Answer:
A) P(t) = 2651.25 [ 1 - cos2wt ] W
B) Real power = 999.79 watts
Reactive power = 2652.86 VA
c) power factor = 0.3526
Explanation:
Given data:
V(t) = 141.4 cos (ωt)
R(t) = 10 Ω
Inductive reactance XL = ωL = 3.77 Ω
Ir(t) = V(t) / R(t) = 14.14
A) Calculate the instantaneous power absorbed by the resistor and by inductor
By resistor :
Pr(t) = V(t) * Ir(t) = 141.4 * 14.14 [tex]cos^{2} wt[/tex] = 1999.396 [tex]cos^{2} wt[/tex]
hence Pr = 999.698 (cos2ωt + 1) w
By Inductor :
Pl(t) = V(t) I'L(t) = 141.4 cosωt * 37.5 cos(ωt - 90)
= 5302.5 [tex]sin^2 wt[/tex]
Hence Pl(t) = 5302.5 [tex]sin^2 wt[/tex] w = 2651.25 [ 1 - cos2wt ] W
B) calculate the real and reactive power
First we have to determine the power factor
Given that : V(t) = 141.4 cosωt v , Ir(t) = 14.14 cosωt A
IL(t) = 37.5 cos (ωt - 90° )
The phasor representation of the above is :
V = [tex]\frac{141.4}{\sqrt{2} } <0^{0} v[/tex] = 141.4 ∠0° , Ir = 10 ∠ 0° , IL = 37. 5 ∠ -90°
Total load current = Ir + IL = 28.35 ∠ -69.35°
power factor = cos -69.35° = 0.3526
Next we will determine the Real and reactive power using the relation below
S = VI = 100 ∠ 0° * 28.35 ∠ -69.35°
= 2835 ∠ 69.35°
S = P + jQ = 999.79 + 2652.85 j
Real power = 999.79 watts
Reactive power = 2652.85 VA
