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MCQEasyJEE 2024LCR Circuits & Resonance

JEE Physics 2024 Question with Solution

A capacitor of capacitance 100 μF100 \, \mu \text{F} is charged to a potential of 12 V12 \, \text{V} and connected to a 6.4 mH6.4 \, \text{mH} inductor to produce oscillations. The maximum current in the circuit would be:

  • A

    3.2 A3.2 \, \text{A}

  • B

    1.5 A1.5 \, \text{A}

  • C

    2.0 A2.0 \, \text{A}

  • D

    1.2 A1.2 \, \text{A}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: C=100×10−6 FC = 100 \times 10^{-6} \, \text{F}, V=12 VV = 12 \, \text{V}, L=6.4×10−3 HL = 6.4 \times 10^{-3} \, \text{H}.

Find: The maximum current ImaxI_{\text{max}} in the LC circuit.

In an LC circuit, the initial energy stored in the capacitor becomes the maximum magnetic energy stored in the inductor.

12CV2=12LImax2\frac{1}{2} C V^2 = \frac{1}{2} L I_{\text{max}}^2

First, calculate the capacitor energy:

Ecap=12×100×10−6×(12)2E_{\text{cap}} = \frac{1}{2} \times 100 \times 10^{-6} \times (12)^2Ecap=12×100×10−6×144=0.0072 JE_{\text{cap}} = \frac{1}{2} \times 100 \times 10^{-6} \times 144 = 0.0072 \, \text{J}

Now equate this to the inductor energy:

12×6.4×10−3×Imax2=0.0072\frac{1}{2} \times 6.4 \times 10^{-3} \times I_{\text{max}}^2 = 0.0072Imax2=0.0072×26.4×10−3=0.01446.4×10−3=2.25I_{\text{max}}^2 = \frac{0.0072 \times 2}{6.4 \times 10^{-3}} = \frac{0.0144}{6.4 \times 10^{-3}} = 2.25Imax=2.25=1.5 AI_{\text{max}} = \sqrt{2.25} = 1.5 \, \text{A}

Therefore, the maximum current is 1.5 A1.5 \, \text{A}. The correct option is B.

The solution labels option D, but its working and final numerical result clearly give 1.5 A1.5 \, \text{A}, which corresponds to option B in the listed options.

Direct Formula Method

Given: C=100×10−6 FC = 100 \times 10^{-6} \, \text{F}, L=6.4×10−3 HL = 6.4 \times 10^{-3} \, \text{H}, V=12 VV = 12 \, \text{V}.

Find: ImaxI_{\text{max}}.

From energy conservation in an LC circuit,

12CV2=12LImax2\frac{1}{2}CV^2 = \frac{1}{2}LI_{\text{max}}^2

So,

Imax=VCLI_{\text{max}} = V\sqrt{\frac{C}{L}}

Substitute the values:

Imax=12100×10−66.4×10−3I_{\text{max}} = 12\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}}Imax=12×164=12×18=1.5 AI_{\text{max}} = 12 \times \sqrt{\frac{1}{64}} = 12 \times \frac{1}{8} = 1.5 \, \text{A}

Therefore, the maximum current is 1.5 A1.5 \, \text{A} and the correct option is B.

Common mistakes

  • Using the capacitor charge formula instead of energy conservation is incorrect because the question asks for the maximum current in an LC oscillation. The correct approach is to equate capacitor energy and inductor energy.

  • Forgetting unit conversion for 100 μF100 \, \mu \text{F} and 6.4 mH6.4 \, \text{mH} gives a wrong numerical answer. Convert them to 100×10−6 F100 \times 10^{-6} \, \text{F} and 6.4×10−3 H6.4 \times 10^{-3} \, \text{H} before substitution.

  • Cancelling the square incorrectly while solving for ImaxI_{\text{max}} leads to an error. After finding Imax2I_{\text{max}}^2, take the positive square root because maximum current is a magnitude.

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