Alternating Current

JEE Main level, single correct, medium

133 of 190

Question 133 of 190

A circuit connected to an ac source of emf e=e0sin⁡⁡(100t)e={e}_{0}\sin ⁡\left(100t\right) with tt in seconds, gives a phase difference of π4\dfrac{\pi }{4} between the emf ee and current ii . Which of the following circuits will exhibit this?

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Answer: (D) RCRC circuit with R=1kΩR=1k\Omega and C=10μFC=10\mu F

ω=100\omega =100 rad/sec
phase difference between ee and ii is given by
tan⁡θ=XR∴θ=π4\tan \theta =\dfrac{X}{R}∴\theta =\dfrac{\pi }{4}
so tan⁡θ=tan⁡π4=1\tan \theta =\tan \dfrac{\pi }{4}=1
in case of RLRL circuit X=XL=ωLX={X}_{L}=\omega L
And in case of RCRC circuit X=Xc=1ωcX={X}_{c}=\dfrac{1}{\omega c}
For option (A):\left(A\right): R=1000ΩR=1000\Omega and
XC=1ωC=1(100)(1×10−6)=10000Ω{X}_{C}=\dfrac{1}{\omega C}=\dfrac{1}{\left(100\right)\left(1\times {10}^{-6}\right)}=10000\Omega
tan⁡⁡θ=XCR≠1\tan ⁡\theta =\dfrac{{X}_{C}}{R}\neq 1
For option (B):\left(B\right): R=103ΩR={10}^{3}\Omega and
XL=Lω=(10−3)(100)=0.1Ω{X}_{L}=L\omega =\left({10}^{-3}\right)\left(100\right)=0.1\Omega
tan⁡⁡θ=XLR≠1\tan ⁡\theta =\dfrac{{X}_{L}}{R}\neq 1
For option (C):R=1000Ω\left(C\right):R=1000\Omega and
XL=ωL=(100)(10)=1000Ω{X}_{L}=\omega L=\left(100\right)\left(10\right)=1000\Omega
tan⁡⁡θ=XLR≠1\tan ⁡\theta =\dfrac{{X}_{L}}{R}\neq 1
For option(D):R=1kΩ=1000Ω\left(D\right):R=1k\Omega =1000\Omega and
XC=1ωC=1(100)(10×10−6)=1000Ω{X}_{C}=\dfrac{1}{\omega C}=\dfrac{1}{\left(100\right)\left(10\times {10}^{-6}\right)}=1000\Omega
tan⁡⁡θ=XCR=1;θ=45o\tan ⁡\theta =\dfrac{{X}_{C}}{R}=1;\theta ={45}^{o}

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