# pls help me solve this problem, imaged included thanks!

Discussion in 'Homework Help' started by DreamFM, Aug 5, 2012.

Aug 5, 2012
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Feb 11, 2012
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3. ### cork_ie Active Member

Oct 8, 2011
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Looks like a doctors prescription to me.

You will get far more help if the sketch was tidied up a little and less ambiguous.
What is the component in the top left hand corner?
Is the lower connection of this component meant to be unconnected?

4. ### absf AAC Fanatic!

Dec 29, 2010
1,898
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Your original diagram looks like fig A and I redrawn it as fig B. In the redrawn diagram, R15 and R13 are in series and you can replace them with just one resistor. R16 & R17 are in parallel and can also be replaced with one resistor.

I hope the final diagram after simplification is easy for you to solve.

Allen

• ###### RES NET.PNG
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Last edited: Aug 5, 2012
5. ### DreamFM Thread Starter New Member

Aug 5, 2012
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Hey thanks, yupp, it is exactly like that.

I have redrawn my diagramme to Figure B, and i cant solve from there.

The problem is that it seems that 18,19,14 are in parallel, and 17 and 16 are in parallel.

But isnt it that 18, 19, 17 and 16 are all in parallel as well? Because i use the current from the source and as long as there is a split or current (node), i assume that the two elements are in parallel, which work perfectly with other circuits.

But not this one..

6. ### mlog Member

Feb 11, 2012
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I think the upper left resistor is 6Ω.

7. ### DreamFM Thread Starter New Member

Aug 5, 2012
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hi, i read the reply from you from the other thread.

the diagramme i have drawn is like the one being posted by the other member.

The main thing is that i cant solve it from there..

Alot of questions.

For example, when the current enters the first node, splitting into right, down(12ohm) and left(6 ohm), the current along the left circuit, does it travel straight to the end voltage? Because it encounters another node right? shouldnt it split too, before it encounter the 4 ohm resistor??

8. ### mlog Member

Feb 11, 2012
276
36
Do you know the technique (equation) for combining 3 resistors in parallel? Combine R14, R18, and R19. Just because there are other parallel combinations doesn't mean you can't combine those 3.

Do you know how to combine series resistors such as R13 and R15? If so, do it and get those completed.

Last edited: Aug 5, 2012
9. ### DreamFM Thread Starter New Member

Aug 5, 2012
7
0
Hi thanks again..

Hmm, i do know how to calculate parallel resistance into effective resistance. but that is only if they are in,let say, two independent loops.

But not someting lke 3 resistors (with different current flowing through them) around one independent loop..

10. ### WBahn Moderator

Mar 31, 2012
23,550
7,206
Components are in parallel if they are connected together at BOTH ends, not just one.

11. ### WBahn Moderator

Mar 31, 2012
23,550
7,206
This is very hard to follow, but I think I deciphered it.

The current through the 6Ω resistor does not necessarily travel straight to the "end voltage" (by which I assume you mean the lower terminal). Because, as you noted, there is a node at the other end of the 6Ω resistor, let's call that Node B, that is connected to a total of five resistors. So the current could go toward any or all of the other four resistors and, likewise, current from any of the other resistors could combine with it. All you know is that whatever current flows into Node B through some resistors must flow out of Node B through the rest.

Look at the marked up diagram I've attached. Which resistors, if any, are connected between the red and blue nodes (nodes A and B)? If there are more than one, they are in parallel. Replace them with their parallel equivalent. Which resistors, if any, are connected between the blue and yellow nodes (nodes B and C)? Again, if there are more than one, they are in parallel. Replace them with their parallel equivalent. What about between A and C? Between A and D? B and D? C and D? Which, if any of the two interior nodes (B and D) are connected to exactly two resistors? These are in series - combine them.

You seem to be hung up on wanting to do a mesh current analysis. Fine. Then do it and you should get the exact same end result. In order to find the equivalent resistance you need to either apply a voltage or a current to the terminals and then analyze the resulting circuit to find the current or the voltage, respectively, at the terminal and then divide the voltage by the current to find the equivalent resistance of the whole thing. If you don't combine any of the resistors, you will have five meshes to analyze. More than I normally like to play with, but these are pretty tame. So feel free to set up your mesh equations and post them and we can take a look and point out where you are going astray. But PLEASE post a diagram defining your mesh currents, as well as the voltage/current source that you drive the circuit with.