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Consider the circuit below.
At node 1,
j 4 3 2
100 20 V 1 V 1 V 1 V 2
2
1 ( 3 j 10 ) j 2 3
100 20 V
V
At node 2,
1 2 - j 2
100 20 V 2^ V 1 V 2 V 2
100 20 - 0. 5 V 1 (^) ( 1. 5 j 0. 5 ) V 2
(2)
From (1) and (2),
2
1
1 j 103 - j 2
V
V
- 1667 j 4. 5 1 j 103 - j 2
100 20 1. 5 j 0. 5
1
- 95 j 364. 5 1 j 103 100 20
- j 2
100 20 V
j 4
V 1 V
Io
2
1 V 1
2 V 2
- 3333 j 9
1 2 1 2 o
V V
I
I (^) o 9.25 -162.12 A
0 0 +20.0000i 1.0000 -18.0000i
V=[12i;0;-8]
V =
0 +12.0000i
0
-8.
I=inv(Z)*V
I =
2.0557 + 3.5651i
0.4324 + 2.1946i
0.5894 + 1.9612i
Vo = –j200(I 2 – I 3 ) = –j200(–0.157+j0.2334) = 46.68 + j31.4 = 56.2633.93˚
vo = 56.26cos(100t + 33.93˚) V.
I 1
V Z
o 120 90
I z
For mesh x,
ZI (^) x ZIzj 120 (1)
For mesh y,
ZI ZI 120 30 103. 92 j 60
o y ^ z (2)
For mesh z,
ZI (^) x ZIy 3 ZIz 0 (3)
Putting (1) to (3) together leads to the following matrix equation:
AI B
- 92 j 60
j 120
I
I
I
( 80 j 35 ) ( 80 j 35 ) ( 240 j 105 )
0 ( 80 j 35 ) ( 80 j 35 )
( 80 j 35 ) 0 ( 80 j 35 )
z
y
x
Using MATLAB, we obtain
I inv(A)*B
1 0.^26412.^3662.^3896.^37 A
o I Ix j
2 1.^91671.^41162.^38143.^63 A
o I Iy Ix j
3 2.^1810.^9542.^3823.^63 A
o I Iy j
Consider the network shown below.
I (^) o I 2
S 2
I +
S 2 1. 2 j 0. 8 kVA
sin(cos ( 0. 9 )) 4 j 1. 937 kVA
- 9
4 j
Let S 4 S 2 S 3 5. 2 j 1. 137 kVA
But
S (^) 4 V o I 2
3
j
o
V
S
I
I 2 11. 37 j 52
Similarly, sin(cos ( 0. 707 )) 2 ( 1 j)kVA
- 707
2 j
But
S 1 (^) V o I 1
3
j
Vo
S
I
I 1 – 14.142 + j14.
I o I 1 I 2 -2.772j66.14 66.2 92.4° A
S o VoI o
S o ( 100 90 )( 66. 2 -92.4) VA
S (^) o 6.62 –2.4° kVA
66.2 92.4° A, 6.62 –2.4° kVA
S 1 S 3
Vo
1
So
Consider the circuit below.
0.2 + j0.04
sin(cos ( 0. 8 )) 15 j 11. 25
- 8
15 j
But
S (^) 2 V 2 I 2
15 j 11. 25
2
*^2 2
V
S
I
I 2 0. 125 j 0. 09375
V 1 V 2 I 2 ( 0. 3 j 0. 15 )
V 1 120 ( 0. 125 j 0. 09375 )( 0. 3 j 0. 15 )
V 1 120. 02 j 0. 0469
sin(cos ( 0. 9 )) 10 j 4. 843
- 9
10 j
But
S 1 (^) V 1 I 1
1
*^1 1 V
S
I
I 1 0. 093 -25.82 0. 0837 j 0. 0405
I I 1 I 2 0. 2087 j 0. 053
V s V 1 I ( 0. 2 j 0. 04 )
V s ( 120. 02 j 0. 0469 )( 0. 2087 j 0. 053 )( 0. 2 j 0. 04 )
V s 120. 06 j 0. 0658
V s 120. 06 0. 03 V
I
I 1
I 2 0.3 +^ j0.15^
V (^) s V 1 V 2
(a) 15 mH 2 601510 5. 655
3 j x x x j
We apply mesh analysis as shown below.
I 1
I x
120<
o V 10
I n
30
I z
o V I y
j5.655
I 2
For mesh x,
120 = 10 I x - 10 I z (1)
For mesh y,
120 = (10+j5.655) I y - (10+j5.655) I z (2)
For mesh z,
0 = -10 I x –(10+j5.655) I y + (50+j5.655) I z (3)
Solving (1) to (3) gives
I x = 20, I y = 17.09-j5.142, I z = 8
Thus,
I 1 =I x =20 A
I 2 =-I y =- 17.09+j5.142 = 17. 85 163. 26 A
o
I n =I y - I x = –2.91 –j5.142 = (^5). 907 119. 5 A o
(b) S 1 ( 120 )I x 120 x 20 2400 , S 2 ( 120 )I y 2051 j 617
S S 1 S 2 [ 4. 451 j 0. 617 ] kVA
(c ) pf = P/S = 4451/4494 = 0.9904 (lagging)
