Answer :

Given,

Inductance, L = 100 mH = 100×10^{-3} H,

Capacitance, C = 2μ F = 2× 10^{-6} F

Resistance, R = 400 Ω

ω = 1000

(A) For phase difference,

Where, ω = angular speed of cycles

L = inductance

C = Capacitance

R = resistance

Thus,

∴

= 135^{0}

And current is in leading phase since

(B) Unit power factor,

Where,

R = resistance

L = Inductance

C’ = total capacitance

Given, unit power factor, cosϕ = 1

⇒

⇒

∴

= 10 μF

Thus, additional capacitance needed in parallel,

C_{a} = C’ – C

= 10 μF – 2 μF

= 8 μF

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