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Alternating Current question

2023 · 25 Jan · Shift 1 · Q70
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Alternating Current question

2023 · 25 Jan · Shift 1 · Q70

JEE MainPhysicsAlternating CurrentNumerical+4 / −1
An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω\OmegaΩ, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is ‾\underline{\hspace{2cm}}​ mH. (Take π2=10\pi^2=10π2=10)
Numerical answer
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Correct answer: 100

  1. For a series LCR circuit, the current amplitude is maximum at resonance.

  2. Resonance condition: XL=XCX_L = X_CXL​=XC​ ωL=1ωC\omega L = \frac{1}{\omega C}ωL=ωC1​ So, L=1ω2CL = \frac{1}{\omega^2 C}L=ω2C1​

  3. Given: f=2.0 kHz=2000 Hzf = 2.0\ \text{kHz} = 2000\ \text{Hz}f=2.0 kHz=2000 Hz C=62.5 nF=62.5×10−9 FC = 62.5\ \text{nF} = 62.5 \times 10^{-9}\ \text{F}C=62.5 nF=62.5×10−9 F

  4. Angular frequency: ω=2πf=2π(2000)=4000π\omega = 2\pi f = 2\pi(2000) = 4000\piω=2πf=2π(2000)=4000π

  5. Substitute into formula: L=1(4000π)2×62.5×10−9L = \frac{1}{(4000\pi)^2 \times 62.5\times 10^{-9}}L=(4000π)2×62.5×10−91​

  6. Simplify: (4000π)2=16×106π2(4000\pi)^2 = 16\times 10^6 \pi^2(4000π)2=16×106π2 Using π2=10\pi^2 = 10π2=10, (4000π)2=16×107(4000\pi)^2 = 16\times 10^7(4000π)2=16×107

Now, ω2C=16×107×62.5×10−9\omega^2 C = 16\times 10^7 \times 62.5\times 10^{-9}ω2C=16×107×62.5×10−9 =16×62.5×10−2= 16 \times 62.5 \times 10^{-2}=16×62.5×10−2 =1000×10−2=10= 1000 \times 10^{-2} = 10=1000×10−2=10

Hence, L=110=0.1 HL = \frac{1}{10} = 0.1\ \text{H}L=101​=0.1 H

  1. Convert to mH: 0.1 H=100 mH0.1\ \text{H} = 100\ \text{mH}0.1 H=100 mH

Therefore, the required inductance is: 100\boxed{100}100​

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