Alternating Current

JEE Main level, single correct, medium

26 of 190

Question 26 of 190

A sinusoidal voltage V(t)=210sin⁡3000tV\left(t\right)=210\sin 3000t volt is applied to a series LCR circuit in which L=10mH,C=25μFL=10mH,C=25\mu F and R=100ΩR=100\Omega. The phase difference (Φ)\left(\Phi \right) between the applied voltage and resultant current will be

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Answer: (A) tan⁡−1(0.17){\tan }^{-1}\left(0.17\right)

Capacitive reactance is XC=1ωC=13000×25×10−6=403Ω{X}_{C}=\dfrac{1}{\omega C}=\dfrac{1}{3000\times 25\times {10}^{-6}}=\dfrac{40}{3}\Omega
Inductive reactance is XL=ωL=3000×10×10−3=30Ω{X}_{L}=\omega L=3000\times 10\times {10}^{-3}=30\Omega
Now, total reactance is X=XL−XCX={X}_{L}-{X}_{C}
=30−403=503Ω=30-\dfrac{40}{3}=\dfrac{50}{3}\Omega
tan⁡ϕ=XR=503100=16\tan \phi =\dfrac{X}{R}=\dfrac{\dfrac{50}{3}}{100}=\dfrac{1}{6}
tan⁡ϕ=16=0.17\tan \phi =\dfrac{1}{6}=0.17
Hence, the phase difference between the applied voltage and resultant current is tan⁡−1(0.17){\tan }^{-1}\left(0.17\right).

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