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NVAEasyJEE 2023LCR Circuits & Resonance

JEE Physics 2023 Question with Solution

An oscillating LC circuit consists of a 75 mH75\,mH inductor and a 1.2 μF1.2\,\mu F capacitor. If the maximum charge to the capacitor is 2.7 μC2.7\,\mu C. The maximum current in the circuit will be:

Answer

Correct answer:9

Step-by-step solution

Standard Method

Given: L=75×10−3 HL = 75 \times 10^{-3} \, \text{H}, C=1.2×10−6 FC = 1.2 \times 10^{-6} \, \text{F}, and maximum charge qmax=2.7×10−6 Cq_{\text{max}} = 2.7 \times 10^{-6} \, \text{C}.

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

Using energy conservation in LC circuits, maximum energy stored in the capacitor equals maximum energy stored in the inductor:

12LImax2=12qmax2C\frac{1}{2} L I_{\text{max}}^2 = \frac{1}{2} \frac{q_{\text{max}}^2}{C}

Solving for ImaxI_{\text{max}}:

Imax=qmaxLCI_{\text{max}} = \frac{q_{\text{max}}}{\sqrt{LC}}

Substituting the given values:

Imax=2.7×10−675×10−3⋅1.2×10−6I_{\text{max}} = \frac{2.7 \times 10^{-6}}{\sqrt{75 \times 10^{-3} \cdot 1.2 \times 10^{-6}}}

Simplifying:

Imax=9×10−3 A=9 mAI_{\text{max}} = 9 \times 10^{-3} \, \text{A} = 9 \, \text{mA}

Therefore, the maximum current in the circuit is 9 mA9 \, \text{mA}.

Energy Conservation in LC Oscillation

Given: The circuit is an ideal LC oscillator with inductor LL and capacitor CC.

Find: The maximum current when the capacitor carries maximum charge.

At one extreme of oscillation, the capacitor has maximum charge and all the energy is electric:

UC=qmax22CU_C = \frac{q_{\text{max}}^2}{2C}

At the other extreme, the current is maximum and all the energy is magnetic:

UL=12LImax2U_L = \frac{1}{2} L I_{\text{max}}^2

Since total energy remains constant in an ideal LC circuit:

qmax22C=12LImax2\frac{q_{\text{max}}^2}{2C} = \frac{1}{2} L I_{\text{max}}^2

Rearranging gives:

Imax=qmaxLCI_{\text{max}} = \frac{q_{\text{max}}}{\sqrt{LC}}

Now substitute qmax=2.7×10−6 Cq_{\text{max}} = 2.7 \times 10^{-6} \, \text{C}, L=75×10−3 HL = 75 \times 10^{-3} \, \text{H}, and C=1.2×10−6 FC = 1.2 \times 10^{-6} \, \text{F}.

This gives the final value:

Imax=9×10−3 AI_{\text{max}} = 9 \times 10^{-3} \, \text{A}

So the numerical answer is 9.

Common mistakes

  • Using the capacitor energy formula incorrectly as 12CV2\frac{1}{2}CV^2 without converting the given maximum charge into voltage. This is wrong because the given quantity is qmaxq_{\text{max}}, not VV. Use qmax22C\frac{q_{\text{max}}^2}{2C} directly instead.

  • Substituting milli and micro units without converting them to SI units. This gives an incorrect numerical value. Convert 75 mH75\,mH to 75×10−3 H75 \times 10^{-3} \, \text{H} and 1.2 μF1.2\,\mu F to 1.2×10−6 F1.2 \times 10^{-6} \, \text{F} before calculation.

  • Taking Imax∝qmaxI_{\text{max}} \propto \sqrt{q_{\text{max}}} from the incorrect substitution shown in some worked solutions. The formula is linear in charge: Imax=qmaxLCI_{\text{max}} = \frac{q_{\text{max}}}{\sqrt{LC}}, so use qmaxq_{\text{max}} directly, not its square root.

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