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Chapter 2 Additional Problems

X2.1 Three circuit elements, 10R = , 50 FC = , and 140 HL = , are connected in series

and excited by a 60 Hz sinusoidal source. Identify and find the value of two series-connected elements that could be substituted without change in the source current.

The network impedance is

1R j L j

C

= + Z

( ) ( )( ) ( )

6

6

110 2 60 140 10

2 60 50 10j j

= +

Z

10 53j= Z

The network can be modeled by a series-connected R and Cwhere

10R =

( ) ( )

1 150 F

2 60 53C

X

= = =

The contribution of the inductor is negligible for the frequency of this problem.

X2.2 Three circuit elements, 10R = , 50 FC = , and 140 HL = , are connected in paralleland excited by a 60 Hz sinusoidal source. Identify and find the value of two parallel-

connected elements that could be substituted without change in the source current.

The network admittance is

1 1j j C

R L

= +Y

( ) ( )( ) ( )6

6

1 12 60 50 10

10 2 60 140 10j j

= +

Y

0.1 18.928j= Y

The network can be modeled by a parallel-connected R and L where

1 110

0.1R

G= = =

( ) ( )

1 1140.141 H

2 60 18.928L

B

= = =

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X2.3 Three circuit elements, 10R = , 50 FC = , and 140 HL = , are connected in seriesand excited by a 60 Hz sinusoidal source. Identify and find the value of two parallel-

connected elements that could be substituted without change in source current.

The network impedance is1

R j L jC

= + Z

( ) ( )( ) ( )

6

6

110 2 60 140 10

2 60 50 10j j

= +

Z

10 53 53.935 79.32j= = Z

Whence, the network admittance is

1 10.00344 0.01822 S

53.935 79.32j= = = +

Y

Z

The network can be modeled by a parallel-connected R and Cwhere

1 1290.7

0.00344R

G= = =

( )

0.0182248.33 F

2 60

BC

= = =

X2.4 For the circuit of Fig. 2.20, let 1 2 120 0 VV V= = , 10x y= = Z Z , and 10z j= Z .

(a) Determine the value of current 1I and (b) the values of 1P and 1Q (average andreactive power supplied to the network by source 1V ).

(a)

1120 0

12 0 A10

xx

VI

= = =

Z

1 2240 0

24 90 A10 90

zz

V VI

+ = = =

Z

By KCL,

1 12 24 26.83 63.43 Ax zI I I j= + = =

(b)

( ) ( )*1 1 1 1 1 120 0 26.83 63.43V P jQ V I = + = = S

( ) ( )1 120 0 26.83 63.43 3219.6 63.43 VAV = = S

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So,

( )1 1 1cos 3219.6cos 63.43 1440.1 WVP S = = =

( )1 1 1sin 3219.6sin 63.43 2879.6 VARsVQ S = = =

X2.5 For the circuit of Fig. 2.20, let 1 22 240 0 VV V= = and 10x y z= = = Z Z Z .

(a) Determine the value of current gI and (b) the value of average power supplied by

source 2V .

(a) By Ohm's law,

1240 0

24 0 A10

xx

VI

= = =

Z

2120 0

12 0 A10y y

V

I

= = = Z

1 2360 0

36 0 A10

zz

V VI

+ = = =

Z

By KCL,

24 0 12 0 12 0 Ag x yI I I= = =

(b) By KCL,

212 0 36 0 48 0 Ay zI I I= + = + =

The average power flow from source 2V is

( ) ( ) ( )2 2 2 2 120 48 cos 0 5760 WP V I = = =

X2.6 The instantaneous voltage and current for the network of Fig. 2.1 (shown in the phasor

domain are ( ) ( )100sin 120 60 Vv t t= and ( ) ( )10cos 120 30 Ai t t= . (a)Calculate the value of average power (P) flowing into the network and (b) determinethe values of a series-connected R-L that model the network.

