Question

An L-R-C series circuit consists of a 2.60 μF capacitor, a 4.50 mH inductor, and a 55.0 Ω resistor connected across an ac source of voltage amplitude 11.0 V having variable frequency.(a) At what frequency is the average power delivered to the circuit equal to 12VrmsIrms?(b) Under the conditions of part (a), what is the average power delivered to each circuit element?(c) What is the maximum current through the capacitor?

Answer 1

frequency = 1816.14 Hz

power = 0.5225 W

maximum current flow by capacitor is 0.19 A

Step-by-step explanation:

given data

capacitor C = 2.60 μF = 2.60 ×
`10^(-6)` F

inductor L = 4.50 mH = 4.50 ×
`10^(-3)` 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 ×
`10^(-3)`) - ( 1/ω(2.60 ×
`10^(-6)`) ) / 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 =
`\sqrt{R^(2)+(\omega L-((1)/(\omega C)))^2 }`

Z =
`\sqrt{55^(2)+(\omega 4.50*10^(-3)-((1)/(\omega 2.60*10^(-6))))^2 }`

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