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MCQEasyJEE 2024LCR Circuits & Resonance

JEE Physics 2024 Question with Solution

A capacitor of reactance 43 Ω4\sqrt{3} \, \Omega and a resistor of resistance 4 Ω4 \, \Omega are connected in series with an AC source of peak value 82 V8\sqrt{2} \, \text{V}. The power dissipation in the circuit is:

  • A

    2 W2 \, \text{W}

  • B

    3 W3 \, \text{W}

  • C

    5 W5 \, \text{W}

  • D

    4 W4 \, \text{W}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Reactance of capacitor XC=43 ΩX_C = 4\sqrt{3} \, \Omega, resistance R=4 ΩR = 4 \, \Omega, peak voltage V0=82 VV_0 = 8\sqrt{2} \, \text{V}.

Find: The power dissipation in the circuit.

For the series combination, the impedance is

Z=R2+XC2Z = \sqrt{R^2 + X_C^2}

Substituting the values,

Z=42+(43)2=16+48=64=8 ΩZ = \sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \, \Omega

Now convert peak voltage to RMS voltage:

VRMS=V02=822=8 VV_{\text{RMS}} = \frac{V_0}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \, \text{V}

The RMS current is

IRMS=VRMSZ=88=1 AI_{\text{RMS}} = \frac{V_{\text{RMS}}}{Z} = \frac{8}{8} = 1 \, \text{A}

Power is dissipated only in the resistor, so

P=IRMS2R=12×4=4 WP = I_{\text{RMS}}^2 R = 1^2 \times 4 = 4 \, \text{W}

Therefore, the power dissipation is 4 W4 \, \text{W} and the correct option is D.

Direct AC Power Relation

Given: R=4 ΩR = 4 \, \Omega, XC=43 ΩX_C = 4\sqrt{3} \, \Omega, V0=82 VV_0 = 8\sqrt{2} \, \text{V}.

Find: Power dissipated in the circuit.

First find the RMS voltage: VRMS=8 VV_{\text{RMS}} = 8 \, \text{V}. Also,

Z=R2+XC2=8 ΩZ = \sqrt{R^2 + X_C^2} = 8 \, \Omega

Hence,

IRMS=VRMSZ=1 AI_{\text{RMS}} = \frac{V_{\text{RMS}}}{Z} = 1 \, \text{A}

Then directly use

P=IRMS2RP = I_{\text{RMS}}^2 R

which gives

P=12×4=4 WP = 1^2 \times 4 = 4 \, \text{W}

This works because the capacitor does not dissipate average power; only the resistor does. So the correct option is D.

Common mistakes

  • Using the peak voltage directly in the power formula is incorrect because power calculations in AC circuits require RMS values. First convert V0V_0 to VRMSV_{\text{RMS}} using VRMS=V02V_{\text{RMS}} = \frac{V_0}{\sqrt{2}}.

  • Assuming the capacitor also dissipates power is wrong. In an ideal AC circuit, average power is dissipated only in the resistor. Use P=IRMS2RP = I_{\text{RMS}}^2 R, not impedance in place of resistance.

  • Taking impedance as R+XCR + X_C is incorrect because resistance and reactance combine vectorially. For a series RCRC circuit, use Z=R2+XC2Z = \sqrt{R^2 + X_C^2}.

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