CH11DP

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Design Problems 
 
DP 11-1 
( ) ( )1
100| | 125 kVA100 kW 0.8
pf=0.8
| |sin cos 0.8 125sin 36.9 75 kVAR
P
P pf
Q −
⎧ = = == ⎫ ⎪⇒⎬ ⎨⎭ ⎪ = = ° =⎩
S
S
 
 
(a) Now pf = 0.95 so 
1
100 105.3 kVA
0.95
sin (cos 0.95) 105.3sin (18.2 ) 32.9 kVAR
P
pf
Q −
= = =
= = ° =
S
S
 
 
so an additional 125 − 105.3 = 19.7 kVA is available. 
 
(b) Now pf = 1 so 
1
100 100 kVA
1
sin (cos 1) 0
P
pf
Q −
= = =
= =
S
S
 
 
and an additional 125−100= 25 kVA is available. 
 
(c) In part (a), the capacitors are required to reduce Q by 75 – 32.9 = 42.1 kVAR. In 
part (b), the capacitors are required to reduce Q by 75 – 0 = 75 kVAR. 
 
(d) 
 
Corrected power factor 0.95 1.0 
Additional available 
apparent power 
19.7 kVA 25 kVA 
Reduction in reactive power 42.1 kVAR 75 kVAR 
 
 
 
 
 
 
 
 
 
 
 
 
 
DP 11-2 
This example demonstrates that loads can be specified either by kW or kVA. The procedure is 
as follows: 
 
First load: ( )( )1 11
1
1 1
50 0.9 45 W50 VA
sin (cos 0.9) 50sin (25.8 ) 21.8 kVAR0.9
P pf
Qpf −
⎧ = = == ⎫ ⎪⇒⎬ ⎨ = = ° == ⎪⎭ ⎩
SS
S
 
Second load: 2
22
1
2 2
45 49.45 kVA45 W 0.91
0.91 sin (cos 0.91) 49.45sin (24.5 ) 20.5 kVAR
P
P pf
pf Q −
⎧ = = == ⎫ ⎪⇒⎬ ⎨= ⎭ ⎪ = = ° =⎩
S
S
Total load: (45 45) j(21.8+20.5) 90 42.3 kVAj= = + + = +L 1 2S S + S 
 
Specified load: 
s
ss
1
s s
90 92.8 kVA90 W 0.97
0.97 sin (cos 0.97) 92.8sin (14.1 ) 22.6 kVAR
P
P pf
pf Q −
⎧ = = == ⎫ ⎪⇒⎬ ⎨= ⎭ ⎪ = = ° =⎩
S
S 
 
The compensating capacitive load is c 42.3 22.6 19.7 kVARQ = − = . 
 
The required capacitor is calculated as 
 
2 3 2
c
c 3
c
(7.2 10 ) 12626 1.01 F
19.7 10 377 (2626)
X C
Q
μ×= = = Ω ⇒ = =×
V
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
DP 11-3 
Find the open circuit voltage: 
 
 
 
oc10 5 10 0.5 0j− + + − =I I V 
 
and 
oc10 
5
−= VI 
so 
 
oc 8 36.9 6.4 4.8 Vj= ∠ ° = +V 
 
Find the short circuit current: 
 
The short circuit forces the controlling voltage to be 
zero. Then the controlled voltage is also zero. 
Consequently the dependent source has been replaced 
by a short circuit. 
 
sc
10 0 2 0 A
5
∠ °= = ∠ °I 
 
The the Thevenin impedance is: 
 
oc
t
sc
 3.2 2.4 j= = +VZ
I
Ω 
 
(a) Maximum power transfer requires L t * 3.2 2.4 j= = −Z Z Ω . 
 
(c) ZL can be implemented as the series combination of a resistor and a capacitor with 
 
( )
1 3.2 and 4.17 mF
100 (2.4)
R C= Ω = = . 
 
(b) 
( )
2
oc
max
| | 64 2.5 W
8 8 3.2
P
R
= = =V 
 
 
 
 
 
 
 
 
 
 
DP 11-4 
 
 
When n is selected to deliver maximum power to Z3, the value of the maximum power is given 
as 
2
s
2
2 2
2 2
2
33 4
R
nP
R
n n
⎛ ⎞⎜ ⎟⎝ ⎠= ⎛ ⎞ ⎛+ + +⎜ ⎟ ⎜⎝ ⎠ ⎝
V
⎞⎟⎠
 
When R = 4 Ω, 
22
s
4 225 48 25
n R
P
n n
= + +
V
 
 
4 2 2 3
2
s 4 2 2
5 4
2 (25 48 25) (100 96 )0 R
(25 48 25)
50 50 0 1 1
dP n n n n n n
dn n n
n n n n
⎡ ⎤+ + − += = ⎢ ⎥+ +⎣ ⎦
⇒ − + = ⇒ = ⇒ =
V
 
 
When R = 8 Ω, a similar calculation gives n = 1.31. 
 
 
DP 11-5 
 
 
Maximum power transfer requires 
 
( )1210 6.28 1 0.628 *j jn
+ = = +Z 
Equating real parts gives 
2
10 1 3n
n
= ⇒ = .16 
 
Equating imaginary parts requires 
 
2 0.628 6.283.16
j X j X= − ⇒ = − 
 
This reactance can be realized by adding a capacitance C in series with the resistor and inductor 
that comprise Z2. Then 
 
( ) ( )( )5 5
1 16.28 6.28 0.1267
2 10 2 10 12.56
X C F
C
μπ π− = = − + ⇒ = =× × 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
DP 11-6 
 
 
Maximum power transfer requires 
 
( )7 67
7
1 || 100 10 10 *
10
100 10
1 10
R j
j C
R j
j RC
−= + ×
= −+
 
 ( ) ( )7 8 9(100 10) 1 10 100 10 10 10R j j RC RC j RC= − + = + + − 
 
Equating real and imaginary parts yields 
 
8 9100 10 and 10 10 0R RC RC= + − = 
then 
8 8
8 8 10 1010 100 10 99 0.101 nF
99
RC R R C
R
− −
− ⎛ ⎞= ⇒ = + = Ω ⇒ = =⎜ ⎟⎝ ⎠
 
 
 
	Design Problems 
	DP 11-1 
	DP 11-2 
	DP 11-3 
	DP 11-4 
	DP 11-5 
	DP 11-6