(a) The phasor terminal voltage and current are100 100

150 30 V2 2

V = =

1030

2I =

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Thus,

( )100 10

cos cos 60 250 W2 2

P VI = = =

(b)

( )2 2

2505

10 / 2

PR

I= = =

100 / 210

10/ 2

VZ

I= = =

( ) ( )2 22 2 10 5 8.66X Z R= = =

8.6622.97 mH

120

XL

= = =

X2.7 For the one-port network of Fig. 2.1, 100 VV V= = , 10 0 AI = , and the average

power flowing into the port is 500 WP= . Determine the values of a parallel-connected R and L that model the network if the source frequency is 60 Hz.

( )22 100

20500

VR

P= = =

The reactive power to the network is

( ) ( ) ( )2 2 22 2 2 1000 500 866.03 VARsQ S P VI P = = = =

( )22 100

11.547866.03

VX

Q= = =

( )

11.54730.63 mH

2 60

XL

= = =

X2.8 Determine the phase angle of voltage V in Problem X2.7.

( ) ( )1 1 500cos cos 60

100 10

PV I

VI

= = = = Since

0 , 60 , or I V = =

100 60 VV =

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X2.9 For the network of Fig. X2.1, the two loads are

Load 1: 20 kW @ 0.8 PF lagging

Load 2: 10 kW @ 0.707 PF leading

The loads are fed from the 60 Hz source SV through a feeder line with impedance

0.2j= lZ . Determine currents 1I , SI , and source voltage SV if the load voltage is480 0 VLV = .

With load average powers known and LV on the reference,

( )1

1

1

20,00052.083 A

480 0.8L

PI

V PF= = =

( )11 1 1cos 52.083 36.87 AI I PF= =

( )( ) ( )

( )1 122 22

10,000cos cos 0.707 29.467 45 A480 0.707L

PI PFV PF

= = =

By KCL,

1 252.083 36.87 29.467 45 63.36 9.46 ASI I I= + = + =

By KVL,

( ) ( )0.2 90 63.36 9.46 480 0 482.25 1.49 VS S LV I V= + = + = lZ

X2.10 For the network of Fig. X2.1, Load 1 is described by 10 kw @ 0.866 PF lagging andLoad 2 is a 6 resistor. Load voltage 240 0 VLV = . The feeder line has an

impedance 0.5 90= lZ . Determine (a) the source current SI , (b) the average

power SP supplied by the source, and (c) the value of source voltage SV .

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(a)

( )( ) ( )

( )1 111 11

10,000cos cos 0.866 48.11 30 A

240 0.866L

PI PF

V PF

= = =

2

240 040 0 A

6 6

LVI

= = =

By KCL,

1 2 48.11 30 40 0 85.13 16.4 ASI I I= + = + =

(b)

( )22

1

24010,000 19.6 kW

6 6

LS

VP P= + = + =

(c) By KVL,

( ) ( )0.5 90 85.13 16.4 240 0S S LV I V= + = + lZ

231.6 10.15 VSV =

X2.11 Describe a single equivalent load for the two loads of Problem X2.9.

Using the value of SI from Problem X2.9,

( ) ( )* 480 0 63.36 9.46eq L S S V I= =

5068.8 9.46 4999.9 833.1 VAeqS j= = +

The PF of the equivalent load is

( )cos 9.46 0.986 laggingeqPF = =

Thus,

@ @: 4999.9 W 0.986 PF laggingeq eq eqLoad P PF =

X2.12 Determine the value of a capacitorCthat if placed across the parallel-connected loads

of Problem X2.9 would correct the combined load PF to unity.

From the solution of Problem X2.11, the total lagging VARs for the combinedloads is 833.1 VARseqQ = . Thus,

( )22 480

276.56833.1

LC

eq

VX

Q= = =

( ) ( )

1 19.59 F

2 60 276.56CC

X

= = =

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X2.13 The three-phase network of Fig. X2.2 is balanced with a b c phase sequence. If10 40ABI = , determine the three line currents.

