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Alternating Current question

2021 · 27 Jul · Shift 2 · Q52
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Alternating Current question

2021 · 27 Jul · Shift 2 · Q52

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
A 100 Ω\OmegaΩ resistance, a 0.1 μ\muμ F capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.
  1. A
    0.70 H
  2. B
    70.3 mH
  3. C
    7.03 ×\times× 10 −-− 5 H
  4. D
    70.3 H
View written solutionFree

Correct answer: D

  1. Condition for resonance in a series RLCRLCRLC circuit

At resonance,

rac{1}{2\pi\sqrt{LC}} = f

So,

L=1(2πf)2CL = \frac{1}{(2\pi f)^2 C}L=(2πf)2C1​
  1. Given data
  • Resistance: R=100 ΩR = 100\,\OmegaR=100Ω (not needed for resonance condition)
  • Capacitance:
C=0.1 μF=0.1×10−6F=10−7FC = 0.1\,\mu F = 0.1 \times 10^{-6} F = 10^{-7} FC=0.1μF=0.1×10−6F=10−7F
  • Resonant frequency:
f=60 Hzf = 60\,Hzf=60Hz
  1. Substitute into the formula
L=1(2π⋅60)2⋅10−7L = \frac{1}{(2\pi \cdot 60)^2 \cdot 10^{-7}}L=(2π⋅60)2⋅10−71​

First calculate:

2π⋅60≈376.992\pi \cdot 60 \approx 376.992π⋅60≈376.99

Then,

(376.99)2≈1.4212×105(376.99)^2 \approx 1.4212 \times 10^5(376.99)2≈1.4212×105

Now,

L=11.4212×105×10−7L = \frac{1}{1.4212 \times 10^5 \times 10^{-7}}L=1.4212×105×10−71​ L=11.4212×10−2L = \frac{1}{1.4212 \times 10^{-2}}L=1.4212×10−21​ L≈70.3 HL \approx 70.3\,HL≈70.3H
  1. Check options
  • A: 0.70 H0.70\,H0.70H ✗
  • B: 70.3 mH70.3\,mH70.3mH ✗
  • C: 7.03×10−5 H7.03 \times 10^{-5}\,H7.03×10−5H ✗
  • D: 70.3 H70.3\,H70.3H ✓

Therefore, the correct answer is Option D.

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