Question

A series RLC circuit with a resistance of 490 Ω has capacitive and inductive reactances of 330 Ω and 460 Ω , respectively.

a)What is the power factor of the circuit?

b)If the circuit operates at 60 Hz, what additional capacitance should be connected to the original capacitance to give a power factor of unity? (F)

c)How should the capacitors be connected?

Answer 1

A. for the RLC circuit

power factor = cos (phi) = R/Z

R = 490 ohm

XL = 460 ohm

XC = 330 ohm

Z = sqrt(R^2 + (XL - XC)^2)

Z = sqrt(490^2 + (460 - 330)^2)

Z = 506.95 ohm

power factor = R/Z = 490/506.95

P.f = 0.966

B.

P.f = 1

R/Z = 1

Z = R

Now, Z is given by

Z = sqrt(R^2 + (XL - XC)^2)

R = sqrt(R^2 + (XL - XC)^2)

which gives

XL = XC

initially

XL = 460 ohm

XC = 330 ohm

we need additional 130 ohm capacitive inductance

Xc = 130 = 1/2*pi*f*C

C = 1/(2*3.14*60*130)

C = 20.41*10^-6 F

C = 20.41 uF

C. Since we have to increase the capacitance so capcitor should be connected in parallel comb.