CBSE Class 12 Physics 2017 Delhi Set 1 Paper

Question : 11 of 26

Marks: +1, -0

SECTION -C

(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) Calculation of phase difference between current and voltage
Name of quantity which leads
(ii) Calculation of value of '

C_1

', is to be connected in parallel
(i)

\;X_{ {\L}}=ω L= (1000 × 10^{-3}) Ω=100 Ω

\;X_{ {\c}}=\;{1}/{ω_{ {\c}}}= (\;{1}/{1000 × 2 × 10^{-6}}) Ω=500 Ω

Phase angle

\tan ϕ=\;{X_L-X_C}/{R}

\tan ϕ\;=\;{100-500}/{400}=-1

ϕ\;=-\;{π}/{4}

X_C>X_L

(

ϕ

phase angle is negative), hence current leads voltage
(ii) To make power factor unity

X_{ {\C}^{'}}\;=X_{ {\L}}

\;{1}/{ω C^{'}}\;=100

C^{'}\;=10 µ {\F}

C^{'}\;=C+C_1

10\;=2+C_1

C_1\;=8 µ {\F}

Go to Question:

Prev Question

Next Question