DP:(p175-19)k2300V,=25A,PF=0.815(lag).Find(a)P;(b)Q; (c)s;(d)S;(e)Z. (a):P=46.9kW;(b):Q-333kvar;(c):57.5/35.4VA; (d):S=57.5kVA;(e):z=92235.4° Example: (p176-20)Find l and p (×5+40000021+(400002 =(120002 400kW 12k 007(g)=478(2366) P=5×4782+400000-=4114kWDP:(p175-19) V=2300V,I=25A,PF=0.815(lag).Find (a)P; (b) Q; (c) ;(d) S; (e) Z. _ S Example:( p176-20) Find I and Ps . 5 0.707( ) 400 lag kW − + 12kV i 2 2 2 2 (12000 ) ( 5 400000) (400000) I I =  + + Ps = 547.8 2 + 400000 = 411.4kW I = 47.8 ( 2366 ) (a):P=46.9kW; (b): Q=33.3kVar; (c): 57.5 35.4 oVA; (d): S=57.5kVA;(e):Z=92 35.4o