Page 171 - Solutions Manual to accompany Electric Machinery Fundamentals

jX R  jX 1   R    j X  X M  
Z  M 1 1 1
TH R    X  j X     R  X  j X  
 1 1 M   1 1 M 

  RX X  RX X  RX M  2   j R X    2 X X  2 X X M  2 
Z  1 1 M 1 1 M 1 1 M 1 M 1
TH 2 2
R  1  1 X M  X 
RX 2 R X  2 X X  2 X X 2
Z  R  jX  1 M  j 1 M 1 M 1 M
TH TH TH 2 2 2 2
R  1  1 X M  X  R  1  1 X M  X 
RX 2 2 2
The Thevenin resistance is R TH  1 M . If X M  R , then R  1 2  X  1 X M    X  1 X M  ,
1
R  1 2  1 X M  X  2
so

 X  2
R  R 1  M 
 TH X  X  M
1
R X  2 X X  2 X X 2
The Thevenin reactance is X TH  1 M 1 M 2 1 M .
R  1 2  1 X M  X 

2
2
2
If X M  R and X M  X then X X M 2  R X  X X and  X  1 X M   2 X M 2   R , so
1
M
1
1
1
M
1
1
XX 2
X  1 M  X
TH
X M 2 1
6-14. Figure P6-1 shows a simple circuit consisting of a voltage source, two resistors, and two reactances in
parallel with each other. If the resistor R L is allowed to vary but all the other components are constant, at
what value of R L will the maximum possible power be supplied to it? Prove your answer. (Hint: Derive
an expression for load power in terms of V, R S , X S , R L and X L and take the partial derivative of that
expression with respect to R L .) Use this result to derive the expression for the pullout torque [Equation
(6-54)].

SOLUTION The current flowing in this circuit is given by the equation
V
I 
R  S jX  S R  L jX L
L
V
I 
L
 S L R  2   R X  X L  S 2

The power supplied to the load is

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