path needs to be followed to understand your thinking and reasoning.

But notice that we can glance at these answers and KNOW that they are wrong because we know that a source current of only 96 mA is not possible.

Moving on to page three doesn't help much. But it looks like page 4 is where you actually started your work.

Your L1 and L2 work looks fine (except for tracking units), but I have NO idea where your starting equation in the L3 part comes from. It uses I2, I4, and I5, which aren't part of the same loop or attached to the same node. Nor do I have any idea where the 22.4 and 25.5 values come from.

Remember what I said about if your setup isn't right, you can't catch it after that? Every time you start a new section of work, get in the habit of asking your self if your initial starting point really is correct. Just double check it after you've written down and verify each term and each sign in it.

I'm not going to go any further checking your work because it would be pointless. So take what you've learned and fix up what you've got.

10. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,548
430
ace2015, your mistake happened when you moved equation L3 from page 1 to page 2. Perhaps you substituted the relation labeled N3: on page 1 into equation L3 from page 1. If that's what you did, you didn't do it correctly. At any rate, equation L3 on page 1 is correct, but it isn't correct on page 2.