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Alternating Current question

2018 · 15 Apr · Shift 1 · Q71
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Alternating Current question

2018 · 15 Apr · Shift 1 · Q71

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
An ideal capacitor of capacitance 0.2 μF0.2\,\mu F0.2μF is charged to a potential difference of 10V.10V.10V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5mH.0.5mH.0.5mH. The current at a time when the potential difference across the capacitor is 5V,5V,5V, is :
  1. A
    0.34  A0.34\,\,A0.34A
  2. B
    0.25  A0.25\,\,A0.25A
  3. C
    0.17  A0.17\,\,A0.17A
  4. D
    0.15  A0.15\,\,A0.15A
View written solutionFree

Correct answer: C

  1. Given data

    C=0.2 μF=0.2×10−6 F=2×10−7 FC = 0.2\,\mu F = 0.2 \times 10^{-6}\,F = 2 \times 10^{-7}\,FC=0.2μF=0.2×10−6F=2×10−7F L=0.5 mH=0.5×10−3 H=5×10−4 HL = 0.5\,mH = 0.5 \times 10^{-3}\,H = 5 \times 10^{-4}\,HL=0.5mH=0.5×10−3H=5×10−4H Initial capacitor voltage: V0=10 VV_0 = 10\,VV0​=10V At some later time: V=5 VV = 5\,VV=5V

  2. Use conservation of energy in the ideal LC circuit

    Since the battery is disconnected and both capacitor and inductor are ideal, total energy remains constant.

    Initial energy stored in the capacitor: U0=12CV02U_0 = \frac{1}{2} C V_0^2U0​=21​CV02​

    At the instant when capacitor voltage is 5 V5\,V5V:

    • Energy in capacitor: UC=12CV2U_C = \frac{1}{2} C V^2UC​=21​CV2
    • Energy in inductor: UL=12Li2U_L = \frac{1}{2} L i^2UL​=21​Li2

    Therefore, 12CV02=12CV2+12Li2\frac{1}{2} C V_0^2 = \frac{1}{2} C V^2 + \frac{1}{2} L i^221​CV02​=21​CV2+21​Li2

    Cancelling 12\frac{1}{2}21​: CV02=CV2+Li2C V_0^2 = C V^2 + L i^2CV02​=CV2+Li2

    So, Li2=C(V02−V2)L i^2 = C(V_0^2 - V^2)Li2=C(V02​−V2)

  3. Substitute values

    Li2=2×10−7(102−52)L i^2 = 2 \times 10^{-7}(10^2 - 5^2)Li2=2×10−7(102−52) =2×10−7(100−25)= 2 \times 10^{-7}(100 - 25)=2×10−7(100−25) =2×10−7×75= 2 \times 10^{-7} \times 75=2×10−7×75 =1.5×10−5= 1.5 \times 10^{-5}=1.5×10−5

    Hence, i2=1.5×10−55×10−4i^2 = \frac{1.5 \times 10^{-5}}{5 \times 10^{-4}}i2=5×10−41.5×10−5​ =3×10−2= 3 \times 10^{-2}=3×10−2

    i=3×10−2≈0.173 Ai = \sqrt{3 \times 10^{-2}} \approx 0.173\,Ai=3×10−2​≈0.173A

  4. Match with options

    i≈0.17 Ai \approx 0.17\,Ai≈0.17A

    So the correct option is: C: 0.17 A0.17\,A0.17A

  5. Comparison with stored answer

    Stored correct answer: C

    Our derived answer: C

    They match.

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