# Finding Total resistance in a circuit.

1. Feb 24, 2013

### doc2142

This is my first engineering course of many to come and I was hopping someone can shed some light on how to add some of those resistors in this network.

1. The problem statement, all variables and given/known data

See attached image, I must find the power of the 10 ohm resistor.

2. Relevant equations

3. The attempt at a solution

This class is circuit analysis. So all I know is KCL, KVL, CDR, VDR, and of course V=IR

Now I tried to Find R total, but I don't really know how to. Here is what I tried.

Starting from Right side, the 10 and 2 are in serial. The total which is 12 is in parallel to the 24. The Total is serial to 2. Now I am stuck here I don't think the total is serial to the 12 because the short circuit at the bottom. The only thing I came up with is that both 12's on the bottom are in parallel? is that possible since they are both connected using 2 points. If so how would I redraw the circuit, would both combined be before the 4 ohm or after the 4 ohm?

If finding Rt isn't needed if you can push me in a different direction please say so. I went through it, and I don't think KCL would work here. So I have no idea of any other way to approach this then finding KCL and then using CDR to find the current and then finding the power of the 10 ohm resistor.

Thank you.

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2. Feb 24, 2013

### jsgruszynski

Several ways to do it. You could write loop equations and solve for the current in the loop that contains the resistor in question. Then P = I^2 R.

You could also reduce the circuit to find the total current coming from the battery, and then current divide back to the target resistor. Then P = I^2 R.

3. Feb 24, 2013

### doc2142

What do you mean by loop it? do you mean just using the resistors on the outside lines? like the 7 - 2 - 2 - 10 with battery only? that will work?

Also I am trying to reduce it to find the current and then divide the current up, however, I don't know how to find total resistance.

4. Feb 24, 2013

### jsgruszynski

5. Feb 24, 2013

### doc2142

I am sorry this is my first 3 weeks in the class. I am having hard time applying any of those into my question since mine is a bit different. Can you elaborate a bit more on this.

6. Feb 24, 2013

### jsgruszynski

Per mesh analysis:

1. Circuit loops
2. Draw current loops for each circuit loop
3. Write KVL/Ohm's law relationships for each loop
4. Each loop will result in an equation that is part of a system of simultaneous equations
5. Solve set of equations with your favorite method (simply algebraic manipulation or matrix math)
6. Mesh currents pop out as the answer
7. Pick the specific mesh current for the resistor and calculate the power

Can't get much more specific that this (without simply solving it for you). If you follow the procedure mechanically it will just work and give the answer.

Alternatively you can avoid some of the algebra/matrix math by reducing the circuit to a single equivalent resistor and the voltage source. Calculate the total current. Then apply current divider rules to successively added back reduced circuit elements (reverse your circuit reduction, step by step, splitting off the current as you approach your original circuit and the target resistor).

7. Feb 24, 2013

### doc2142

Ok thank you, one quick question though for the first loop.

CW loop -24 + 7(I 1) + 4 (I 1 - I 2) + 12 (I ?) The last part, does the current split again after going through both 12 ohms? so would it be 12 (I 1 - I 2 - I 2 )?
or would it just be 12 (I 1 - I 2)?

8. Feb 24, 2013

### jim hardy

You're doing fine so far. That is a valid approach, simplifying your circuit one step at a time. Sometimes it is easier than writing equations but usually not .
Yes they are.

good observation. They cannot possibly have different voltages across them.

Remember your definition of a node. Put your finger on the wire and everywhere you can move it without entering a circuit element is one node.
Look at that big node - circuit common symbol, battery's negative, outboard end of both Mr 12's, bottom of Mr 10 and bottom of Mr 24 are one node. (Sorry for the slang - i name things as if i were talking to them, it helps me think straight)

So redraw those two 12's as a single 6 straight down from bottom of Mr 4 to that circuit common node.
They threw a confusion factor at you by drawing them horizontal. It it helps, first just swing them down vertical by dragging their outboard ends around the node to just below their inboard ends.. it's all the same node...
you'll say to yourself 'it cant be that simple...' but it is. Learning is largely just believing what you already know...
KCL and KVL will work. And they will give same answer you get by simplifying the circuit in the iterative approach you started. I suggest working it both ways, it'll do wonders for your confidence in both methods.

redraw the circuit after each simplification. Plan on using quite a bit of paper, and be very neat about your sketches.

1st simplification - replace those two horizontal 12's with a vertical 6, as we discussed earlier
2nd simplification - replace that vertical 6 and the vertical 4 above it with a vertical 10.

now start working from right , like you did.
3nd simplification - replace the rightmost vertical 10 and horizontal 2 with a vertical 12
4th simplification -that vertical 12 and vertical 24 are in parallel, replace them with a single vertical 8
5th simplification - That 8 and the horizontal 2 are in series, replace them with a vertical 10.
6th simplification - replace the two vertical 10's with a vertical 5

Now you have a simple circuit - two resistors in series, totalling 12 ohms, across a 24 volt source.

So you know voltage at the junction of the two resistors, looks to me like ten volts which means two amps through Mr 7..

Now work backward, un-simplifying your circuit in reverse order we simplified it.
You will see voltage at every point appear unambiguously as you go.

Having done that, it would be wise to solve the problem using both node voltage and mesh current approaches, just to build your skill. You'll be expected to be able to do that when you go to work in industry.

When you get to point all 3 methods give you the same answer on first try, you are getting very good. Practice!

old jim

9. Feb 24, 2013

### doc2142

Wow I can't thank you enough you just showed me a very neat trick to simplifying some of those circuits. I wish my professor gives us pointers like this.

Those 2 12's were really giving me the problem but after I did what you told me, made the problem so much easier.

As for the Mesh I don't think we have gone over that yet, we are still in our first 2 weeks of our circuit analysis so I don't believe we need to know that yet. However, I will try to go over it and learn it.

I am at the end of my 2nd year in EE bachelor degree, and my next 2 years I heard the hardest so I should be coming here more often in the next 2 years to try and get help. The first 2 years where the calculus, electricity and magnetism and DFQ were easy, now off to the good hard stuff!

Again thanks again, you just added a whole new way for me to look at a problem.