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Alternating Current question

2022 · 25 Jun · Shift 2 · Q57
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  5. /2022 · 25 Jun · Shift 2 · Q57

Alternating Current question

2022 · 25 Jun · Shift 2 · Q57

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 μ\muμ F and R = 100 Ω\OmegaΩ. The phase difference (Φ\PhiΦ) between the applied voltage and resultant current will be :
  1. A
    tan −-− 1(0.17)
  2. B
    tan −-− 1(9.46)
  3. C
    tan −-− 1(0.30)
  4. D
    tan −-− 1(13.33)
View written solutionFree

Correct answer: A

  1. Given data

    The applied voltage is V(t)=210sin⁡3000tV(t)=210\sin 3000tV(t)=210sin3000t so the angular frequency is ω=3000 rad/s\omega=3000\ \text{rad/s}ω=3000 rad/s

    Also, L=10 mH=10×10−3 H=0.01 HL=10\ \text{mH}=10\times 10^{-3}\ \text{H}=0.01\ \text{H}L=10 mH=10×10−3 H=0.01 H C=25 μF=25×10−6 FC=25\ \mu\text{F}=25\times 10^{-6}\ \text{F}C=25 μF=25×10−6 F R=100 ΩR=100\ \OmegaR=100 Ω

  2. Reactances

    Inductive reactance: XL=ωL=3000×0.01=30 ΩX_L=\omega L=3000\times 0.01=30\ \OmegaXL​=ωL=3000×0.01=30 Ω

    Capacitive reactance: XC=1ωC=13000×25×10−6X_C=\frac{1}{\omega C}=\frac{1}{3000\times 25\times 10^{-6}}XC​=ωC1​=3000×25×10−61​

    First compute: 3000×25×10−6=0.0753000\times 25\times 10^{-6}=0.0753000×25×10−6=0.075

    Hence, XC=10.075=13.33 ΩX_C=\frac{1}{0.075}=13.33\ \OmegaXC​=0.0751​=13.33 Ω

  3. Net reactance

    For a series LCR circuit, X=XL−XC=30−13.33=16.67 ΩX=X_L-X_C=30-13.33=16.67\ \OmegaX=XL​−XC​=30−13.33=16.67 Ω

  4. Phase difference

    In a series LCR circuit, phase angle is given by tan⁡ϕ=XL−XCR\tan\phi=\frac{X_L-X_C}{R}tanϕ=RXL​−XC​​

    Therefore, tan⁡ϕ=16.67100=0.1667≈0.17\tan\phi=\frac{16.67}{100}=0.1667\approx 0.17tanϕ=10016.67​=0.1667≈0.17

    So, ϕ=tan⁡−1(0.17)\phi=\tan^{-1}(0.17)ϕ=tan−1(0.17)

  5. Match with options

    This corresponds to: tan⁡−1(0.17)\boxed{\tan^{-1}(0.17)}tan−1(0.17)​

    Therefore, the correct option is A.

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