Answer:
a) ZT = [ 38.786 ∠-4.28° ] ohms
b) ZT = [ 8.673 ∠4.06° ] ohms
Explanation:
Given that;
Voltage V = 480V
three loads of 1) 10Kw, 8 kVAR
2) 5 kVA, 0.9 power factor lagging
3) 12kW, 0.95 pf leading
Now for load1
Active power P = V^2 / R
Resistance R1 = V^2 / P
= (480 * 480) / 10*1000 = 23.04 OHMS
Reactive power Q = 8kVAR
let x be reactance
Q = V^2 / x
8 * 1000 = (480 * 480) / x
x1 = j28.8 ohms (inductive)
Load 2
Active power = kVA * nf = 5 * 9 = 4.5 Kw
Reactive power = sqrt(S^2 - P^2)
= Sqrt (5^2 - 4.5^2)
= 2.179 kVAR
now Resistance R2 = V^2 / P2 = (480 * 480) / 4.5*10^3
= 51.2 ohms
Reactance X2 = V^2 / Q2 = (480 * 480) / 2.178*10^3
X2 = 105.75 ohms
x2 = -j105.75 ohms (capacitive)
Load 3
Active power = 12kW AND PF = 0.95
so
kVA rating = 12/0.95 = 12.63 kVA
reactive power Q3 = sqrt (12.63^2 - 12^2 )
= 3.94 kVAR
Resistance R3 = V^2/P3 = (480*480) / 12*10^3
= 19.2 ohms
Reactance x3 = V2 / Q3 = (480 * 480) / 3.94*10^3
x3 = 58.47 ohms
x3 = -j58.47 ohms (capacitive)
NOW
a)
series combination of R and X
1/ZT = [1/(R1 + jx1)] + [1/(R2-jx2)] + [1/(R3-jx3)]
we substitute
1/ZT = [1/(23.04 + j28.8)] + [1/(51.2 - j105.74)] + [1/(19.2-58.48)]
1/ZT = 0.02571 + j1.92688*10^-3
ZT = 38.67 -j2.897 (total impedance)
so ZT = [ 38.786 ∠-4.28° ] ohms
b)
parallel combination of R AND X
1/ZT = 1/R1 + 1/jx1 + 1/R2 + 1/(-jx2) + 1/R3 + 1/(-jX3)
1/ZT = 1/23.04 + 1/j28.8 + 1/51.2 + 1/(-j105.74) + 1/19.2 + 1/(-58.48)
1/ZT = 0.1150 - j0.008164
ZT = 8.65 +j0.6141
so ZT = [ 8.673 ∠4.06° ] ohms
