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February 5, 2006
CHAPTER 11
P.P.11.1 i( t)= 15 sin( 10 t+ 60 °)= 15 cos( 10 t− 30 °)
v( t)= 80 cos( 10 t+ 20 ° )
p( t)=v(t)i(t)=( 80 )( 15 )cos( 10 t+ 20 °)cos( 10 t− 30 ° )
8015 [cos( 20 t 20 30 ) cos( 20 - 30 )] 2
p( t)= ⋅ ⋅ + °− ° + − °
p( t) = 600 cos( 20 t − 10 ° ) + 385. 7 W
= V I cos(θ −θ) = 2
P (^) m m v i 385. 7 W
P.P.11.2 V = IZ = 200 ∠ 8 °
V I cos( ) 2
P = m m θv−θi
= ( 200 )( 10 )cos( 8 °− 30 °) = 2
P 927. 2 W
P.P.11.
3 j
I
For the resistor,
I R = I = 2. 53 ∠ 26. 57 °
V R = 3 I = 7. 59 ∠ 26. 57 °
V I
P (^) R m m 9. 6 W
−
8 ∠ 45 ° V I^ j 1 Ω
For the inductor,
I L= 2. 53 ∠ 26. 57 °
V L = j I L= 2. 53 ∠( 26. 57 °+ 90 °)= 2. 53 ∠ 116. 57 °
= ( 2. 53 ) cos( 90 °) = 2
P
2 L^0 W
The average power supplied is
= ( 8 )( 2. 53 )cos( 45 °− 26. 57 °) = 2
P 9. 6 W
P.P.11.4 Consider the circuit below.
For mesh 1,
- 40 + ( 8 −j 2 ) I 1 +(-j 2 ) I 2 = 0
( 4 − j) I (^) 1 −j I 2 = (^20) (1)
For mesh 2,
- j 20 + (j 4 −j 2 ) I 2 +(-j 2 ) I 1 = 0
- j j j 10 1 2
I + I = (2)
In matrix form,
j 10
4 j -j
2
1
I
I
Δ = 2 +j 4 , Δ 1 =-10+j20, Δ 2 = 10 +j 60
1 I (^) 1 and = ∠ ° Δ
2 I 2
For the 40-V voltage source,
V s= 40 ∠ 0 °
I 1 = 5 ∠ 53. 14 °
= ( 40 )( 5 )cos(- 53. 14 °) = 2
Ps - 60 W
j 4 Ω
- j 2 Ω
−
40 V j20 V
−
I 1 I 2
13 j 6
( 10 )( 8 j 4 ) V Th 5 I
Z (^) L Z Th 3. 415 − j 0. 7317 Ω
8 R
P
2
L
2 Th max
V
1. 429 W
P.P.11.6 We first find Z (^) Thand V Th across RL.
Let Z 1 = 80 +j 60
9 ( 1 j 3 ) 90 j 30
( 90 )(-j 30 ) 2 90 ||(-j^30 ) = − −
Z = =
= = 17. 181 j 24. 57 80 j 60 9 j 27
( 80 j 60 )( 9 j 27 ) Z Th Z 1 || Z 2
89 j 33
( 9 )( 1 j 3 ) ( 120 60 ) 1 2
2 Th ∠ °
Z Z
Z
V
V Th= 35. 98 ∠- 31. 91 °
R (^) L= Z (^) Th = 30 Ω
The current through the load is
- 181 j 24. 57
Th R^ L
Th
Z
V
I
The maximum average power absorbed by RL is
R
P
2 L
2 max I^6.^863 W
P.P.11.
