Tài liệu Bài giải mạch P12 doc

Chapter 12, Solution 1.

(a) If 400 , then Vab =

= ∠ - 30° = 3

400

Van 231∠ - 30° V

Vbn = 231∠ - 150° V

Vcn = 231∠ - 270° V

(b) For the acb sequence,

= − = V ∠0°− V ∠120° Vab Van Vbn p p

 = ∠ ° 











 = + − V 3 - 30

2

3

j 2

1

Vab Vp 1 p

i.e. in the acb sequence, lags by 30°. Vab Van

Hence, if 400 , then Vab =

= ∠30° = 3

400

Van 231∠30° V

V V bn = 231∠150°

V V cn = 231∠ - 90°

Chapter 12, Solution 2.

Since phase c lags phase a by 120°, this is an acb sequence.

Vbn = 160∠(30°+120°) = 160∠150° V

Chapter 12, Solution 3.

Since V leads by 120°, this is an bn Vcn abc sequence.

V V an = 208∠(130°+120°) = 208∠250°

Chapter 12, Solution 4.

Vbc = Vca∠120° = 208∠140° V

Vab = Vbc∠120° = 208∠260° V

= ∠ °

∠ ° = ∠ ° = 3 30

208 260

3 30

ab

an

V

V V 120∠230°

Vbn = Van∠ -120° = 120∠110° V

Chapter 12, Solution 5.

This is an abc phase sequence.

= 3 ∠30° Vab Van

or = ∠ °

∠ ° = ∠ ° = 3 30

420 0

3 30

ab

an

V

V V 242.5∠ - 30°

V V bn = Van∠ -120° = 242.5∠ - 150°

Vcn = Van∠120° = 242.5∠90° V

Chapter 12, Solution 6.

= 10 + j5 = 11.18∠26.56° Z Y

The line currents are

= ∠ °

∠ ° = = 11.18 26.56

220 0

Y

an

a Z

V

I 19.68∠ - 26.56° A

I A b = I a∠ -120° = 19.68∠ - 146.56°

I c = I a∠120° = 19.68∠93.44° A

The line voltages are

Vab = 200 3 ∠30° = 381∠30° V

Vbc = 381∠ - 90° V

Vca = 381∠ - 210° V

The load voltages are

V V AN = I a Z Y = Van = 220∠0°

VBN = Vbn = 220∠ - 120° V

VCN = Vcn = 220∠120° V

Chapter 12, Solution 7.

This is a balanced Y-Y system.

+

− 440∠0° V ZY = 6 − j8 Ω

Using the per-phase circuit shown above,

= −

∠ ° = 6 j8

440 0 a I A 44∠53.13°

I A b = I a∠ -120° = 44∠ - 66.87°

I c = I a∠120° = 44∠173.13° A

Chapter 12, Solution 8.

V 220 V , L = Z Y = 16 + j9 Ω

= ∠ ° + = = = 6.918 - 29.36

3 (16 j9)

220

3

V V

Y

L

Y

p

an Z Z

I

IL = 6.918 A

Chapter 12, Solution 9.

= +

∠ ° = + = 20 j15

120 0

L Y

an

a Z Z

V

I 4.8∠ - 36.87° A

I A b = I a∠ -120° = 4.8∠ - 156.87°

I c = I a∠120° = 4.8∠83.13° A

As a balanced system, I n = 0 A

Chapter 12, Solution 10.

Since the neutral line is present, we can solve this problem on a per-phase basis.

For phase a,

= ∠ ° −

∠ ° = + = 6.55 36.53 27 j20

220 0

2 A

an

a Z

V

I

For phase b,

= ∠ ° ∠ ° = + = 10 -120 22

220 -120

2 B

bn

b Z

V

I

For phase c,

= ∠ ° +

∠ ° = + = 16.92 97.38 12 j5

220 120

2 C

cn

c Z

V

I

The current in the neutral line is

-( ) n a b c I = I + I + I

or n a b c -I = I + I + I

- (5.263 j3.9) (-5 j8.66) (-2.173 j16.78) I n = + + − + +

I n = 1.91− j12.02 = 12.17∠ - 81° A

Chapter 12, Solution 11.

∠ °

∠ ° = ∠ ° = ∠ ° = 3 - 90

220 10

3 - 90 3 - 90

bc BC

an

V V

V

Van = 127∠100° V

VAB = VBC∠120° = 220∠130° V

VAC = VBC∠ -120° = 220∠ -110° V

If = 30∠60°, then bB I

= 30∠180° aA I , = 30∠ - 60° cC I

= ∠ ° ∠ °

∠ ° = ∠ ° = 17.32 210

3 - 30

30 180

3 - 30

aA

AB

I

I

= 17.32∠90° BC I , = 17.32∠ - 30° CA I

I AC = -ICA = 17.32∠150° A

BC Z VBC I =

= ∠ °

∠ ° = = 17.32 90

220 0

BC

BC

I

V

Z 12.7∠ - 80° Ω

Chapter 12, Solution 12.

Convert the delta-load to a wye-load and apply per-phase analysis.

Ia

110∠0° V +

− ZY

= = ∠ ° Ω ∆ 20 45 3 Y

Z

Z