In equation form: If Ea and Eb are each 100 volts,

at figure 3-27.) For example, in view A, the positive

then

maximum voltage generated in coils A and B act in the

direction of the arrows, and B leads A by 120°. This

arrangement may be obtained by assuming coils A and

B to be two armature windings located 120° apart. If

The value of Er may be derived as follows:

each voltage has an effective value of 100 volts, the total

voltage is Er = 100 volts, as shown by the polar vectors

1. Erect a perpendicular to Er divides the isosceles

in view B.

triangle into two equal right triangles.

If the connections of coil B are reversed (view C)

2. Each right triangle has a hypotenuse of 100 volts

with respect to their original connections, the two

and abase of 100 cos 30°, or 86.6 volts.

voltages are in opposition. You can see this by tracing

3. The total length of Er is 2 x 86.6, or 173.2 volts.

the circuit in the direction of the arrow in coil A.

To construct the line voltage vectors E1,2, E2,3, and

1. The positive direction of the voltage in coil B is

E3,1, in figure 3-26, it is first necessary to trace a path

opposite to the direction of the trace

around the closed circuit that includes the line wires,

armature windings, and one of the three voltmeters. For

2. The positive direction of the voltage generated

example, in figure 3-21, consider the circuit that

in coil A is the same as that of the trace.

includes the upper and middle wires, the voltmeter

Therefore, the two voltages are in opposition.

connected across them, and the ac generator phases a

3. This effect is the same as though the positive

and b. The circuit trace is started at the center of the

maximum value of Eb were 60° out of phase

wye, proceeds through phase a of the ac generator, out

with that of Ea, and Eb acted in the same

line 1, down through the voltmeter from line 1 to line 2,

direction as& when the circuit trace was made

and through phase b of the ac generator back to the

(view D) to vector.

starting point. Voltage drops along line wires are

disregarded. The voltmeter indicates an effective value

4. Ea is accomplished by reversing the position of

equal to the vector sum of the effective value of voltage

Eb from that shown in view B, to the position

in phases a and b. This value is the line voltage, E1,2.

shown in view D, which completes the

According to Kirchhoff's law, the source voltage

parallelogram.