Alternating Current Analysis 243
ANALYTICAL SolutionZRjXTL 34 34j ^4
3(^22) ∠ 1
tan
ZT = 5 ∠53.13°
I V
ZT
10 0
55313
25313
∠
∠
∠
.
. (in RMS values)
P (active power) = I^2 ⋅^ R = (^22) 3 = 12 W
QL (reactive power) = I^2 XL = (^22) * 4 = 16 var
S (apparent power) = P + jQL
Sj 12 16 12 16 ^16 va
12
(^22) ∠ 1
tan
S = 20 ∠53.13° va
V = 10 0 ° L = 2 H
R = 3 Ω
i(t)
XL = 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.