Alternating Current Analysis
243
ANALYTICAL Solution
Z
R jX
j
T
L
ϭ ϩ
ϭ ϩ ϭ
ϩ
Ϫ
3
4
3
4
4
3
2
2
1
∠
tan
Z T = 5 ∠53.13°
I
V
Z T
ϭ
ϭ
Њ
Њ
ϭ Ϫ
Њ
10 0
5 53 13
2
5313
∠
∠
∠
.
.
(in RMS values)
P (active power) = I
2
⋅ R = 2
2 * 3 = 12 W
Q L (reactive power) = I
2 * X L = 2
2 * 4 = 16 var
S (apparent power) = P + jQ L
S
j
v a
ϭ ϩ
ϭ
ϩ
Ϫ
12
16
12
16
16
12
2
2
1
∠
tan
S = 20 ∠53.13° va
V = 10 0°
L = 2 H
R = 3 Ω
i(t )
X L = 4 Ω
v (t ) = (14.1) cos(2t ) V
FIGURE 3.15
Network of R.3.47.
S (apparent power
in va)
P (active power in w)
Q (reactive power
in var)
FIGURE 3.14
Power triangle.
CRC_47760_CH003.indd 243
CRC_47760_CH003.indd 243
7/23/2008 1:27:35 PM
7/23/2008 1:27:35 PM
243
ANALYTICAL Solution
Z
R jX
j
T
L
ϭ ϩ
ϭ ϩ ϭ
ϩ
Ϫ
3
4
3
4
4
3
2
2
1
∠
tan
Z T = 5 ∠53.13°
I
V
Z T
ϭ
ϭ
Њ
Њ
ϭ Ϫ
Њ
10 0
5 53 13
2
5313
∠
∠
∠
.
.
(in RMS values)
P (active power) = I
2
⋅ R = 2
2 * 3 = 12 W
Q L (reactive power) = I
2 * X L = 2
2 * 4 = 16 var
S (apparent power) = P + jQ L
S
j
v a
ϭ ϩ
ϭ
ϩ
Ϫ
12
16
12
16
16
12
2
2
1
∠
tan
S = 20 ∠53.13° va
V = 10 0°
L = 2 H
R = 3 Ω
i(t )
X L = 4 Ω
v (t ) = (14.1) cos(2t ) V
FIGURE 3.15
Network of R.3.47.
S (apparent power
in va)
P (active power in w)
Q (reactive power
in var)
FIGURE 3.14
Power triangle.
CRC_47760_CH003.indd 243
CRC_47760_CH003.indd 243
7/23/2008 1:27:35 PM
7/23/2008 1:27:35 PM