The art of mesh analysis

[Cabdullahi]

Because i cant be arsed to revise for the exam tomorrow ive come up with an idea that could benefit me and also you guys


This is mesh analysis

A mesh is a closed loop(starting at a node and returning along that same node without passing through an intermediary node

Basically mesh analysis is a powerful way to calculate and analysis circuits

ok first and foremost we need a circuit to work from

and here it is sorry about the quality of the drawing Ive used paint



to clarify the values of the components:

R1 = 4 Ohms
R2 = 4 Ohms
R3 = 6 Ohms
R4 = 8 Ohms
V = 12 volts
I = 1 Ampere


basically we will divide up the circuit into three loops

loop one is the box containing the voltage source and the 2 resisters

that's V12 , R3 and R4

so using the algebraic formula

we put all the values together

so ......v12=R3(I2-I3) - R4(12-13)

what we want to find is the unknown value of the current we've got the resistance and the voltage

v12=R3(I2-I3) - R4(12-13)
^
^
^
we rewrite it but this time with the values in

R3 = 6 Ohms
R4 = 8 Ohms
V = 12 volts
I1,I2,I3...Unknown

V12=R2(I2-I3) - R4

we then move everything we know to the right and everything we dont know to the left

PUT THE VALUES IN and this is what you get
v12 = 6(I2)-6(I3) - 8(12) - 8

we then move everything we know to the right and everything we dont know to the left

so 20 = 14I2 - 6 I3

Thats the equation complete for loop 1

now loop 2



the values
R1 = 4 Ohms
R2 = 4 Ohms
R3 = 6 Ohms

because there isn't any source we take it as 0

0=R2(I3) + R1 (I3-I1) + R3 (I3-I2)
PUT THE VALUES IN and we get
0=4(I3) + 4(I3) - 4 + 6(I3) - 6(I2)
the left over 4 from I3 will be treated like a source because we cannot have 0 as a value it wouldn't work

move all the known values to the right and the unknown to the left

you get

4 = 14(13) - 6(I2)........ loop 2



now this bit is where i always struggle because my maths is pants .....to find the current through R2 we simply use simultaneous equation to derive the current through resistor 3


20 = 14 I2 - 6 I3
4 = - 6 I2 - 14 I2
----------------

14/6 gives you 2.3333333333

you multiply the bottom to get it to look like the top

4 = - 6 I2 - 14 I2 >>>>>will become this >>>>>9.333333332 = 14 - 32.66666666

20 = 14 I2 - 6 I3
9.333333332 = 14 I2 - 32.66666666I3

canceled

20 = - 6 I3
9.333333332 = - 32.66666666I3
---------------------------------
10.7 = 26.7

divide 10.7 by 26.7 to get I3

10.7/26.7=0.4007ampere

I3=0.4007

and that's mesh analysis

other members who are doing the same or similar can you guys assist me or the those who are good at maths maybe you guys can help me with the simultaneous bit

[Sahabiyaat]

the only thing i understand, is the drawing of the smiley face. Needs to comb hair and get plactic surgery on the nose , otherwise very pretty.

[GreyKode]

hey abdullahi, if it might interest you I am an electrical engineer. I think I can help u if you want.

[Cabdullahi]

hey abdullahi, if it might interest you I am an electrical engineer. I think I can help u if you want.

fantastic ! can you check if my mesh analysis is correct thanks

[GreyKode]

v12=R3(I2-I3) - R4(12-13)

I think something is wrong here
correction :
v12 = R3(I2-I3)+(I2-I1)R4

[Cabdullahi]

yes thanks thats right my mistake.....

[GreyKode]

loop1
v12 = R3(I2-I3)+(I2-I1)R4
loop2
0=I3*R2 + (I3-I1)*R1 + (I3-I2)*R3

loop3
I1=1A.

