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Zorbas/Electric Machines, 2e

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5

Chapter 1W Basic Electromagnetic Concepts

1W Basic Electromagnetic Concepts 1W.1 Examples and Problems on Electric Circuits 1W.2 Examples on Magnetic Concepts

This chapter includes additional examples and problems that will complement the

understanding of the corresponding concepts found in the chapter on Basic

Electromagnetic Concepts in the book.



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1W.1 Examples and Problems on Electric Circuits

EXAMPLE 1W-1

Find the polar form of

a) 3 + j 4

b) –1 – j1

Solution:

Refer to Fig. 1W-1.

a) 3 + j 4 = r ∠θ

! = 3! + 4! = 5

θ = arc tan !! = 53°

Thus,

3 + j 4 = 5 ∠53°

b) −1 − j1 = r ∠θ

! = 1! + 1! = 2

θ = arc tan !!!!

= –135°

Thus,

–1 – j1 = 2 ∠−135°



7

FIG. 1W-‐1

EXAMPLE 1W-2

Find the rectangular form of

a) 10 ∠37° b) 5 ∠135°

Solution:

a) 10 ∠37° = 10 (cos 37° + j sin 37°)

= 8 + j 6

b) 5 ∠135° = 10 (cos 135° + j sin 135°)

= −7.07 + j 7.07



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EXAMPLE 1W-3

Find Z1 + Z2 , Z1 Z2 and Z1Z2

if

Z1 = 5 ∠37° Ω

Z2 = 10 ∠53° Ω

Solution:

Z1 = 5 ∠37° = 4 + j3 Ω

Z2 = 10 ∠53° = 6 + j8 Ω

I) Z1 + Z2 = 4 + 6 + j(3 + 8) = 10 + j11 Ω

II) Z1 × Z2 = 5 × 10 ∠37°+ ∠53°

= 50 ∠90°

= 0 + j50

= j50

III) Z1Z2

= 510 ∠37°− ∠53° = 0.5 ∠16°

EXAMPLE 1W-4

In Fig. 1W-4(a) find the current Ιx by using

a) KVL

b) KCL

c) Thévenin’s theorem

d) Current-divider technique



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FIG. 1W-‐4 (a)

Solution:

a) KVL Fig. 1W-4(b)

Loop #1 20 − Ι1 (10 + 10) + 10 Ι2 = 0 …(1)

Loop #2 0 − Ι1 (10) + Ι2 (10 + 5) = 0 …(2)

From Eq. (2)

Ι1 = 1.5 Ι2 …(3)

and from (3) and (1)

20 – 1.5 (20) Ι2 + 10 Ι2 = 0

Ι2 = 1.0 A

= Ιx



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FIG. 1W-‐4 (b)

b) Refer to Fig. 1W-4(c).

As reference voltage is selected that of Node Z2, that is

Vz 2 = 0

Writing the nodal equations, we obtain

Node Z1

!!!10 +

!!!5 +

!!! − 2010 = 0

Vz 1 = 5 volts

and

!! =!!!5 = 1 A



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FIG. 1W-‐4 (c)

c) Using Thévenin’s (Fig. 1W-4(d)) to find:

Rth (Fig. 1W-4(e)):

!!! = 5+10 1010+ 10

= 10Ω

Vth (Fig. 1W-4(f)):

Vth= I 10

=20

10+10 10

=10 volts

!! =!!!!!!

=1010 = 1 A

!! = !!



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FIG. 1W-‐4 (d)

FIG. 1W-‐4 (e) and (f)

d) Using the current-divider concept (Fig. 1W-4(a)):

!! = !10

10+ 5

=20

10+ 10//5×10

10+ 5

= 1.0 A



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EXAMPLE 1W-5

Find ι in Figs. 1W-5(a) and 1W-5(b).

FIG. 1W-‐5 (a) and (b)

Solution:

Refer to Fig. 1W-5(a)

a) ! = !" + !!∫ ! !"

Taking Laplace transforms

!! = ! ! ! +

1!"

!(!) =!

! 1!" + !

Inverting

! ! =!! !

!! !"

= 10!!!/! m A

See Fig. 1W-5(c).



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b) Refer to Fig. 1W-5(b).

! = ! ! + !!"!"

Transforming

!! = ! ! ! + ! !

! ! =!

