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MHT CET 2021 (23 Sep, Shift 2) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from MHT CET 2021 (23 Sep, Shift 2)

Q94MHT CET 2021 · 23 SEP · SHIFT 2Medium

Resonance, Bandwidth & Q Factor

A series L-C-R circuit containing a resistance of 120 Ω120\ \Omega has angular frequency 4×105 rad s−14 \times 10^{5}\text{ rad s}^{-1}. At resonance the voltage across resistance and inductor are 60 V60\text{ V} and 40 V40\text{ V} respectively, then the value of inductance will be

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Answer: (a)

  1. Current I=VRR=60120=0.5 AI = \frac{V_R}{R} = \frac{60}{120} = 0.5\text{ A}.
  2. XL=VLI=400.5=80 ΩX_L = \frac{V_L}{I} = \frac{40}{0.5} = 80\ \Omega
  3. L=XLω=804×105=2×10−4 HL = \frac{X_L}{\omega} = \frac{80}{4\times10^{5}} = 2\times10^{-4}\text{ H}
  4. So the answer is (a) 0.2 mH0.2\text{ mH}.
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