Question

(0.75,5.598) i(t) in A p(t) in W (2.167,0) t(s) (0.25,-0.402) (0.333,0) (0.167.0) Figure 1: Time-domain plots of y(t), i(t) and p(t)

Answer 1

SOLUTION) (0.75,5.598) _pct) AVA t(sec) (0.167,0) (2.363,0) VA C0.25-0,402) (0,333,0) act) ict) an giren figure, voltage wareform leads the evareform. so, the impedance is panductive (Ane) current 2)Apparent power (5)=WP2 +Q2 where, p=active power and Q = reactive power. Now, P = vac acoso and Q = visi ino. where, vand I are RMS are RMS rabres of voltage and current wareform and ø is phase-angle between voltage and curent wareform. 0 = Wot = 2480 (0.333-0,163)=21 +(0.533-0,167) = * = 27 cost Ś ) [as, Plt) = 0 at t=0,333 sec U(t)=0 at t=0.167 sec and, I =2se sec from graph) - 076 Again, P = va cosø= 2 2 matha cosø [As where um and am are maximum valve e of voltage and current ware form and, Pm = Vmam and, v=vm/rz and 1=2m/r2]. Pm coso 2

Now, Pm=5.598-(-0.40 2) = 6.4 wast (From graph) row, p= p m cosø= 6.4 cos 7/6 = 2.77 watt. and Q = Pm sing tas, v2 = Vm= 2 = PM = Q = bot sina = 1.6 VAR. so,s=√ P2+Q2 = √2.72241.6² = 3.2 VA. (Ane) 2 Average poer CP)=2,77 watt - (Ana) pourer wer factor cosø=cos = 0,866 (lag). (An) -) com compler power (s)= P + jQ = 324(2.92451.6)va =3.2 L300 VA. Cons)