Practice Example Solutions Chapter 1 1 Alexander Sadiku [PDF]

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Monday, June 27.2011



CHAPTER 11 P.P.11.1



i (t )  33 sin(10t  60)  33 cos(10t  30) v(t )  330 cos(10t  20)



p(t )  v(t ) i (t )  (330)(33) cos(10t  20) cos(10t  30) 1 p(t )  10890[cos(20t  20  30)  cos(20  (30))] 2 p( t )  (3.5 + 5.445cos(20t–10˚)) kW 1 P  Vm I m cos( v   i )  3.5 kW 2 P.P.11.2



V  I Z  13208



1 V I cos(v  i ) 2 m m 1 P  (1320)(33) cos(8  30)  20.19 kW 2



P



P.P.11.3 3



32045 V



I



+ 



I



j1 



32045  101.1926.57 3 j



For the resistor, I R  I  101.1926.57 VR  3I  303.626.57 1 1 PR  Vm I m  (303.6)(101.19)  15.361 kW 2 2



For the inductor,



I L  101 .1926.57 VL  j I L  101.19( 26.57  90)  101.19116.57 1 PL  (101.19) 2 cos(90)  0 W 2



The average power supplied is 1 P  (320)(101.19) cos(45  26.57)  15.361 kW 2 P.P.11.4



Consider the circuit below. 8



40 V



+ 



I1



j4 



I2



-j2 



+ 



j20 V



For mesh 1, - 40  (8  j2) I 1  (- j2) I 2  0 ( 4  j) I 1  j I 2  20



(1)



- j20  ( j4  j2) I 2  (- j2) I 1  0 - j I 1  j I 2  j10



(2)



For mesh 2,



In matrix form,  4  j - j I 1   20    -j j  I 2   j10     2  j4 , I1 



 1  -10  j20 ,



1  553.14 and 



I2 



 2  10  j60



2  13.6 17.11 



For the 40-V voltage source, Vs  400 I 1  5 53.14 -1 Ps  (40)(5) cos(-53.14)  - 60 W 2 For the j20-V voltage source,



Vs  2090 I 2  13.6 17.11 -1 Ps  (20)(13.6) cos(90  17.11)  - 40 W 2



For the resistor, I  I1  5 V  8 I 1  40 1 P  (40)(5)  100 W 2



The average power absorbed by the inductor and capacitor is zero watts. P.P.11.5 We first obtain the Thevenin equivalent circuit across Z L . Z Th is obtained from the circuit in Fig. (a). -j4 



j10 



8



Z th



5



(a) Z Th  5 || (8  j4  j10) 



(5)(8  j6)  3.415  j0.7317 13  j6



VTh is obtained from the circuit in Fig. (b).



-j4 



j10  I



8



5



12 A



+ V th 



(b) By current division, I



8  j4 (12) 8  j 4  j10  5



VTh  5 I 



(60)(8  j 4)  37.5 - 51.34 13  j 6



Z L = (Z Th )* = [3.415–j0.7317] Ω Pmax 



P.P.11.6 Let



VTh



2



8 RL







(37.5) 2  51.47 W (8)(3.415)



We first find Z Th and VTh across R L . Z 1  80  j60



(90)(- j30)  9 (1  j3) 90  j30 (80  j60)(9  j27)  Z1 || Z 2   17.181  j24.57  80  j60  9  j27



Z 2  90 || (- j30)  Z Th



Z2 (9)(1  j3) (12060)  (12060) Z1  Z 2 89  j33  35.98  - 31.91



VTh  VTh



R L  Z Th  30 



The current through the load is VTh 35.98 - 31.91   0.6764 - 4.4 I Z Th  R L 47.181  j24.57 The maximum average power absorbed by R L is 1 1 2 Pmax  I R L  (0.6764) 2 (30)  6.863 W 2 2 P.P.11.7



0  t 1  16t i(t )   32  16t 1  t  2



2 1 T 2 1 1 i dt    (16t ) 2 dt   (32  16t ) 2 dt   1  2  0 T 0 2 256  1 2  t dt   (4  4t  t 2 ) dt    1 2  0 1  t 3   256  128    4t  2t 2   12   3  3 3 



2  I rms 2 I rms



2 I rms



T2



I rms 



256  9.238 A 3











2 P  I rms R  9.238 2 (9)  768 w



P.P.11.8



T   , v(t )  100 sin(t ), 0  t  



1 T 2 1  v dt   (100 sin(t )) 2 dt   0 T 0 4 10  1  1  cos(2t ) dt  5000  0 2



2 Vrms 



2 Vrms



Vrms  70.71 V P



P.P.11.9



2 Vrms 5000   833.3 W R 6



The load impedance is Z  60  j40  72.1133.7 



The power factor is pf  cos(33.7)  0.8321 lagging Since the load is inductive V 32010 I   4.438 - 23.69 A Z 72.1133.7 The apparent power is S = V rms (I rms )* = 0.5(320)(4.438)(10˚–(–23.69˚)) = 71033.69˚ VA P.P.11.10



