Respuesta :
Answer:
frequency = 1816.14 Hz
power = 0.5225 W
maximum current flow by capacitor is 0.19 A
Explanation:
given data
capacitor C = 2.60 μF = 2.60 × [tex]10^{-6}[/tex] F
inductor L = 4.50 mH = 4.50 × [tex]10^{-3}[/tex] H
resistor R = 55.0 Ω
voltage amplitude V = 11.0 V
to find out
frequency and average power and maximum current
solution
we apply power formula that is
power = V × I × cos∅ ..............1
and given
avg power = 1/2 ×V rms × I rms
cos∅ = 1/2
∅ = 60°
so
tan∅ = ωL - ( 1/ωC ) / R
tan60 = ω(4.50 × [tex]10^{-3}[/tex]) - ( 1/ω(2.60 × [tex]10^{-6}[/tex]) ) / 55
ω = 11405.41 rad/s
so frequency = ω / 2π
frequency = 11405.41 / 2π
frequency = 1816.14 Hz
and
current = V/Z
here Z is impedance
Z = [tex]\sqrt{R^{2}+(\omega L-(\frac{1}{\omega C}))^2 }[/tex]
Z = [tex]\sqrt{55^{2}+(\omega 4.50*10^(-3)-(\frac{1}{\omega 2.60*10^(-6)}))^2 }[/tex]
Z = 57.74 Ω
so
current = 11 / 57.74
current = 0.19 A
so
power = 1/2 ×I×V × cos60
power = 1/2 ×0.19×11× cos60
power = 1/2 ×0.19×11× cos60
power = 0.5225 W
and
maximum current flow by capacitor is 0.19 A
