Physics Circle home Student Login

Home›Chapters›Alternating Current PYQs›Resonance, Bandwidth & Q Factor›Question 74

JEE Main 2022 (29 Jun, Shift 2) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2022 (29 Jun, Shift 2)

Q74JEE Main 2022 · 29 JUN · SHIFT 2Easy

Resonance, Bandwidth & Q Factor

An inductor of 0.5 mH, a capacitor of 200 μ\muF and a resistor of 2 Ω\Omega are connected in series with a 220 V ac source. If the current is in phase with the emf, the frequency of ac source will be ____________ ×\times 102 Hz.

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

Show answer and solution

Answer: 5

  1. Current in phase with emf means resonance: f=12πLCf = \dfrac{1}{2\pi\sqrt{LC}}
  2. LC=(0.5×10−3)(200×10−6)=10−7⇒LC=3.16×10−4 sLC = (0.5\times10^{-3})(200\times10^{-6}) = 10^{-7} \Rightarrow \sqrt{LC} = 3.16\times10^{-4}\text{ s}
  3. f=12π×3.16×10−4≈503 Hz≈5×102 Hzf = \dfrac{1}{2\pi \times 3.16\times10^{-4}} \approx 503\text{ Hz} \approx 5\times10^{2}\text{ Hz}
  4. So the answer is 5 (the '102' in the question means 10210^2).
Report a mistake

More questions on Resonance, Bandwidth & Q Factor