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Alternating Current question

2022 · 24 Jun · Shift 2 · Q54
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  5. /2022 · 24 Jun · Shift 2 · Q54

Alternating Current question

2022 · 24 Jun · Shift 2 · Q54

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
Given below are two statements : Statement I : The reactance of an ac circuit is zero. It is possible that the circuit contains a capacitor and an inductor. Statement II : In ac circuit, the average power delivered by the source never becomes zero. In the light of the above statements, choose the correct answer from the options given below.
  1. A
    Both Statement I and Statement II are true.
  2. B
    Both Statement I and Statement II are false.
  3. C
    Statement I is true but Statement II is false.
  4. D
    Statement I is false but Statement II is true.
View written solutionFree

Correct answer: C

  1. Analyze Statement I

    The net reactance of an AC circuit is X=XL−XCX = X_L - X_CX=XL​−XC​ where XL=ωL,XC=1ωC.X_L = \omega L, \qquad X_C = \frac{1}{\omega C}.XL​=ωL,XC​=ωC1​.

    If the circuit contains both an inductor and a capacitor, it is possible that XL=XC,X_L = X_C,XL​=XC​, so that X=XL−XC=0.X = X_L - X_C = 0.X=XL​−XC​=0.

    This happens in resonance.

    Therefore, Statement I is true.

  2. Analyze Statement II

    The average power delivered in an AC circuit is Pavg=VrmsIrmscos⁡ϕ,P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi,Pavg​=Vrms​Irms​cosϕ, where ϕ\phiϕ is the phase difference between voltage and current.

    If the circuit is purely inductive or purely capacitive, then ϕ=90∘,\phi = 90^\circ,ϕ=90∘, so cos⁡90∘=0.\cos 90^\circ = 0.cos90∘=0.

    Hence, Pavg=0.P_{\text{avg}} = 0.Pavg​=0.

    So the statement “the average power delivered by the source never becomes zero” is false.

    Therefore, Statement II is false.

  3. Choose the correct option

    • Statement I: True
    • Statement II: False

    Hence, the correct option is C\boxed{\text{C}}C​

  4. Comparison with stored answer

    Stored correct answer: C

    My derived answer: C

    So they agree.

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