# Audel electrical course for apprentices and journeymen - Rosenberg P.

ISBN: 0-764-54200-1

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OC = 10 amperes so neutral current = 0 amperes

As before, the portion of OC diametrically opposite to ON is

subtracted from ON, giving the neutral current for any unbalances.

272 Chapter 24

Figure 24-13 Three-phase system with unbalanced loads.

Figure 24-14 Vector diagram of currents of the circuit in Figure 24-13.

The same method will be used for unbalanced loads of all phases (see Figure 24-15). Draw a parallelogram AOCN with line NC parallel to line AO, and NO will be measured to find the current in ON. In this case, ON is 5.2 amperes.

All of this may be worked out by simple trigonometry. From the law of cosines,

ON2 = OC2 + OA2 - 2 X OA X OC X cos NCO

To find angle NCO, observe that angles AOC and CNA are equal and 120° each. Since the interior angles of a quadrilateral

Power in Polyphase Circuits 273

A

Figure 24-15 Finding current in three-phase unbalanced load.

total 360°, the sum of angles NAO and NCO must be 360° -2(120°) = 120°. But angles NAO and NCO are equal; therefore, angle NCO is 120°/2 = 60°. Thus

ON2 = 62 + 32 - 2 X 6 X 3 cos 60°

= 45 - 36 X 0.5 = 45 - 18 = 27

So

ON = 227 = 5.2 A

In similar vector fashion, ON may be added to OB.

The same method may be used for the other phase currents which were just covered.

Formulas

V/A = sin 120°/sin 30° = sin 60°/sin 30° = 1.732 = 23 For Wye Connections

E = e 23

i = I

274 Chapter 24

e = E/ 23

p = el

p = JE/ 23

p = 3p = 3IE/ 23 = IE 23

For less than 100% PF,

P = 23 X IE X cos f

For Delta Connections

I = i 23 i = 123

e = E p = ei P = Ei

P = 3p = ^ = IE 23 23

For less than 100% PF,

P = 23 X IE cos f

R = P/I2

Law of Cosines:

a2 = b2 + c2 — 2bc cos A

where sides a, b, and c of triangle abc are opposite angles A, B, and

C, respectively.

Questions

1. In a wye system, if the line voltage is known, give the formula for the phase voltage.

2. In a wye system, give the formula for the line current.

Power in Polyphase Circuits 275

3. In a delta system, give the formula for the line current.

4. In a wye system, give the formula for the phase voltage.

5. In a delta system, give the formula for the phase voltage.

6. Give the formula for total power in VA for three-phase circuits.

7. Give the formula for total power in watts for three-phase circuits.

8. Knowing the VA, how do you find the kVA?

9. Knowing the watts, how do you find the kilowatts?

Chapter 25

Transformer Principles

Chapter 16 covered electromagnetic induction. In transformers, the theory of mutual induction is the basis of their operation.

Induction Coil

Before discussing transformers as they are commonly thought of, the induction coil may be covered. It has the same fundamental principle as the transformer, except the input is usually DC, which is interrupted by a vibrator or some other means.

A bundle of soft iron wires usually forms the core, as in Figure

25-1. The primary winding is composed of relatively few turns of wire, and the secondary conductor is wound outside over the primary conductors and consists of many turns of a smaller-size wire than the primary.

LOW-VOLTAGE DC Figure 25-1 Typical induction coil.

Battery B supplies the electricity to the induction coil. Since DC causes induction for the initial charge only, a means must be supplied for interrupting the supply from the battery very quickly. This may be accomplished by a magnetic vibrator and points of V. When

277

278 Chapter 25

switch S1 is closed, the primary is energized, and the armature of vibrator V is pulled to the core, opening the points of V, deenergizing the primary; the capacitor C absorbs the current that would cause sparking at the points, causing a more rapid collapse of the magnetic field and thus a higher induced emf in the secondary.

Figure 25-2 illustrates the action that takes place. When the primary is energized, the primary current rises from A to B. This in turn induces a small emf in the secondary in the opposite direction, as shown by AFC. As the time AC for the rising of current in AB is considerable, the magnitude of the induced emf, GF, in the secondary is small. When the current rising against the emf of selfinduction reaches its maximum at B, and is momentarily stationary, the induced emf falls to C. The interrupter opens and breaks the circuit, and the current collapses from B to E, and with it the magnetic flux. The capacitor ensures a very short time interval from C to E, during which the flux collapses. The magnitude of the induced emf in a negative direction will soar to a great height as from C to P, and when the current finally reaches zero at point E, the secondary emf has collapsed from P to zero.

p

A

Figure 25-2 Relation of primary current and secondary emf in an induction coil.

Transformer Principles 279

In an induction coil, it is not necessary to apply an AC emf, as the vibrator or interrupter causes a pulsating DC. The secondary emf, AFC, is of little value as it is so feeble.

There are many variations of the induction coil. The intent here was merely to show the principle. The induction coil with which everyone is most familiar is probably the ignition coil on the automobile. (It would be good to observe that there is no such item as a DC transformer.)

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