A resistance of 8 Ω, a capacitive reactance of 10 Ω and an inductive reactance of 4 Ω are connected to form a series RLC circuit. The impedance and power factor of the circuit will be,

Concept:

For a series RLC circuit, the net impedance is given by:

Z = R + j (XL - XC)

XL = Inductive Reactance given by:

XL = ωL

XC = Capacitive Reactance given by:

XL = 1/ωC

The magnitude of the impedance is given by:

\(|Z|=\sqrt{R^2+(X_L-X_C)^2}\)

\(\tan \phi = \frac{{{X_L} - {X_C}}}{R}\)

\(\therefore \cos \phi = \frac{R}{{\sqrt {{{\left( {{X_L} - {X_C}} \right)}^2} + {R^2}} }} = \frac{R}{Z}\)

 Power factor \( = \frac{R}{Z}\)

Calculation:

R = 8 Ω

X_C = 10 Ω

X_L = 4 Ω

\(|Z|=\sqrt{R^2+(X_L-X_C)^2}\)

\(|Z|=\sqrt{8^2+(10-4)^2}\)

|Z| = 10 Ω

Power factor \( = \frac{R}{Z}\)\( = \frac{8}{10}\)

= 0.8 lead (Because X_C > X_L)