Based on [2.26],

( ) ( )303 10 40 3 30 17.32 10 AjaA ABI I e = = =

By three-phase symmetry,

17.32 110 A 17.32 130 AbB cC I I= =

X2.14 Rework Problem X2.13 if the phase sequence is a c b and all else is unchanged.

Since there is no applicable formula developed in the text for a c b phasesequence, aAI will be determined by use of KCL. For a c b sequence,

10 160 A 10 80 ABC CAI I= =

But,

10 40 10 80 17.32 70 AaA AB CAI I I= = =

By three-phase symmetry,

17.32 170 A 17.32 50 AbB cC I I= =

X2.15 The three-phase network of Fig. X2.2 is balanced with a b c phase sequence. Let0=

lZ and 3 4L j= + Z . If 48 20 AABI = , determine the value of source voltage

anV .

Since 0=lZ ,

( ) ( ) ( )48 20 5 53.13 240 73.13 Vab AB AB LV V I= = = = Z

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X2.16 Rework Problem X2.15 except let 0.1j= lZ .

Based on [2.26],

( ) ( )303 48 20 3 30 83.14 10 AjaA ABI I e = = =

By three-phase symmetry,83.14 130bBI =

By KVL and using ABV as determined in Problem X2.15,

ab aA AB b BV I V I = + l lZ Z

( ) ( ) ( ) ( )83.14 10 0.1 90 240 73.13 83.14 130 0.1 90 251.67 75.1 VabV = + =

Based on [2.16],30

145.31 45.1 V3

jab

an

V eV

= =

X2.17 The three-phase network of Fig. X2.2 is balanced with a b c phase sequence. Let0.1j= lZ and 3 4L j= + Z . If 120 0 VanV = , determine (a) the value of load

current ABI and (b) load voltage BCV .

(a) By delta-wye conversion, phase a of the network can be modeled as shown in Fig.X2.3. Whence,

( )

120 068.66 55.1 A

/ 3 0.1 3 4 / 3

anaA

L

VI

j j

= = =

+ + +lZ Z

Based on [2.26], the required load current is

( ) ( )30 68.67 55.1 1 3039.64 25.1 A

3 3

jaA

AB

I eI

= = =

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(b) By Ohm's law,

( ) ( )39.64 25.1 5 53.13 198.2 28.03 VAB AB LV I= = = Z

From three-phase symmetry,

198.2 91.97 VBCV =

2.18 For the network of Problem X2.16, calculate (a) the total complex power supplied bythe three-phase source and (b) the total complex power delivered to the load.

(a) Based on [2.66], the source-supplied complex power is

( ) ( )3 3 251.67 83.14 45.1 10TS ab aAV I = = + S

36.24 55.1 kVATS = S

(b) Using the result for ABV from Problem X2.15 and [2.55], the load complex power

is

( ) ( )*3 3 240 73.13 48 20 34.56 53.13 kVATL AB ABV I= = = S

X2.19 Calculate the total average power supplied to the delta-connected load for Problem

X2.17.

Since the line impedance is pure reactive, the average power may be calculatedat either the source or load. The average power delivered to the load using the results

of Problem X2.17 is

( ) ( ) ( ) ( )3 cos 3 198.2 68.66 cos 53.13 14.142 kWTL AB aA LP V I= = =Z

X2.20 The three-phase network of Fig. X2.4 is balanced with a b c phase sequence. Let10R X= = . If the load voltage 480 0ABV = , find (a) the value of source current

baI and (b) the total average power supplied to the load.

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(a) Based on [2.16],

480 0277.14 30 V

3 30 3 30

AB

AN

VV

= = =

Then,

277.14 3019.6 75 A

10 10

ANaA

VI

R jX j

= = =

+ +

Based on [2.29],

19.6 7511.32 45 A

3 30 3 30

aAba

II

= = =

(b) The load power is

( ) ( )23 3 19.6 10 588 WTL aAP I R= = =

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