Mesh Analysis: Solving for I1 with Loop Equations

In summary: I didn't know how to express it at the beginning but that's what I had in mindarctan -1 is the same as arctan 0. It's not directly related to what you're doing.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
  • #1
shaltera
90
0

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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  • #2
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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  • #3
40-12=28
Well spotted
 
  • #4
shaltera said:
40-12=28
Well spotted

Right. Now finish, isolating i1 and writing it in a standard form.
 
  • #5
Thanks.I decided to use matrices to solve this problem.I hope matrices is correct
 

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  • #6
For i1 I have -0.0557<84.63deg

Is it correct?
 
  • #7
shaltera said:
For i1 I have -0.0557<84.63deg

Is it correct?

No, it doesn't match what I find.
 
  • #8
(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
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.
 
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  • #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
 
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  • #31
i2=(-100i1-j10)/25=-10(10i1-j)/25=-2(10i1-j)/5

i2=-2(10(0.093<-98.86)-j)/5

Honestly I don't know how to continue
 
  • #32
shaltera said:
i2=(-100i1-j10)/25=-10(10i1-j)/25=-2(10i1-j10)/5

Okay, but simpler:

##i_2 = -4 i_1 - j\frac{10}{25}##

EDIT: and use the rectangular version of i1 to plug in.
 
  • #33
i2=-4(0.093<-98.86)-j10/25=-4(-0.0143+j0.091)-j0.4=0.052-j0.364-j0.4=0.052-j0.764
 
  • #34
shaltera said:
Equation above have two unknown,if I replace i1 with 0.093<-98.86 look mess

Only i2 is unknown. Convert i1 to rectangular form before plugging it in.

Note that you should make a habit of keeping more digits in intermediate values! Otherwise repeated conversions and calculations will cause rounding and truncation errors to creep into your results (especially for angular values). Always keep extra guard digits!
 
Last edited:
  • #35
0.052-j0.764=sqr(0.0522+(-0.764)2tan -1(-(0.764/0.052)=0.765<-86.106
 

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