Page 212 - Solutions Manual to accompany Electric Machinery Fundamentals

(d) There are 8 coils on this machine in two parallel paths, with each coil having a resistance of 0.04 .
Therefore, the total resistance R is
A
    0.04 0.04 0.04 0.04  0.04 0.04 0.04 0.04  


 
R  0.04 0.04 0.04 0.04   0.04 0.04 0.04 0.04 
A






R  A 0.08 
(e) The voltage produced by this machine is 33.9 V, as found in part (c). Therefore, the current
flowing in the machine will be
E 33.9 V
I  A   31.4 A
A
R  A R load 0.08   1.0 
(f) The induced torque is given by Equation 7-46:

ZrlBI 12 cond 0.10 m 0.3 m 1.0 T 33.9 A 
  A 
ind
a 2 current paths
 ind  4.12 N m, CW (opposite to the direction of rotation)

(e) The terminal voltage of this machine is given by
V  T E  A I R A
A

% M-file: prob7_2.m
% M-file to create a plot of the terminal voltage vs load
% current for the machine of problem 7-2.

% First, initialize the values needed in this program.
Ea = 31.4; % Stator resistance
Ra = 0.08; % Stator reactance

% Get the line currents to calculate the anser for
Ia = 0:1:35;

% Calculate the terminal voltage
Vt = Ea - Ia * Ra;

% Plot the terminal voltage
figure(1);
plot(Ia,Vt,'b-','LineWidth',2.0);
xlabel('\bfCurrent (A)');
ylabel('\bf\itV_{T} \rm\bf(V)');
title ('\bfTerminal Voltage vs Load');
axis([ 0 35 0 32]);
grid on;
hold off;
The resulting plot is shown below.

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