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Alternating Current question

2023 · 31 Jan · Shift 1 · Q74
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  5. /2023 · 31 Jan · Shift 1 · Q74

Alternating Current question

2023 · 31 Jan · Shift 1 · Q74

JEE MainPhysicsAlternating CurrentNumerical+4 / −1
An inductor of 0.5 mH0.5 ~\mathrm{mH}0.5 mH, a capacitor of 20 μF20 ~\mu \mathrm{F}20 μF and resistance of 20 Ω20 ~\Omega20 Ω are connected in series with a 220 V220 \mathrm{~V}220 V ac source. If the current is in phase with the emf, the amplitude of current of the circuit is x\sqrt{x}x​ A. The value of xxx is ‾\underline{\hspace{2cm}}​
Numerical answer
View written solutionFree

Correct answer: 242

  1. Condition for current to be in phase with emf

    In a series RLCRLCRLC circuit, current is in phase with the applied emf at resonance.

    So, the impedance is purely resistive: Z=R=20 ΩZ = R = 20\,\OmegaZ=R=20Ω

  2. Given source voltage

    The given ac source is 220 V220\,\text{V}220V, which is the rms voltage.

    Hence, Vrms=220 VV_{\text{rms}} = 220\,\text{V}Vrms​=220V

  3. Find rms current

    Using Ohm’s law for AC at resonance: Irms=VrmsR=22020=11 AI_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{220}{20} = 11\,\text{A}Irms​=RVrms​​=20220​=11A

  4. Find amplitude (peak) current

    Amplitude of current is: I0=2 Irms=2×11=112 AI_0 = \sqrt{2}\, I_{\text{rms}} = \sqrt{2}\times 11 = 11\sqrt{2}\,\text{A}I0​=2​Irms​=2​×11=112​A

    Given that amplitude is x\sqrt{x}x​ A, so x=112\sqrt{x} = 11\sqrt{2}x​=112​

  5. Compute xxx

    Squaring both sides: x=(112)2=121×2=242x = (11\sqrt{2})^2 = 121 \times 2 = 242x=(112​)2=121×2=242

  6. Answer check

    Derived answer: 242242242

    Stored correct answer: 242242242

    They match.

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