# Mesh analysis

1. Nov 13, 2013

### shaltera

1. The problem statement, all variables and given/known data
Use mesh current analysis to determine I1

2. Relevant equations
Loop 1

47i1+j100(i1+i2)-12=0
(47+j100)i1+j100i2=12

Loop 2
-j75i2+j100(i1+i2)-10=0
j100i1+j25i2=10

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

j100i1+j25i2=10 (-4)
-j400i1-j100i2=-40

(47+j100)i1+j100i2=12
-j400i1+(47+j100)i1=-38
j400i1-i1(47+j100)=38
(-47-j300)i1=38

Is it correct?

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Last edited by a moderator: Nov 13, 2013
2. Nov 13, 2013

### Staff: Mentor

A slight problem with the voltage value indicated above.

3. Nov 13, 2013

### shaltera

40-12=28
Well spotted

4. Nov 13, 2013

### Staff: Mentor

Right. Now finish, isolating i1 and writing it in a standard form.

5. Nov 13, 2013

### shaltera

Thanks.I decided to use matrices to solve this problem.I hope matrices is correct

#### Attached Files:

• ###### m1.jpg
File size:
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Views:
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6. Nov 14, 2013

### shaltera

For i1 I have -0.0557<84.63deg

Is it correct?

7. Nov 14, 2013

### Staff: Mentor

No, it doesn't match what I find.

8. Nov 14, 2013

### shaltera

(47+j500)i1=-28
i1=(-28)/(47-j500)=(-28)/(sqroot(472+5002))tan -1(500/47)=
-28/502.2<84.63deg=-0.05557<84.63deg

9. Nov 14, 2013

### Staff: Mentor

Show your algebra beginning from where you left off here in your first post:

$j400 i_1 - (47 + j100) i_1 = 28$

10. Nov 14, 2013

### shaltera

For mesh A
-12+47i1+j100(i1+i2)=0
47+j100i1+j100i2=12
(47+j100)i1+j100i2=12

For mesh B
-10-j752+j100(i1+i2)=10
j100i1+j25i2=10

Therefor
(47+j100)i1+j100i2=12
j100i1+j25i2=10 (x(-4))

(47+j100)i1+j400i1+j100i2-j100i2=12-40
(47+j100)i1-j400i1=-28
47i1+j100i1-j400i1=-28
(47-j300)i1=-28

11. Nov 14, 2013

### shaltera

i1=-0.922<80.09deg

hope is correct

Last edited: Nov 14, 2013
12. Nov 14, 2013

### Staff: Mentor

The final expression is okay.

The final expression is okay.

I think you tried to subtract the mesh B equation from the mesh A equation, but didn't change the sign of the i1 term.

Last edited: Nov 14, 2013
13. Nov 14, 2013

### shaltera

j100i1+j25i2=10 multiply by (-4)
-j400i1-j100i2=-40
then I add equation for mesh A to B
(47+j100)i1-j400i+j100i2-j100i2=12-40
+j100i2-j100i2=0
Therefor
(47+j100)i1-j400i1=-28
47i1+j100i1-j400i1=-28
47i1-j300i1=-28
(47-j300)i1=-28

14. Nov 14, 2013

### Staff: Mentor

Okay, that looks better. Continue.

15. Nov 14, 2013

### shaltera

i1=-=-28/303.65<80.096=-0.0922<80.096

As well should I mention that I have used KVL to solve this problem

16. Nov 14, 2013

### Staff: Mentor

You should incorporate the sign of the numeric part into the angle since the numeric part is the magnitude of the complex value, and magnitudes are always positive (absolute) values. Your angle appears to be off by one degree... typo?

Yes, that was clear from your equations.

17. Nov 14, 2013

### shaltera

i1=-28/(47-j300)=-28/sqr (472-3002)tan-1(-300/47)=-28/303.65<80.096=0.093<-81.10deg

10=j100i1+j25i2 =100x1<90x0.093<-81.10+j25i2
10=9.3<(90-81.10)+j25i2
10=9.3cos(9)+j9.3sin(9)=9.19+j1.439+j25i2
10=9.19+j1.439+j25i2
-0.81+j1.439=-j25i2
1.651<-0.94/-j25i2=1.651<-0.94/25<90=0.066<-90.94

Last edited: Nov 14, 2013
18. Nov 14, 2013

### Staff: Mentor

The angle you've found for the denominator doesn't reflect the sign of the argument of the arctan function. Note that the "-" is associated with the numerator of the argument, -300. So the angle should end up in the [STRIKE]third[/STRIKE] fourth quadrant. Also the angle value is off by a degree (typo?).

Next, the negative sign on the numerator (-28) has disappeared. It should get rolled into the final angle... think of the numerator as 28 ∠ ±180°. Or, perhaps more easily, make the numerator positive to begin with (multiply top and bottom by -1 before you begin).

Last edited: Nov 14, 2013
19. Nov 14, 2013

### shaltera

Calculated differently:

(-28/47-j300)(47+j300)/(47+j300)=(-1316-j8400)(2209-j290000)=(-1316-j8400)/92209
(-1316/92209)-(j8400/92209)=-0.0142-j0.091=0.092<98.86deg

20. Nov 14, 2013

### Staff: Mentor

Good, but the angle should be negative. The two components of the current are both negative, putting the angle in the third quadrant.