Question

The rms current in the circuit below is shown as a function of the frequency of the driving Emf. Capacitor C = 0.240?F and the total resistance R = 229 ?.

Calculate the rms value of the driving Emf. Note that the curve passes through a grid intersection point.

Calculate the maximum average power dissipated in the circuit.

Calculate the inductance in the circuit.

Answer 1

Answer 2

Answer 3

We know that current in a RLC circuit is maximum when at resonance.

So, the peak we see which occurs at frequency , f = 700 Hz corresponds to the resonant frequency.

At resonance frequency, the inductive reactance (Xl) is equal to capacitive reactance (Xc)

So, at resonance , Xl = Xc

So, 2*pi*f*L = 1/(2*pi*f*C)

where L = inductance ,

C= capacitance = 0.24 uF = 0.24*10^-6 F

f = resonance frequency = 700 Hz

So, L = 1/(4*pi^2*f^2*C)

= 1/(4*3.14^2*700^2*0.24*10^-6)

= 0.216 H <----------------- inductance of the circuit

So, Vrms = Irms(at resonance) *R

Irms(at resonance) = 0.35 A

So, Vrms = 0.35*229 = 80.15 V <-------------answer

maximum avg power dissipated = Vrms*Irms

= 80.15*0.35

= 28.05 W <-----------answer