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JEE Main 202120 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
20 July
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2021 · 20 July · Morning Shift · Physics · Alternating Current

In an LCR series circuit, an inductor 30mH and a resistor 1Ω are...

In an LCRLCR series circuit, an inductor 30mH30mH and a resistor 1Ω1Ω are connected to an ACAC source of angular frequency 300rads−1.300rad{s}^{- 1} . The value of capacitance for which the current leads the voltage by 45°45 ° is 1x×10−3F\dfrac{1}{x} \times {10}^{- 3}F. Then the value of xx is

Correct answer

0

Step-by-step solution

tanϕ=xC−xLRtanϕ = \dfrac{{x}_{C} - {x}_{L}}{R}
tan⁡45=xC−xLR\tan 45 = \dfrac{{x}_{C} - {x}_{L}}{R}
xC−xL=R{x}_{C} - {x}_{L} = R
1\omegaC−\omegaL=R\dfrac{1}{\omegaC} - \omegaL = R
1\omegaC−300×0.03=1\dfrac{1}{\omegaC} - 300 \times 0 . 03 = 1
1\omegaC=10\dfrac{1}{\omegaC} = 10
C=110ω=110×300C = \dfrac{1}{10\omega} = \dfrac{1}{10 \times 300}
C=13×10−3C = \dfrac{1}{3} \times {10}^{- 3}
X=3X = 3
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