# Total resistance in series-parallel networks

Discussion in 'Homework Help' started by tux, Dec 2, 2012.

1. ### tux Thread Starter New Member

Nov 11, 2012
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Homework question is attached.
My "solution" is attached. It is not finished.

Is total resistance calculated correctly?

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2. ### JoeJester AAC Fanatic!

Apr 26, 2005
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You have three branches in that circuit ....

A : R1, R2, R3, R4, R5, R6
B : R7, R8, R9
C : R10, R11, R12

When your doing this exercise it's best to do each branch individually and then the three branches in parallel.

You also know from KCL that the total current will be the sum of the three branch currents.

(1) What is the current in the branch (B) left of the battery?

(2) What is the current in the branch (C) right of the battery?

(3) What is the current in the branch (A) moving downward and right of the battery?

Last edited: Dec 2, 2012
3. ### tux Thread Starter New Member

Nov 11, 2012
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R total in branch B is 3 ohms?

4. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Correct ... and that makes the current in that branch?

5. ### WBahn Moderator

Mar 31, 2012
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I'm just gonna comment some on your original work. JoeJester's doing a fine job of helping you along and I won't muddy the waters.

In you original work, you were very, very sloppy with your units. Some places you included them, some places you didn't, some places they were correct, and some places they weren't. Get the in the habit of always, always, always tracking your units throughout your calculations and checking them frequently. If your units don't work out, you KNOW the answer is wrong. Why not stack the deck in your favor and catch mistakes early and also catch the majority of mistakes you make since the majority of mistakes screw up the units>

Second, always, always, always ask if the answer (including intermediate answers) make sense. Do this by estimating (or at least bounding) the answer before you start and then asking if the answer you get is consistent with that.

In this case, your computation of R(7,8,9) had 3Ω in parallel with 24Ω. If it had been two 3Ω resistors you know the answer would have been 1.5Ω. You also know that the answer has to be less than 3Ω. Since 24Ω is significantly greater than 3Ω (heck, is 8x the size), you can expect the parallel combination of them to be closer to 3Ω than to 1.5Ω. So when you got 1.67Ω, warning bells should have gone off.

Also, you would be well served to write your 1's and 4's more carefully. Your 1's often look like carat symbols or upside down v's or u's. Your 4's... well, I don't know what they look like. They don't look like a 4 and I suspect that many people reading your work would be confused by them.

Speaking of people reading your work -- you simply stated the value of R(1,2,4,6) without even setting it up. First, this is confusing because it implies that R3 and R5 aren't part of this equivalent resistance. Second, let's say that you had done something wrong and gotten 5Ω instead of 6Ω. If you've set it up correctly but just made a simple math error or just wrote down the answer wrong (which is what I suspect happened with R(7,8,9), then you can still get partial credit. For instance, I can see what errors you made, namely overlooking the short and then making a math error, and so I can give you some partial credit for what you did right. But if you just put an answer, then it is all or nothing (at best).

6. ### tux Thread Starter New Member

Nov 11, 2012
8
0
@WBahn

Thank you for suggestions. I will try to improve my handwriting.

I have written only part of solution so anyone who is interested in helping me can see what is wrong without taking too much time to figure out what I want.

I still don't understand why is it 3 ohms.
Why don't we use 12 ohms resistors in calculation?

7. ### WBahn Moderator

Mar 31, 2012
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The less of your solution you provide, the longer it takes to figure out what you did wrong because we have to reconstruct what we think you probably did instead of just being able to look at what you actually did.

Notice that there is a wire going that is providing an alternative path for current to flow. Think of that as a 0 ohm resistor. What is the equivalent resistance of 24 ohms in parallel with 0 ohms?

8. ### tux Thread Starter New Member

Nov 11, 2012
8
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(24*0) / (24+0)=0

What if there is 3 ohms resistor?

Then resistors 3 ohms and 3 ohms are in series, the same is true for 12 ohms.
R total = (3 ohms + 3 ohms) parallel to (12 ohms + 12 ohms)=6 parallel to 24?

9. ### WBahn Moderator

Mar 31, 2012
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You mean a 3 ohm resistor instead of a short?

If so, then you have the two 12 ohm resistors in series, giving 24 ohms, in parallel with 3 ohms, giving 2.67 ohms (IIRC), which is then in series with the remaining 3 ohm resistor giving 5.67 ohms.

Again, asking if the answer makes sense if very useful. Without the 12 ohm resistors at all, you have two 3 ohm resistors in series, giving 6 ohms. If you then put an alternate path in parallel to one of the resistors, the total resistance has to go down.

10. ### tux Thread Starter New Member

Nov 11, 2012
8
0
Thank you very much.

11. ### WBahn Moderator

Mar 31, 2012
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Use units!

(24Ω*0Ω)/(24Ω+0Ω) = 0Ω

12. ### JoeJester AAC Fanatic!

Apr 26, 2005
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You need to work on answering how much currient is flowing in that branch and then show your work for the other two branches ...

tux likes this.

Nov 11, 2012
8
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I=2 a
i 10=6 a
i 6=0.5 a

14. ### WBahn Moderator

Mar 31, 2012
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Once again, invoke the "does the answer make sense" test.

Doesn't whatever current is flowing in each branch all have to come from the battery? If so, then doesn't the current I have to be at least as large as any of the individual branch currents?

Also, the unit symbol for amperes is a capital A. A lower case a is the prefix for atto.

Apr 26, 2005
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