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

Thread Starter

mahatma666

Joined Mar 13, 2014
21
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 :/
 

Thread Starter

mahatma666

Joined Mar 13, 2014
21
Please show us what you have done so far, as requested in the forum rules.

Thanks,
Matt

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.

Thanks for the reply.
 

Thread Starter

mahatma666

Joined Mar 13, 2014
21
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..
 

DerStrom8

Joined Feb 20, 2011
2,390
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.
 

Thread Starter

mahatma666

Joined Mar 13, 2014
21
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.
 

amilton542

Joined Nov 13, 2010
497
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.
 

Thread Starter

mahatma666

Joined Mar 13, 2014
21
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.
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)?
 

shteii01

Joined Feb 19, 2010
4,644
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)?
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.
 

amilton542

Joined Nov 13, 2010
497
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)?
Forget currents for a minute and apply KVL to a loop of your choice.
 

shteii01

Joined Feb 19, 2010
4,644
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.
 
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