244 Practical MATLAB® Applications for Engineers
or
SV IRMS RMS*(10 0∠∠)( 2 53 13. )20 53 13∠.va
The power triangle for the circuit diagram of Figure 3.15 is shown in Figure 3.16.
R.3.48 Evaluate the active power P, apparent power S, the reactive power Q, and the PF,
if the impedance of a given circuit is Z = 7. 0 5 + j9.7 Ω and the voltage across it is
Vm = 60 V.
ANALYTICAL Solution
Zj 705 97 705 97
97
705
....tan22 1.
.
∠
Ω
Z = 12 ∠54°
I
V
m Z
m^60
12
5A
VRMS = 0.707 * 60 = 42.42 V
IRMS = 0.707 * 5 = 3.535 A
Taking the current I as reference, the phasor diagram that relates the current I to the
voltage V is shown in Figure 3.17.
PF = cos(54°) = 0.588
sin(54°) = 0.809
then
S = 42.42 * 3.535 = 150 va
P = |S|c o s ( ) = (^150) * 0.588 = 88.1 W
S = 20 va
P = 12 W
QL = 16 var
= 53.13°
FIGURE 3.16
Power triangle of the circuit in Figure 3.15.
V = 42.42 ∠ 54 °
I = 3.535 ∠ 0 °
= 54°
FIGURE 3.17
Phasor diagram of R.3.48.