! ! ! ! + !

! ! =!! 1− !!! ! !

= 2 1− !!!.!×!"! ! m A

See Fig. 1W-5(d).

FIG. 1W-‐5 (c) and (d)



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EXAMPLE 1W-6

The current through the resistor of the circuit in Fig. 1W-6(a) is

!! = 2 sin 377! +!6 !

a) Determine the rms value of the total current.

b) Determine the power factor of the circuit.

c) Draw the phasor diagram of all voltages and currents.

FIG. 1W-‐6 (a)

Solution:

a) !! =!!∠30 = 1.414 ∠30°A

!! = 150(1.414)∠30 = 212.13∠30°A

!! =!"!.!"/!"

!"" !""×!"!! ∠!"= 5.6268 ∠−60°A

! = 1.414∠30°+ 5.6268∠−30°

= 4.038− !4.1658 = 5.8 ∠−45.89°A

b) cos (30º + 45.89º) = cos 75.89º = 0.2437 lagging

c) See Fig. 1W-6(b).



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FIG. 1W-‐6 (b)

PROBLEM 1W-1

A factory operates at a power factor of 0.8 lagging. When a capacitor of 860 kVAR is

connected in parallel to the load, the power factor becomes 0.95 leading and the utility’s

meter indicates an average power of 800 kW.

Determine the factory’s apparent power before and after the hook-up of the

capacitor bank.



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PROBLEM 1W-2

Refer to the partial power distribution of a residence (Fig. P1W-2). Determine the voltage

across Z1 when:

a) The neutral is disconnected (switch A is open and breakers b1 and b2 are

closed).

b) Switch A and breaker b2 are closed, while the breaker b1 is open.

FIG. P1W-‐2

PROBLEM 1W-3

For the circuit shown in Fig. P1W-3, determine an expression for the voltage across the

inductor, following the closing of the switch at t = 0. Sketch it and show all pertinent

values.



18

FIG. P1W-‐3

PROBLEM 1W-4

Refer to Fig. P1W-4.

a) Determine Thévenin’s equivalent across the terminals A–B.

b) Evaluate the above given

R = 8000 Ω, R1 = 1000 Ω, R2 = 500 Ω, R3 = 1000 Ω

K = 20

FIG. P1W-‐4



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PROBLEM 1W-5

a) Determine the voltage VAB, in the circuit of Fig. P1W-5(a), by nodal analysis.

b) Using Thévenin’s theorem, determine the current Ι in Fig. P1W-5(b).

FIG. P1W-‐5



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PROBLEM 1W-6

Refer to circuit in Fig. P1W-6. Determine

a) The current through the 10 Ω resistor.

b) The power consumed by the capacitor.

c) The current through the 20 Ω resistor.

d) The average power, reactive power, and complex power supplied by the

voltage source.

e) The input impedance of the circuit.

f) The current through the capacitor by using Thévenin’s theorem.

g) The voltage across the capacitor.

h) The phasor diagram of voltages and currents in each element.

FIG. P1W-6

PROBLEM 1W-7

a) Briefly explain the reasons that an individual is always capacitively and

magnetically coupled to his or her environment.

b) What are some of the applications of these natural phenomena?



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1W.2 Examples on Magnetic Concepts

EXAMPLE 1W-7

In the electrical power distribution of the 1976 Olympics, the 12 kV monoconductor

power cables were routed under the steel cover plate of a manhole. Upon the system’s

energization, the cables were damaged and the power supply to various buildings was

interrupted. Explain the reasons for it.

Solution:

1) The flux (Φ1) of the current through the monoconductor cables closed its loop

by cutting through the steel cover plate of the manhole. (Magnetic flux lines

form closed loops.)

2) This flux disturbed the magnetic domains of the steel plate and because action

is equal and opposite to reaction, the plate produced its own opposing flux

(Φ2).

3) This flux was generated by the so-called circular (eddy) currents (Ιε).

4) The circular currents in turn produced rate of heat (Ιε2 RE) within the material.

5) The alternating nature of the flux also produced alignment and realignment of

the magnetic domains and thus additional rate of heat produced (hysteresis

losses) due to frictional forces.

6) The generation of eddy current and hysteresis losses overheated the cables,

their insulation was damaged, and a flashover took place between the steel

plates and the monoconductor cables.

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