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JEE Main 20179 AprilOnline
Exam
JEE Main
Year
2017
Date
9 April
Mode
Online
Subject
Physics
Chapter
Alternating Current

JEE Main 2017 · 9 April · Online · Physics · Alternating Current

A sinusoidal voltage of peak value 283V and angular frequency 320 s...

A sinusoidal voltage of peak value 283V283V and angular frequency 320s−1320{s}^{- 1} is applied to a series LCRLCR circuit. Given that R=5Ω,L=25mHR = 5Ω, L = 25mH and C=1000μFC = 1000\mu F.The total impedance and phase difference between the voltage across the source and the current will respectively be-

Options

  1. A10Ωandtan⁡−1(53)10Ωand\tan^{- 1} \left(\dfrac{5}{3}\right)
  2. B7Ωand45o7Ωand{45}^{o}
  3. C7Ωandtan⁡−1(53)7Ωand\tan^{- 1} \left(\dfrac{5}{3}\right)
  4. D10Ωandtan⁡−1(83)10Ωand\tan^{- 1} \left(\dfrac{8}{3}\right)

Correct answer

B. 7Ωand45o7Ωand{45}^{o}

Step-by-step solution

ε0=283volt,ω=320s−1{ε}_{0} = 283volt, \omega = 320{s}^{- 1}

XL=320×25×10−3=8Ω{X}_{L} = 320 \times 25 \times {10}^{- 3} = 8Ω

XC=1\omegaC=1320×1000×10−6{X}_{C} = \dfrac{1}{\omegaC} = \dfrac{1}{320 \times 1000 \times {10}^{- 6}}

XC=10000320=3.1Ω{X}_{C} = \dfrac{10000}{320} = 3.1Ω

R=5ΩR = 5Ω

Z=R2+(XL−XC)2=50≈7ΩZ = \sqrt{{R}^{2} + {\left({X}_{L} - {X}_{C}\right)}^{2}} = \sqrt{50}\approx 7Ω

tanϕ=XL−XCR=1tan ϕ = \dfrac{{X}_{L} - {X}_{C}}{R} = 1

⟹ϕ=45o⟹ ϕ = {45}^{o}
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