So the equation of I2 should be something like this?

R2(I2-I1) + E2 + R3(I2-I3) + E3 + R5(I2-I4) + E5=0

Why did you put minus in those two places?

 

Why did you put minus in those two places?

Because they were going from positive to negative.. D;?


... and so the final equation should be

R2(I2-I1)+R3(I2-I3)+R5(I2-I4)=-42?

 

in the instructions say to get only Current from R3 and potency from R4... Shouldn't I ignore the others meshes/nodes?, since I got 2 power supplies in the mesh and 1 on the node..

 

Because they were going from positive to negative.. D;?


... and so the final equation should be

R2(I2-I1)+R3(I2-I3)+R5(I2-I4)=-42?

Yep, that equation looks good.

Resistors are passive devices, they don't care for signs so for the sake of simplicity we simply assume they are "positive" or have no sign.

You have 4 meshes. You must have 4 equations in order to solve the whole thing. It is the only way to find the specific item you want.

 

You have 4 meshes. You must have 4 equations in order to solve the whole thing. It is the only way to find the specific item you want.

Ohh.. dam okei ill keep posting my other equations .. thanks for ur help <3

 

Ohh.. dam okei ill keep posting my other equations .. thanks for ur help <3

Are you familiar with Simultaneous Equations?

This is what you are doing, you are setting up the Simultaneous Equations. Once you are done setting them up, solve them.

http://en.wikipedia.org/wiki/System_of_linear_equations

 

#3 -E3 + R3(I3-I2) + R4 + E4 = 0

#4 R5(I4-I2) + (-E6) + R6 + (-E5) = 0

Am i Correct?

 

#3 -E3 + R3(I3-I2) + R4(I3) + E4 = 0

#4 R5(I4-I2) + (-E6) + R6(I4) + (-E5) = 0

Am i Correct?

You forgot the currents in two places. I inserted them.

That is it. You have setup your equations. Now solve them.

 

ok so the equations are

2 4 0 0 0 0 = 16
0 4 10 0 0 0 = -42
0 0 10 8 0 0 = 4
0 0 0 0 1 6 = 28

???

 

with the -42

I1 = 31 Amps
I2 = -23/2 Amps
I3 = 2/5 Amps
I4 = 25 Amps

with +42

I1 = -11 Amps
I2 = 19/2 Amps
I3 = 2/5 Amps
I4 = 28 Amps

But i think they are incorrect because if i add them all i dont get 0... [I did the equations with Mathlab so theres no way they are wrong]


Any idea where it went wrong?

 

ok so the equations are

2 4 0 0 0 0 = 16
0 4 10 0 0 0 = -42
0 0 10 8 0 0 = 4
0 0 0 0 1 6 = 28

???

That makes no sense.

Just write the equations properly. Once you do some organizing, they should look like this:

A*I1+B*I2+C*I3+D*I4=H

where A, B, C, D, are numbers representing resistors and H represents the voltages. I1, I2, I3, I4 are the currents that you are trying to find out.

 

But i dont get it -frustration-.. ive been here since like 3pm trying to do all the exercises the teacher left us and this is the last one omfg so hard...

i got so far

2(I1)+4(I1-I2)=16
4(I2-I1)+10(I2-I3)+1(I2-I4)=-42
10(I3-I2)+8(I3)=4
1(I4-I2)+6(I4)=28

Wah i do from now?....

 

But i dont get it -frustration-.. ive been here since like 3pm trying to do all the exercises the teacher left us and this is the last one omfg so hard...

i got so far

2(I1)+4(I1-I2)=16
4(I2-I1)+10(I2-I3)+1(I2-I4)=-42
10(I3-I2)+8(I3)=4
1(I4-I2)+6(I4)=28

Wah i do from now?....

Equation 1.
2I1+4(I1-I2)=16
2I1+4I1-4I2=16
6(I1)-4(I2)=16

Equation 2.
10(I3-I2)+8(I3)=4
10I3-10I2+8I3=4
18(I3)-10(I2)=4

Equation 3.
1(I4-I2)+6(I4)=28
1I4-1I2+6I4=28
7(I4)-1(I2)=28

Now look how interesting? Did you notice that I1, I3 and I4 can be represented in terms of I2?

6(I1)-4(I2)=16
6(I1)=4(I2)+16
\(I1=\frac{4(I2)+16}{6}\)

18(I3)-10(I2)=4
18(I3)=10(I2)+4
\(I3=\frac{10(I2)+4}{18}\)

7(I4)-1(I2)=28
7(I4)=I2+28
\(I4=\frac{I2+28}{7}\)

Now. Take the last equation: 4(I2-I1)+10(I2-I3)+1(I2-I4)=-42
and replace I1, I3, and I4 with:
\(I1=\frac{4(I2)+16}{6}\)
\(I3=\frac{10(I2)+4}{18}\)
\(I4=\frac{I2+28}{7}\)
Solve for I2.

Once you know I2, find I3. I3 is current through R4. Use Ohm's Law (V=I*R or I think you use E=I*R) to find voltage (potency? potential?) across R4.

Then use I2 and I3 to find current through R3. Remember how we represented the voltage across R3? V=R3*(I2-I3) So the current through R3 is I2-I3. You know I2, you know I3, subtract one from the other. If you do I2-I3 and you get positive answer, than the current is in direction of I2. If I2-I3 gives you a negative answer, then the current is in direction of I3.

 

Ohhh.. i think i get it

 

Do I have to substract (I2) as a common factor from the equation?

 

Do I have to substract (I2) as a common factor from the equation?

I2 is the mesh current of the mesh in the center.

Whenever you deal with a resistor that is shared by two meshes, then you need to consider the mesh current in both meshes that share that resistor.

For example. R3 is shared by two meshes. So you can not say: The current though R3 is I2, or, The current though R3 is I3. The current through R3 will be some portion of current from I2 and some portion from I3.

Now look at the R1. This resistor belong to only one mesh. So there is only one current passes through it. In this case I1 passes though R1. So it is simple.

 

I1 = .14354
I2 = 3.4593
I3 = -1.8804
I4 = -3.7847

its still wrong :/

 

its still wrong :/

I1=0.1435
I2=-3.784
I3=-1.88
I4=3.4593
These are mesh currents.
The negative numbers indicate that my original choice of direction was wrong. Remember, I chose that all the mesh currents will be in clockwise direction. I1 and I4 are in clockwise direction. I2 and I3 are negative, that means they are in counter clockwise direction.

Now. Potency of R4. I assume that means potential across R4. You use E for potential?
E=R4(I3)=8(1.881)=15.04 volt

Next. Current on R3. I assume that means current through R3.
I2-I3=-3.78-(-1.88)=-1.9 ampere
I originally assumed that the current will be moving clockwise. In this case from left to right. The negative sign indicate that my assumption is wrong. So if I change direction of the current, then the sign change too. The current though R3 is 1.9 ampere and this current is from right to left.

I am off to sleep. Good luck.

 

But if i add all those number i dont get 0.. id be breaking Kirchhoff's law

 

Thank you for your awesome help