Mesh Analysis: Solving for I1 with Loop Equations

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Discussion Overview

The discussion revolves around using mesh current analysis to determine the current I1 in a circuit. Participants explore various loop equations and attempt to solve for I1 through algebraic manipulation and matrix methods. The conversation includes technical details, calculations, and corrections related to the equations derived from the mesh analysis.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants present loop equations for mesh analysis, including complex numbers and impedance.
  • Multiple attempts to isolate I1 are made, with varying results and methods, including matrix approaches.
  • There are corrections regarding the voltage values and signs in the equations, with some participants questioning earlier calculations.
  • Discrepancies in the calculated angles for I1 are noted, with participants providing different methods for determining the angle.
  • Some participants express uncertainty about the correctness of their calculations and seek validation from others.
  • There is discussion about the use of arctan and the importance of considering the signs of the arguments to determine the correct quadrant for angles.
  • Participants explore alternative methods for calculations, such as using the atan2 function for angle determination.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct value of I1, as different calculations yield varying results. There are multiple competing views on the correct approach and the accuracy of the calculations presented.

Contextual Notes

Some calculations appear to have missing assumptions or unresolved mathematical steps, particularly regarding the handling of complex numbers and angle determination. The discussion reflects a range of interpretations and methods without resolving the discrepancies.

Who May Find This Useful

Students and practitioners interested in circuit analysis, particularly those learning about mesh current analysis and complex impedance in electrical engineering.

shaltera
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Homework Statement


Use mesh current analysis to determine I1

Homework 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

The Attempt at a Solution


Homework Statement


Homework 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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shaltera said:

Homework Statement


Use mesh current analysis to determine I1

Homework 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

The Attempt at a Solution


Homework Statement





Homework Equations



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

(47+j100)i1+j100i2=12
-j400i1+(47+j100)i1=-38 <---- check that value
j400i1-i1(47+j100)=38
(-47-j300)i1=38

Is it correct?

A slight problem with the voltage value indicated above.
 
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Likes   Reactions: 1 person
40-12=28
Well spotted
 
shaltera said:
40-12=28
Well spotted

Right. Now finish, isolating i1 and writing it in a standard form.
 
Thanks.I decided to use matrices to solve this problem.I hope matrices is correct
 

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For i1 I have -0.0557<84.63deg

Is it correct?
 
shaltera said:
For i1 I have -0.0557<84.63deg

Is it correct?

No, it doesn't match what I find.
 
(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
 
shaltera said:
(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

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

##j400 i_1 - (47 + j100) i_1 = 28##
 
  • #10
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
i1=-0.922<80.09deg

hope is correct
 
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  • #12
shaltera said:
For mesh A
-12+47i1+j100(i1+i2)=0
47+j100i1+j100i2=12 ##~~~~##<--- lost your parentheses here...
(47+j100)i1+j100i2=12 ##~~~~##<--- but recovered them here
The final expression is okay.

For mesh B
-10-j752+j100(i1+i2)=10 ##~~~~~##<--- Voltage source appears twice!
j100i1+j25i2=10 ##~~~~~##<---- But now the duplicate's gone!
The final expression is okay.

Therefor
(47+j100)i1+j100i2=12
j100i1+j25i2=10 (x(-4)) = -j400i1 - j100i2 = -40

(47+j100)i1+j400i1+j100i2-j100i2=12-40 <--- What happened here?
(47+j100)i1-j400i1=-28
47i1+j100i1-j400i1=-28
(47-j300)i1=-28
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:
  • #13
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
shaltera said:
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

Okay, that looks better. Continue.
 
  • #15
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
shaltera said:
i1=-=-28/303.65<80.096=-0.0922<80.096
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?

As well should I mention that I have used KVL to solve this problem
Yes, that was clear from your equations.
 
  • #17
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
 
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  • #18
shaltera said:
i1=-28/(47-j300)=-28/sqr (472-3002)tan-1(-300/47)=-28/303.65<80.096=0.093<-81.10deg
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).
 
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  • #19
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
shaltera said:
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

Good, but the angle should be negative. The two components of the current are both negative, putting the angle in the third quadrant.
 
  • #21
I simple used Pythagorean theorem and arctan

sqr((-0.0142)2+(-0.091)2)tan -1(-0.091/-0.0142)=
0.092<-98.86deg
 
  • #22
CASIO fx-83ES, calculated arctan (0.091/0.0142) as 81.25deg. Online calculator gave me result:98.86

Which one is correct?
 
  • #23
shaltera said:
I simple used Pythagorean theorem and arctan

sqr((-0.0142)2+(-0.091)2)tan -1(-0.091/-0.0142)=
0.092<-98.86deg

Arctan can't "see" the two negative signs of its argument; they cancel when you do the division. It's up to you, the user, to make sure that the angle ends up in the right quadrant. I'm sure you covered the domain and range of the trig functions in one of your previous courses dealing with trigonometry.

As an alternative to arctan, if your calculator has an atan2(y,x) function, then it takes both arguments separately and always returns the right result.
 
  • #24
gneill said:
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).

I didn't know how to express it at the beginning but that's what I had in mind

arctan -1 (300/47)
 
  • #25
Look like online calculator is right :)

0.092<-98.86deg
 
  • #26
Ok.How about calculations in a second equation (i2) in P17?Is it correct?
 
  • #27
shaltera said:
CASIO fx-83ES, calculated arctan (0.091/0.0142) as 81.25deg. Online calculator gave me result:98.86

Which one is correct?

81.13° for that argument to arctan. The 98.86 value looks like the result for -28/(47 - j300).
 
  • #28
10=j100i1+j25i2
10=9.3<(90-98.86)-j25i2=100x1<90x0.093<-98.86+j25i2
10=-9.3cos(9)-j9.3sin(9)=-9.19-j1.439+j25i2
10=-9.19-j1.439+j25i2
 
  • #29
shaltera said:
10=j100i1+j25i2
10=9.3<(90-98.86)-j25i2=100x1<90x0.093<-98.86+j25i2
10=-9.3cos(9)-j9.3sin(9)=-9.19-j1.439+j25i2
10=-9.19-j1.439+j25i2

It's getting messy plugging in numbers and then rearranging while calculating. Why not start by clearing out the j's from the right hand side? Multiply through by -j.

##-j10 = 100 i_1 + 25 i_2##

Isolate ##i_2## and then then start plugging in values.
 
  • #30
gneill said:
It's getting messy plugging in numbers and then rearranging while calculating. Why not start by clearing out the j's from the right hand side? Multiply through by -j.

##-j10 = 100 i_1 + 25 i_2##

Isolate ##i_2## and then then start plugging in values.

i2=(-100i1-j10)/25=-10(10i1-j)/25=-2(10i1-j)/5
 
Last edited:

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