8 4 t 1 t 2
4 t 0 t 1 i( t) T = 2
= ∫ =^ [^ ∫ +∫ − ]
2
1
2
1
0
2
T
0
2 2 rms (^4 t) dt (^84 t) dt 2
i dt T
I
= [^ ∫ +∫ − + ]
2
1
2 1
0
2 2 rms t dt (^44 t t )dt 2
I
t 4 t 2 t 3
I 8
2 1
3 2 2 rms ⎥= ⎦
I (^) rms 2. 309 A
P I R
2 rms^48 W
P.P.11.8 T =π,v(t)= 8 sin(t), 0 <t<π
∫ ∫
π
π
0
T 2
0
2 2 rms (^8 sin(t)) dt
v dt T
V
[ 1 cos( 2 t)] dt 32 2
V
0
2 rms − = π
∫
π
Vrms = 5. 657 V
R
V
P
2 rms 5. 333 W
P.P.11.9 The load impedance is
Z = 60 +j 40 = 72. 11 ∠ 33. 7 ° Ω
The power factor is
pf = cos( 33. 7 °) = 0. 832 (lagging)
Since the load is inductive
2. 08 - 23. 7 A
Z
V
I
The apparent power is
S Vrms Irms 156 VA
P.P.11.10 The total impedance as seen by the source is
8 j 2
(j 4 )( 8 j 6 ) 10 j 4 ||( 8 j 6 ) 10 −
Z = + − = +
Z = 12. 69 ∠ 20. 62 °
The power factor is
pf = cos( 20. 62 °) = 0. 936 (lagging)
Z
V
I
rms rms
Let I 2 be the current through the 60-Ω resistor.
R
P
P I R I
2 2
2 = 2 ⎯⎯→ = = =
I 2 = 2 (rms)
V o = I 2 ( 60 +j 20 )= 120 +j 40
- 2 j 2. 4 30 j 10
o 1 = + −
V
I
I = I 1 + I 2 = 5. 2 +j 2. 4
V = 20 I + V o =( 104 +j 48 )+( 120 +j 40 )
V = 224 +j 88 = 240.7 ∠ 21.45˚ Vrms
For the 20-Ω resistor,
V = 20 I = 204 +j 48 = 114. 54 ∠ 24. 8 °
I = 5. 2 +j 2. 4 = 5. 727 ∠ 24. 8 °
S = VI = ∠ ° ∠ °
S = 656 VA
For the (30 – j10)-Ω impedance,
V o= 120 +j 40 = 126. 5 ∠ 18. 43 °
I 1 = 3. 2 +j 2. 4 = 4 ∠ 36. 87 °
S 1 = V o I 1 = ∠ ° ∠ °
S 1 = 506 ∠-18.44° = 480 − j 160 VA
For the (60 + j20)-Ω impedance,
I 2 = 2 ∠ 0 °
S 2 = V o I 2 = ∠ ° ∠ °
S (^) 2 = 253 ∠ 18. 43 ° = 240 + j 80 VA
The overall complex power supplied by the source is
( 240. 67 21.45)( 5. 727 - 24.8 )
S T = VI = ∠ ° ∠ °
S (^) T= 1378. 3 ∠-3.35° = 1376 − j 80 VA
P.P.11.
For load 1,
P 1 = 2000 , pf = 0. 75 =cosθ 1 ⎯⎯→ θ 1 =- 41. 41 °
cos
P
P S cos S 1
1 1 1 1 1 = θ
= θ ⎯⎯→ =
Q 1 =S 1 sinθ 1 =- 176. 85
S 1 =P 1 +jQ 1 = 2000 −j 1763. 85 (leading)
For load 2,
P 2 = 4000 , pf = 0. 95 =cosθ 2 ⎯⎯→ θ 2 = 18. 19 °
cos
P
S
2
2 2 = θ
Q 2 =S 2 sinθ 2 = 1314. 4
S 2 =P 2 +jQ 2 = 4000 +j 1314. 4 (lagging)
The total complex power is
S = S 1 + S 2 = 6 − j 0. 4495 kVA
P 6000
pf S
0. 9972 (leading)
P.P.11.15 pf= 0. 85 =cosθ ⎯⎯→ θ= 31. 79 °
- 8 kVA sin( 31. 79 )
sin
Q
Q Ssin S = °
θ
= θ ⎯⎯→ =
P =Scosθ= 225. 93 kW
For pf = 1 =cosθ 1 ⎯⎯→ θ 1 = 0 °
Since P remains the same,
cos
P
P P Scos S 1
1 1 1 1 1 = θ
= = θ ⎯⎯→ =
Q 1 =S 1 sinθ 1 = 0
The difference between the new Q 1 and the old Q isQ .c
2 Q (^) c = 140 kVAR=ωCVrms
π
×
3
C 30. 69 mF
