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CBSE Class 12 Physics 2017 Delhi Set 1 Paper

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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 C1C_1, to be connected in parallel with the capacitor CC, 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 ' C1C_1 ', is to be connected in parallel
(i)   XL=ωL=(1000×10−3)  Ω=100  Ω\;X_{L} = \omega L = (1000 \times 10^{-3}) \; \Omega = 100 \; \Omega
  Xc=1ωc=(11000×2×10−6)  Ω=500  Ω\;X_{c} = \frac{1}{\omega_{c}} = \left(\frac{1}{1000 \times 2 \times 10^{-6}}\right) \; \Omega = 500 \; \Omega
Phase angle tan⁡ϕ=XL−XCR\tan \phi = \frac{X_L - X_C}{R}
tan⁡ϕ=100−500400=−1\tan \phi = \frac{100-500}{400} = -1
ϕ=−π4\phi = -\frac{\pi}{4}
As XC>XLX_C > X_L ( ϕ\phi phase angle is negative), hence current leads voltage
(ii) To make power factor unity
XC′=XLX_{C'} = X_L
1ωC′=100\frac{1}{\omega C'} = 100
C′=10  μFC' = 10 \; \mu \mathrm{F}
C′=C+C1C' = C + C_1
10=2+C110 = 2 + C_1
C1=8  μFC_1 = 8 \; \mu \mathrm{F}
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