Question

An AC voltage with an amplitude of 130 V is applied to a series combination of a 224 μF capacitor, a 114 mH inductor, and a 15.7 Ω resistor.

Calculate the power dissipated by the circuit at a frequency of 50.0 Hz.

Calculate the power factor at this frequency.

Calculate the power dissipation at a frequency of 60.0 Hz.Calculate the power factor at this frequency.

Calculate the power factor at this frequency.

Answer 1

at f = 50 Hz

inductive reactance, XL = 2*pi*f*L

= 2*pi*50*114*10^-3

= 35.81 ohm

capacitive reactance, XC = 1/(2*pi*f*C)

= 1/(2*pi*50*224*10^-6)

= 14.21 ohm

Impedance,
z = sqrt(R^2 +(XL -XC)^2)

= sqrt(15.7^2 + (35.81 - 14.21)^2)

= 26.7 ohms

so, Current Amplitude through ckt, I = V/Z

= 130/26.7

= 4.87 A

Average power dissipated, Pavg = (1/2)*I^2*R

= 0.5*4.87^2*15.7

= 186.2 W ---------->>>>>>>>>>>>>>>>>>>>Answer

Phase angle, phi = tan^-1( (XL-XC)/R)

= tan^-1( (35.81-14.21)/15.7)

= 54 degrees

so, power factor = cos(phi)

= cos(54)

= 0.588 ---------->>>>>>>>>>>>>>>>>>>>Answer

at f = 60 Hz

inductive reactance, XL = 2*pi*f*L

= 2*pi*60*114*10^-3

= 43 ohm

capacitive reactance, XC = 1/(2*pi*f*C)

= 1/(2*pi*60*224*10^-6)

= 11.84 ohm

Impedance,
z = sqrt(R^2 +(XL -XC)^2)

= sqrt(15.7^2 + (43 - 11.84)^2)

= 34.9 ohms

so, Current Amplitude through ckt, I = V/Z

= 130/34.9

= 3.72 A

Average power dissipated, Pavg = (1/2)*I^2*R

= 0.5*3.72^2*15.7

= 108.6 W ---------->>>>>>>>>>>>>>>>>>>>Answer

Phase angle, phi = tan^-1( (XL-XC)/R)

= tan^-1( (43-11.84)/15.7)

= 63 degrees

so, power factor = cos(phi)

= cos(63)

= 0.45 ---------->>>>>>>>>>>>>>>>>>>>Answer