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Student Neighbour - Matrix Algebra
9 Answers
Neighbour asks if someone could please present the answer using matrix algebra to the transformer question in the link. TIA!!
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Try this... Only peak values of current in the primary and secondary circuits will be considered. Given (omega) w = 2000 and mutual inductance = 8 mH then; XL1 = j8 Ω, XL2 = j40 Ω and the reactance M = j16 Ω Kirchhoff's Voltage Loop - Primary circuit; -6 + 2.i1 + j8.i1 - j16.i2 = 0 rearranging becomes; (2 + j8).i1 - j16.i2 = 6 ---> eq 1 KVL - Secondary circuit; j40.i2 +...
17:29 Mon 08th May 2023
I am sure kuiperbelt appreciates any posts giving the solutions to the questions, showing how they were arrived at or providing techniques for dealing with them.
This is a Question and Answer forum, they are valid questions and doubting his motives for asking them is not helpful.
If you don't want to answer kuiperbelt's questions, that's fine, feel free to ignore them.
This is a Question and Answer forum, they are valid questions and doubting his motives for asking them is not helpful.
If you don't want to answer kuiperbelt's questions, that's fine, feel free to ignore them.
-- answer removed --
Try this...
Only peak values of current in the primary and secondary circuits will be considered.
Given (omega) w = 2000 and mutual inductance = 8 mH then;
XL1 = j8 Ω, XL2 = j40 Ω and the reactance M = j16 Ω
Kirchhoff's Voltage Loop - Primary circuit;
-6 + 2.i1 + j8.i1 - j16.i2 = 0 rearranging becomes;
(2 + j8).i1 - j16.i2 = 6 ---> eq 1
KVL - Secondary circuit;
j40.i2 + 100.i2 - j16.i1 = 0 --- rewriting;
-j16.i1 + (100 + j40).i2 = 0 ----> eq 2
eq 1 and eq 2 form a simultaneous pair thus i1 and i2 can be found
by the application of matrix algebra. Due to AB syntax issues, see link below.
https:/
Only peak values of current in the primary and secondary circuits will be considered.
Given (omega) w = 2000 and mutual inductance = 8 mH then;
XL1 = j8 Ω, XL2 = j40 Ω and the reactance M = j16 Ω
Kirchhoff's Voltage Loop - Primary circuit;
-6 + 2.i1 + j8.i1 - j16.i2 = 0 rearranging becomes;
(2 + j8).i1 - j16.i2 = 6 ---> eq 1
KVL - Secondary circuit;
j40.i2 + 100.i2 - j16.i1 = 0 --- rewriting;
-j16.i1 + (100 + j40).i2 = 0 ----> eq 2
eq 1 and eq 2 form a simultaneous pair thus i1 and i2 can be found
by the application of matrix algebra. Due to AB syntax issues, see link below.
https:/
emeritus\ It may well be that you will get assistance\
My dear fellow you are mistaken. I have posted approx two dozen academic type questions and without exception, all have
been answered by OPs here on AB!!
The three students (that I am looking out for) have continually expressed their thanks!! In turn, my gratitude goes out in abundance to all those OPs whom have made time to contribute an answer. Such has been the response at times, it has been very difficult not knowing who to award BestAnswer!!
Seven days from now my grandson begins the gruelling task of sitting through a plethora of GCSE exams over several weeks.
The good news is, I'll not be bothering this board with anymore math GCSE questions after next Monday LOL. The bad news is, my nephew will be in his second year of his Math A-Level come September!!
My dear fellow you are mistaken. I have posted approx two dozen academic type questions and without exception, all have
been answered by OPs here on AB!!
The three students (that I am looking out for) have continually expressed their thanks!! In turn, my gratitude goes out in abundance to all those OPs whom have made time to contribute an answer. Such has been the response at times, it has been very difficult not knowing who to award BestAnswer!!
Seven days from now my grandson begins the gruelling task of sitting through a plethora of GCSE exams over several weeks.
The good news is, I'll not be bothering this board with anymore math GCSE questions after next Monday LOL. The bad news is, my nephew will be in his second year of his Math A-Level come September!!
FAO >14:35
The method below uses the current flowing in or away from the (reference) dot.
Consider mutually induced voltage in the primary circuit given by j16.i2 volts.
Assuming conventional current flow, i2 is in a clockwise direction and enters L2 from the bottom of the coil. This means bottom of L2 is +ve. It follows the top of L2 (dot) must be -ve. Therefore the dot in the primary circuit (top of L1) must also be negative.
Since i1 is flowing into the dot, this mutually induced voltage is negative and will oppose those voltages in the primary being generated by i1. Therefore j16.i2 becomes -j16.i2
Had i1 been flowing away from the dot (anticlockwise),
then the mutually induced voltage would have been aiding with those generated by i1. Thus j16.i2 would remain +j16.i2
Now see if the student can determine the induced voltage in the secondary circuit using the method above.
Part b.) From the circuit symbol the transformer is Ideal.
The method below uses the current flowing in or away from the (reference) dot.
Consider mutually induced voltage in the primary circuit given by j16.i2 volts.
Assuming conventional current flow, i2 is in a clockwise direction and enters L2 from the bottom of the coil. This means bottom of L2 is +ve. It follows the top of L2 (dot) must be -ve. Therefore the dot in the primary circuit (top of L1) must also be negative.
Since i1 is flowing into the dot, this mutually induced voltage is negative and will oppose those voltages in the primary being generated by i1. Therefore j16.i2 becomes -j16.i2
Had i1 been flowing away from the dot (anticlockwise),
then the mutually induced voltage would have been aiding with those generated by i1. Thus j16.i2 would remain +j16.i2
Now see if the student can determine the induced voltage in the secondary circuit using the method above.
Part b.) From the circuit symbol the transformer is Ideal.
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