Now substituting you need to solve the following 2 eqns
12=6*(I2-I3) + 8*(I2-1)
0=4*I3 + (I3-1)*4 + (I3-I2)*6

20 = 14*I2 -6*I3
4 = 14*I3 - 6*I2

[Cabdullahi]

loop1

Now substituting you need to solve the following 2 eqns
12=6*(I2-I3) - 8*(I2-I3)
0=4*I3 + (I3-1)*4 + (I3-I2)*6

which gives us this
v12 = 6(I2)-6(I3) - 8(12) - 8
0=4(I3) + 4(I3) - 4 + 6(I3) - 6(I2)
.
.
.
.
20 = 14 I2 - 6 I3
4 = - 6 I2 - 14 I2

[GreyKode]

I am afraid youre mistaken, maybe its my equations.
But check again.

[Cabdullahi]

12=6*(I2-I3) - 8*(I2-I3)

once you add the value this is what you get:

v12 = 6(I2)-6(I3) - 8(12) - 8

thats how i did it..i hope its clear because you cant really explain it in words

[GreyKode]

sorry..check my edited post

[Cabdullahi]

oh ok so how did you get 14 on the bottom line did you times 6 by 2.33333333333

[GreyKode]

starting from here

12=6*(I2-I3) + 8*(I2-1)
0=4*I3 + (I3-1)*4 + (I3-I2)*6

expanding brackets

12=6*I2-6*I3 + 8*I2 - 8 ----> 12+8 = 6*I2-6*I3 + 8*I2 ----> 20 = 14*I2 - 6I3

0=4*I3 + I3*4 - 4 + I3*6 - I2*6 ----> 4 = 4*I3 + I3*4 + I3*6 - I2*6 --->
4 = 14*I3-I2*6

[GreyKode]

starting from here

12=6*(I2-I3) + 8*(I2-1)
0=4*I3 + (I3-1)*4 + (I3-I2)*6

expanding brackets

12=6*I2-6*I3 + 8*I2 - 8 ----> 12+8 = 6*I2-6*I3 + 8*I2 ----> 20 = 14*I2 - 6I3

0=4*I3 + I3*4 - 4 + I3*6 - I2*6 ----> 4 = 4*I3 + I3*4 + I3*6 - I2*6 --->
4 = 14*I3-I2*6

[Cabdullahi]

ok now what about the simultaneous equation part..............

you have to change this

20 = 14*I2 -6*I3
4 = 14*I3 - 6*I2
to this
20 = 14 I2 - 6 I3
4 = - 6 I2 - 14 I2

and then finding I3 by multiplying the bottom line by 2.333333333

[GreyKode]

ok now what about the simultaneous equation part..............

you have to change this

20 = 14*I2 -6*I3
4 = 14*I3 - 6*I2
to this
20 = 14 I2 - 6 I3
4 = - 6 I2 + 14 I3

and then finding I3 by multiplying the bottom line by 2.333333333

20 = 14 I2 - 6 I3
4 = - 6 I2 + 14 I3 * 14/6

20 = 14 * I2 - 6* I3
56/6 = -14 *I2 + 14*(14/6)* I3 adding

[Cabdullahi]

20 = 14 I2 - 6 I3
4 = - 6 I2 + 14 I3 * 14/6

20 = 14 * I2 - 6* I3
56/6 = -14 *I2 + 14*(14/6)* I3 adding

wow thats just confused me

this is what i did

20 = 14 I2 - 6 I3
4 = - 6 I2 + 14 I3<<<<<<<<<time by 2.333333333333

outcome is 9.3333333 = 14 + 32.666666666

then 20 = 14 I2 - 6 I3
9.3333333 = 14 + 32.666666666
---------------------------------------------
10.6 0 -26.7

10.7/26.7 = 0.4007 ampere

I3 = 0.4007 ampere

[Cabdullahi]

i had the test today inshallah ive done enough...you guyz just make dua for me

[GreyKode]

How was it? Anything challenging?

[Cabdullahi]

phasor diagrams, i have know clue ....i just guessed.....and opamp.... those two i dont really understand because i never went for lectures,i kinda slacked off