244 Practical MATLAB® Applications for Engineers
orSV IRMS RMS*(10 0∠∠)( 2 53 13. )20 53 13∠.vaThe 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 SolutionZj 705 97 705 97
97
705....tan22 1.
.∠
ΩZ = 12 ∠54° I
V
m Z
m^60
125AVRMS = 0.707 * 60 = 42.42 VIRMS = 0.707 * 5 = 3.535 ATaking 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°) = 0.588sin(54°) = 0.809thenS = 42.42 * 3.535 = 150 vaP = |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.