#### Question

(i) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase : current or voltage ?

(ii) Without making any other change, find the value of the additional capacitor C_{1}, to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity.

#### Solution

(i) Given V = V_{0}sin(1000t + ϕ)

ω = 1000 s^{-1}

Given,

L = 100 mH

C = 2 μF

R = 400 Ω

Phase difference ϕ= tan-1((XL−XC)/R)

X_{L} = ωL = 1000 × 100 × 10^{-3} = 100 Ω

`X_C=1/(omegaC)=1/(1000xx2xx10^-6)=500Omega`

`ϕ= tan^-1((100−500)/400)= tan^-1(-1)`

ϕ = -450 and the current is leading the voltage.

(ii)

For power factor to be unity, R = Z

or X_{L} = X_{C}

`ω^2 =1/(LC)` (C = resultant capacitance)

`10^6 = 1/(100xx10^-3xxC')`

⇒C' = 10^{-5} F

For two capacitance in parallel, resultant capacitance C'= C + C1

10^{-5} = 0.2 x 10^{-5} + C_{1}

⇒C_{1} = 8 μF