CBSE Class 12 Physics 2017 Delhi Set 1 Paper

Question : 11

Total: 26

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 C1, 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 ' C1 ', is to be connected in parallel
(i) ‌XL=ωL=(1000×10−3)Ω=100Ω
‌Xc=‌

ωc

=(‌

1000×2×10−6

)Ω=500Ω
Phase angle tan‌ϕ=‌

XL−XC

tan‌ϕ‌=‌

100−500

400

=−1
ϕ‌=−‌

As XC>XL ( ϕ phase angle is negative), hence current leads voltage
(ii) To make power factor unity
XC′‌=XL
‌

ωC′

‌=100
C′‌=10µF
C′‌=C+C1
10‌=2+C1
C1‌=8µF

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