Respuesta :
Answer:
impedance Z = 416.66 ohm
voltage across inducance V = 346.99 V
power factor = 0.720
Explanation:
given data
resistor R = 300
voltage amplitude = 500 V
resistor = 216 W
to find out
impedance and amplitude of the voltage and power factor
solution
we apply here average power formula that is
average power = I²×R ............1
I = [tex]\frac{Vrms}{Z}[/tex]
so
average power = ([tex]\frac{Vrms}{Z}[/tex])²×R
Vrms = [tex]\frac{1}{\sqrt{2} }[/tex] × Vmax
Z = V × [tex]\sqrt{\frac{R}{2P} }[/tex]
Z = 500 × [tex]\sqrt{\frac{300}{2*216} }[/tex]
impedance Z = 416.66 ohm
and
we know voltage across inductor is here express as
V = I × X .............2
so here X will be by inductance
Z² = R² + (X)²
(X)² = 416.66² - 300²
X = 289.15 ohm
and I = [tex]\frac{V}{Z}[/tex]
I = [tex]\frac{500}{416.66}[/tex]
I = 1.20 A
so from equation 2
V = 1.20 × 289.15
voltage across inducance V = 346.99 V
and
average power = Vmax × Imax × cos∅
tan∅ = [tex]\frac{289.15}{300}[/tex]
tan∅ = 43.95°
so power factor is
power factor = cos43.95°
power factor = 0.720
