3 phase system electrical question

[darwinharianto]

Homework Statement

https://www.physicsforums.com/attachments/question-jpg.73386/ the current go from a A N and n
the 2nd current go from b B N and n

the first thing, i make the matrix
27-4j -3 i1 230 <0
-3 27-4j x i2 = 230<0

x means the matrix multiplied to the next
< means the degree
then with determinant from the first matrix
i find 767.04 <16.35

then to find i
i1 = 230<0 /767.04<16.35 27-4j -3 1
i2 x -3 27-4j x 1

is this right?
how to count 230<0 / 767.04<16.35
i don't understand

Homework Equations

The Attempt at a Solution

[/B]
the first thing, i make the matrix
27-4j -3 i1 230 <0
-3 27-4j x i2 = 230<0

x means the matrix multiplied to the next
< means the degree
then with determinant from the first matrix
i find 767.04 <16.35

then to find i
i1 = 230<0 /767.04<16.35 27-4j -3 1
i2 x -3 27-4j x 1

is this right?
how to count 230<0 / 767.04<16.35
i don't understand

 

[gneill]

Your matrix looks okay (although its formatting has been punished by the font!), but your solution doesn't look correct. Here it is rendered with LaTeX:

$$\left( \begin{array}{cc} {27 - j4} & -3 \\ -3 & {27 - j4} \end{array} \right)
\left( \begin{array}{c} I1 \\ I2 \end{array} \right) =
\left( \begin{array}{c} 230 \\ 230 \end{array} \right)$$

Can you check your determinant calculation? I see different magnitude and angle values.

 

[darwinharianto]

Your matrix looks okay (although its formatting has been punished by the font!), but your solution doesn't look correct. Here it is rendered with LaTeX:

$$\left( \begin{array}{cc} {27 - j4} & -3 \\ -3 & {27 - j4} \end{array} \right)
\left( \begin{array}{c} I1 \\ I2 \end{array} \right) =
\left( \begin{array}{c} 230 \\ 230 \end{array} \right)$$

Can you check your determinant calculation? I see different magnitude and angle values.

sorry about that :D
and thank you for the matrix format and the reply

the determinan is 704-j216?
so it means 736.39 with -17.05 degree?

so the matrix go like this?
$$\left( \begin{array}{c} I1 \\ I2 \end{array} \right) = 230<0 / 736.39 < -17.05
\left( \begin{array}{cc} 27-j4 & -3 \\ -3 & 27-j4 \end{array} \right)
\left( \begin{array}{c} 1 \\ 1 \end{array} \right)$$

what is that $$\left( \begin{array}{c} 1 \\ 1 \end{array} \right)$$ mean?
i get the formula from the ppt but i still don't understand why it is there
and i don't understand how to calculate 230<0 / 736.39 < -17.05 manually

 

[gneill]

sorry about that :D
and thank you for the matrix format and the reply

the determinan is 704-j216?
so it means 736.39 with -17.05 degree?

Yes, that's the correct value for the determinant.

so the matrix go like this?
$$\left( \begin{array}{c} I1 \\ I2 \end{array} \right) = 230<0 / 736.39 < -17.05
\left( \begin{array}{cc} 27-j4 & -3 \\ -3 & 27-j4 \end{array} \right)
\left( \begin{array}{c} 1 \\ 1 \end{array} \right)$$

what is that $$\left( \begin{array}{c} 1 \\ 1 \end{array} \right)$$ mean?
i get the formula from the ppt but i still don't understand why it is there

I'm not sure. I haven't seen that particular method used before. The $\left( \begin{array}{c} 1 \\ 1 \end{array} \right)$ is a column vector. I presume that it's meant to be multiplied with the impedance matrix in order to yield a new column vector. That would make sense since there's a column vector to the left of the equals also. But I don't see how that equation is meant to solve for the currents.

I would use Cramer's Rule to solve the problem (look it up). It uses determinants, and you've already got the value of one of them.

and i don't understand how to calculate 230<0 / 736.39 < -17.05 manually

That's just division of one complex number by another.

 

[darwinharianto]

Yes, that's the correct value for the determinant.

I'm not sure. I haven't seen that particular method used before. The $\left( \begin{array}{c} 1 \\ 1 \end{array} \right)$ is a column vector. I presume that it's meant to be multiplied with the impedance matrix in order to yield a new column vector. That would make sense since there's a column vector to the left of the equals also. But I don't see how that equation is meant to solve for the currents.

I would use Cramer's Rule to solve the problem (look it up). It uses determinants, and you've already got the value of one of them.

That's just division of one complex number by another.

OIC
it is just an inverse
moved the left 1st matrix to the right side
then it make i1 = 7.59< 0.31
and i2 = 7.59<0.31
is this right?
then IaA = I1
InN = -(I1+I2)
IbB = I2

 

[gneill]

OIC

Please don't use text-speak abbreviations here on PF. They're not allowed.

it is just an inverse
moved the left 1st matrix to the right side
then it make i1 = 7.59< 0.31
and i2 = 7.59<0.31
is this right?

It doesn't work for me; I can't see how that math could be correct. If I wanted to "move" the impedance matrix from the left side to the right side of the equation, I'd pre-multiply both sides with the inverse of the matrix. But then what is moved to the right side would be the inverse, not the original matrix. Take a generic equation of this type; let I be the unknown current column vector, V the known voltage column vector, and Z the impedance matrix. The equation is written:

$Z \; I = V$
$Z^{-1}Z \; I = Z^{-1}V$
$I = Z^{-1}V$

The problem I have with the equation that you've quoted is that I don't see the inverse of the impedance matrix coming about by simply dividing the matrix by its determinant. What you'd want is to use Cramer's rule which divides the adjoint matrix by the determinant, and solve for the I's individually.

then IaA = I1
InN = -(I1+I2)
IbB = I2

Yes, once you've solved for the mesh currents then then branch currents are obtained as you've shown.

 

[darwinharianto]

Please don't use text-speak abbreviations here on PF. They're not allowed.

It doesn't work for me; I can't see how that math could be correct. If I wanted to "move" the impedance matrix from the left side to the right side of the equation, I'd pre-multiply both sides with the inverse of the matrix. But then what is moved to the right side would be the inverse, not the original matrix. Take a generic equation of this type; let I be the unknown current column vector, V the known voltage column vector, and Z the impedance matrix. The equation is written:

$Z \; I = V$
$Z^{-1}Z \; I = Z^{-1}V$
$I = Z^{-1}V$

The problem I have with the equation that you've quoted is that I don't see the inverse of the impedance matrix coming about by simply dividing the matrix by its determinant. What you'd want is to use Cramer's rule which divides the adjoint matrix by the determinant, and solve for the I's individually.

Yes, once you've solved for the mesh currents then then branch currents are obtained as you've shown.

sorry about that
and thanks for replying

okay
and the result is I1 = 9.44<9.45 and I1=I2 or I1=9.31+J1.55
is this correct?

 

[gneill]

Yes, that's correct.

 

[darwinharianto]

sorry about that
and thanks for replying

okay
and the result is I1 = 9.44<9.45 and I1=I2 or I1=9.31+J1.55
is this correct?

and for the phasor diagram just draw a line with 9.44 and 9.45 degrees right?

 

[gneill]

and for the phasor diagram just draw a line with 9.44 and 9.45 degrees right?

Presumably you'll want to show the source voltage phasors along with the phasors for all the currents that you were asked to find.

 

[darwinharianto]

Presumably you'll want to show the source voltage phasors along with the phasors for all the currents that you were asked to find.

oohhh
so there will be 3 line
thx a lot