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JEE Main 2025 (23 Jan, Shift 2) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2025 (23 Jan, Shift 2)

Q27JEE Main 2025 · 23 JAN · SHIFT 2Easy

Resonance, Bandwidth & Q Factor

In a series LCR circuit, a resistor of 300 Ω300\ \Omega, a capacitor of 25 nF and an inductor of 100 mH are used. For maximum current in the circuit, the angular frequency of the ac source is _________ ×104\times 10^{4} radians s−1\text{s}^{-1}

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

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Answer: 2

  1. Maximum current at resonance: ω0=1LC\omega_0 = \dfrac{1}{\sqrt{LC}}
  2. LC=(0.1)(25×10−9)=2.5×10−9=25×10−10⇒LC=5×10−5 sLC = (0.1)(25\times10^{-9}) = 2.5\times10^{-9} = 25\times10^{-10} \Rightarrow \sqrt{LC} = 5\times10^{-5}\text{ s}
  3. ω0=15×10−5=2×104 rad/s\omega_0 = \dfrac{1}{5\times10^{-5}} = 2\times10^{4}\text{ rad/s}
  4. So the answer is 2.
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