The total impedance as seen by the source is ( j4)(8  j6) Z  10  j4 || (8  j6)  10  8  j2 Z  12.6920.62



The power factor is pf  cos(20.62)  0.936 (lagging) V 1650 I rms  rms   13.002 - 20.62 Z 12.6920.62 The average power supplied by the source is equal to the power absorbed by the load. 2 P  I rms R  (13.002) 2 (11.88)  1,062 W = 2.008 kW or



P  Vrms I rms pf  (165)(13.002)(0.936)  2.008 kW



P.P.11.11 (a)



S  Vrms I *rms  (110 85)(0.4  - 15) S  4470 VA S  S  44 VA



(b)



S  4470  15.05  j41.35



Q  41.35 VAR



P  15.05 W ,



(c)



pf  cos(70)  0.342 (lagging) Vrms 11085   27570 I rms 0.4 - 15 Z  94.06  j258.4  Z



P.P.11.12 (a)



If Z  250 - 75 ,



(b)



Q  S sin 



(c)



S



P.P.11.13



2 Vrms Z



pf  cos(-75)  0.2588 (leading)



  S 



Q  100 kVAR   103.53 kVA sin  sin(-75)



  Vrms  S  Z  (103530 )(250)  5.087 kV



Consider the circuit below. I



20  I1



V



+ 



(30–j10)



+ Vo



 Let I 2 be the current through the 60- resistor. P 240 P  I 22 R   I 22   4 R 60 I 2  2 (rms)



I2 (60+j20) 



Vo  I 2 (60  j20)  120  j40 I1 



Vo  3.2  j2.4 30  j10



I  I 1  I 2  5.2  j2.4



V  20 I  Vo  (104  j48)  (120  j40) V  224  j88  240.721.45˚ V rms For the 20- resistor, V  20 I  204  j48  114.5424.8 I  5.2  j2.4  5.72724.8



S  V I *  (114.5424.8)(5.727 - 24.8) S  656 VA For the (30 – j10)- impedance, Vo  120  j40  126.518.43 I 1  3.2  j2.4  4 36.87 S 1  Vo I 1*  (126 .518.43 )( 4  - 36.87 )



S 1 = 506‫–ס‬18.44° = [480–j160] VA



For the (60 + j20)- impedance, I 2  2 0  S 2  Vo I *2  (126 .518.43)( 2  - 0)



S 2 = 253‫ס‬18.43° = [240+j80] VA



The overall complex power supplied by the source is S T  V I *  (240.6721.45)(5.727 - 24.8)



S T = 1378.3‫–ס‬3.35° = [1376–j80] VA



P.P.11.14



For load 1, pf  0.75  cos 1   1  -41.41 P1 P1  S1 cos 1   S1   2666.67 cos 1 Q1  S1 sin 1  -176.85 S1  P1  jQ 1  2000  j1763.85 (leading) P1  2000 ,



For load 2, P2  4000 , pf  0.95  cos  2    2  18.19 P2 S2   4210.53 cos  2 Q 2  S 2 sin  2  1314.4 S 2  P2  jQ 2  4000  j1314.4 (lagging)



The total complex power is S = S 1 + S 2 = [6–j0.4495] kVA P 6000 pf    0.9972 (leading) 6016.18 S P.P.11.15



pf  0.85  cos      31.79 Q 140 Q  S sin    S    265.8 kVA sin  sin(31.79) P  S cos   225.93 kW



For pf  1  cos 1



  1  0



Since P remains the same,



P  P1  S1 cos 1



  S1 



P1  225.93 cos 1



Q 1  S1 sin 1  0



The difference between the new Q 1 and the old Q is Q c . 2 Q c  140 kVAR  CVrms



C



140  10 3  30.69 mF (2 )(60)(110) 2



P.P.11.16



Let



The wattmeter measures the average power from the source. Z 1  4  j2



Z 2  12 || j9 



(12)( j9)  4.32  j5.76 12  j9



Z  Z 1  Z 2  8.32  j3.76  9.13 24.32 2



V (120 ) 2 S  VI  *   1577 .224.32 VA 9.13 - 24.32 Z *



P  S cos   1.437 kW



P.P.11.17



Demand charge  $5  32,000  $160,000 Energy charge for the first 50,000 kWh  $0.08  50,000  $4,000 The remaining energy  500,000  50,000  450,000 kWh Charge for this bill  $0.05  450,000  $22,500 Total bill  $160,000  $4,000  $22,500  $186,500



P.P.11.18



Energy consumed  800 kW  20  26  416,000 kWh



The power factor of 0.88 exceeds 0.85 by 3  0.01 . Hence, there is a power factor credit which amounts to an energy credit of 0.1 416,000   3  1248 kWh 100 Total energy billed  416,000  1,248  414,752 kWh Energy cost  $0.06  414,752  $24,885.12