Obtain the overall impedance for the following cases a P 10

Obtain the overall impedance for the following cases: (a) P = 1000 W, pf = 0.8 (leading), V_rms = 220 V (b) P= 1500 W, Q = 2000 VAR (inductive), I_rms = 12 A (c) S = 4500 60 degree VA, V = 120 45 degree V

Solution

a) p=1000 watts.

power factor=0.8 leading

v=220 volts.

we know

p=vicos theta=1000

220*i*cos theta=1000

220*i*0.8=1000

i=1000/(220*0.8)=5.68 ampss.

z=v/i=220/5.68=38.72 ohms.

b) from the power triangle

s^2=p^2+q^2

s^2=(1500)^2+(2000)^2

s=2500

s=v*i=2500

v*12=2500

v=208.33 volts.

z=v/i=208.33/12=17.36 ohmss.

c) s=4500<60 va

v=120<45 volts.

s=v*i=4500<60

120<45*i=4500<60

i=37.5<15 amps.

z=v/i=(120<45)/(37.5<15)=3.2<30 ohms.