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JEE Main 2021 (20 Jul, Shift 1) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2021 (20 Jul, Shift 1)

Q84JEE Main 2021 · 20 JUL · SHIFT 1Easy

Resonance, Bandwidth & Q Factor

In an LCR series circuit, an inductor 30 mH and a resistor 1 Ω\Omega are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which, the current leads the voltage by 45∘{}^\circ is 1x×10−3\frac{1}{x} \times 10^{-3} F. Then the value of x is ____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

  1. Current leads voltage, so the circuit is capacitive: tan⁡45∘=XC−XLR=1\tan 45^\circ = \dfrac{X_C - X_L}{R} = 1
  2. XL=ωL=300×0.03=9 Ω⇒XC=9+1=10 ΩX_L = \omega L = 300 \times 0.03 = 9\ \Omega \Rightarrow X_C = 9 + 1 = 10\ \Omega
  3. C=1ωXC=1300×10=13×10−3 FC = \dfrac{1}{\omega X_C} = \dfrac{1}{300 \times 10} = \dfrac{1}{3}\times10^{-3}\text{ F}
  4. So x=3x = 3.
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