# Help me! Newbie in Need :/.. hw help

Discussion in 'Homework Help' started by mahatma666, Mar 13, 2014.

1. ### mahatma666 Thread Starter New Member

Mar 13, 2014
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I got this huge ass circuit and i gotta do it over Nodes and Meshes Analysis.. i cant figure it out at all.. Could somebody please help me?

Mesh Method to calculate Potency on R4
Node Method to calculate current on R3

Please i really need the help :/

2. ### DerStrom8 Well-Known Member

Feb 20, 2011
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Please show us what you have done so far, as requested in the forum rules.

Thanks,
Matt

3. ### mahatma666 Thread Starter New Member

Mar 13, 2014
21
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I can't figure out anything really.. the teacher just gave it to us [we only made circuits with 1 or 2 power sources max in class] and then he just gave us that one like there, make it. I'm really frustrated because he said it will count in the final grade of the class... All i made was put the positive or negative signs on the resistors.

And i mean i know the rules about mesh and nodes, i just dont know how to put them to work.. the circuit its so confusing.

4. ### amilton542 Active Member

Nov 13, 2010
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Explain what you know about KCL and KVL.

5. ### mahatma666 Thread Starter New Member

Mar 13, 2014
21
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well i know the formulas the teacher gave us.. The equations and everything all i dont know is how to start the circuit.. and how to put the + or - signs in the equation..

6. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,329
I think you'd probably find it easiest to use Mesh Analysis for this problem. Try looking up how that works and it should at least give you a start. If you have any specific questions, feel free to ask provided you show us your work so far.

7. ### amilton542 Active Member

Nov 13, 2010
494
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Yes in part, but you didn't answer my question. Explain what you know about KCL and KVL.

8. ### mahatma666 Thread Starter New Member

Mar 13, 2014
21
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well all i want to know is if E1,E2,E5 and E6 affect the 2 resistors i need to solve, as i said i dont know where to start with the circuit.. since it has so many Power sources, and @amilton542 i know enough of KCL and KVL.. im struggling in how to start the analysis.

9. ### amilton542 Active Member

Nov 13, 2010
494
64
Really now? From what I've gathered you don't understand the two theorems you've failed to answer.

10. ### mahatma666 Thread Starter New Member

Mar 13, 2014
21
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I know the rules of the theorems, i do not know how to start to analyze that certain circuit with all those power supplies..

11. ### amilton542 Active Member

Nov 13, 2010
494
64
Let's discuss KVL.

Why do the sum of the voltage drops equate to zero around a closed loop?

Feb 19, 2010
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13. ### mahatma666 Thread Starter New Member

Mar 13, 2014
21
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because the sum of the voltage drops is equal to supply voltage... Now can you please help me understand the circuit? lol..

Mar 13, 2014
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15. ### amilton542 Active Member

Nov 13, 2010
494
64
That then is your problem why more than one energy source is putting you off.

The voltage drops sum to zero because a PD can't exist between a point and itself.

Now traverse a single loop of your choice applying your associated variables convention and show what you get.

16. ### mahatma666 Thread Starter New Member

Mar 13, 2014
21
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I kinda got it now thanks to the mesh shteii01 showed me, but i still do not get how the Power supplies should be put in the equation.. For example E2 is in I1 and I2.. i dont know how to put that into the equation.. I know R2 is R2(I1-I2).. but how about E1 and E2?... should be E2(I1-I2)?

17. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,500
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The voltage sources are treated as short circuits by the current.

Let use use Mesh Current Method for a moment.

You have a current that is moving around the mesh. When current encounters the resistor, it enters the resistor and then exits it. In this situation you apply Ohm's Law, Voltage across the resistor is Current through the resistor times the Resistance of the resistor, V=I*R.

When current encounters the voltage source like the battery you have, it sees a short circuit. But! You are interested in the voltage! And the voltage is already provided for you so you don't really care about the I and R anyway.

18. ### amilton542 Active Member

Nov 13, 2010
494
64
Forget currents for a minute and apply KVL to a loop of your choice.

19. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,500
511
Lets take the left most mesh of your circuit.

We will call the current in that mesh I1.
We will draw the current clockwise.

I1 is leaving the battery E1 and enters R1.
R1(I1)+(-E2)+R2(I1-I2)+(-E1)=0

Now. Why is E2 and E1 are negative? Because when mesh current enters voltage source from the Negative terminal, we assign negative sigh to the voltage source.

The above equation can be rewritten:
R1(I1)+R2(I1-I2)=E1+E2
2(I1)+4(I1-I2)=12+4

I2 is the mesh current from the next mesh to the right. The resistor R2 is shared by both meshes so it will be influenced by currents from both meshes.

20. ### mahatma666 Thread Starter New Member

Mar 13, 2014
21
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So the equation of I2 should be something like this?

R2(I2-I1) + E2 - R3(I2-I3) + E3 - R5(I2-I4) + E5?