This question was previously asked in

ISRO (SDSC) Technician B (Electrical) April 2018 Official Paper

Option 2 : 10 Ω, 0.8 leading

**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})**