Mesh Analysis: Solving for I1 with Loop Equations

[shaltera]

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?

 

[gneill]

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.

 

[shaltera]

40-12=28
Well spotted

 

[gneill]

40-12=28
Well spotted

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

 

[shaltera]

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

 

[shaltera]

For i1 I have -0.0557<84.63deg

Is it correct?

 

[gneill]

For i1 I have -0.0557<84.63deg

Is it correct?

No, it doesn't match what I find.

 

[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

 

[gneill]

(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:

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

 

[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

 

[shaltera]

i1=-0.922<80.09deg

hope is correct

 

[gneill]

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.

 

[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

 

[gneill]

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.

 

[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

 

[gneill]

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.

 

[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

 

[gneill]

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).

 

[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

 

[gneill]

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.

 

[shaltera]

I simple used Pythagorean theorem and arctan

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

 

[shaltera]

CASIO fx-83ES, calculated arctan (0.091/0.0142) as 81.25deg. Online calculator gave me result:98.86

Which one is correct?

 

[gneill]

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.

 

[shaltera]

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)

 

[shaltera]

Look like online calculator is right :)

0.092<-98.86deg

 

[shaltera]

Ok.How about calculations in a second equation (i₂) in P17?Is it correct?

 

[gneill]

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).

 

[shaltera]

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

 

[gneill]

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.

 

[shaltera]

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

 

[shaltera]

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

 

[gneill]

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.

 

[shaltera]

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

 

[gneill]

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!

 

[shaltera]

0.052-j0.764=sqr(0.052²+(-0.764)²tan ^-1(-(0.764/0.052)=0.765<-86.106

 

[gneill]

0.052-j0.764

What is that? If it's i₂ it doesn't look right. Show your steps.

 

[shaltera]

i1=0.093<-98.86

I converted polar form to complex numbers for i1
-0.0143+j0.091 then I entered in

i2=--4i1-j10/25
i2=-4(-0.0143+j0.091)-j0.4=0.052-j0.364-j0.4=0.052-j0.764
i2=0.052-j0.764=sqr(0.052²+(-0.764)²)tan ^-1(-(0.764/0.052)=0.765<-86.106

 

[gneill]

i1=0.093<-98.86

I converted polar form to complex numbers for i1
-0.0143+j0.091 then I entered in

Should be: -0.0143 - j0.0911
Check your conversion. You should note that an angle of -98.86° places it in the third quadrant, so both terms must be negative.

i2=--4i1-j10/25
i2=-4(-0.0143+j0.091)-j0.4=0.052-j0.364-j0.4=0.052-j0.764
i2=0.052-j0.764=sqr(0.052²+(-0.764)²)tan ^-1(-(0.764/0.052)=0.765<-86.106

 

[shaltera]

I calculated with Matlab

for I1=-0.0143-0.0911i=0.0922<-98.9040
for I2= 0.0571-0.0356i=0.0673<-31.9553

I don't get it why we have to calculate by hand?

 

[shaltera]

Can I use Matlab for my assessment?

a=[47+100j 100j;100j 25j]
b=[12;10]
i=a\b

then magn=abs(i)
and then the angle
angle(i)*180/pi

Why do we have to calculate by hand?I'm not living in 20 century

 

[gneill]

I calculated with Matlab

for I1=-0.0143-0.0911i=0.0922<-98.9040
for I2= 0.0571-0.0356i=0.0673<-31.9553

I don't get it why we have to calculate by hand?

Yes, those values look good. As to why you have to calculate by hand, I can only presume that there are "pedagogical reasons"; You'll have to ask your instructor.

 

[The Electrician]

Can I use Matlab for my assessment?

a=[47+100j 100j;100j 25j]
b=[12;10]
i=a\b

then magn=abs(i)
and then the angle
angle(i)*180/pi

Why do we have to calculate by hand?I'm not living in 20 century

Who says you have to calculate by hand? Your instructor? Many instructors will require you to calculate by hand for a while so that you learn how complex arithmetic works. But, after a while, you should be able to use modern software or calculators.

 

[shaltera]

Apart of Matlab, I do not have a calculator with atan2 function.I was looking for a formula to convert polar to rectangular form but all examples are with positive values

 

[gneill]

Many calculators have built-in polar to rectangular (and back) functions. They will accomplish the same thing.