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NEET 2026 (Re-neet) · Series LCR Circuit

Series LCR Circuit question from NEET 2026 (Re-neet)

Q5NEET 2026 · RE-NEETEasy

Series LCR Circuit

An ac voltage V=220sin⁡(2×103t)V = 220 \sin\left(2 \times 10^{3} t\right) Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is:

(Given: L=10 mH, C=25 μF, R=100 ΩL = 10\text{ mH},\ C = 25\ \mu\text{F},\ R = 100\ \Omega)

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Answer: (b)

  1. ω=2×103\omega = 2 \times 10^3 rad/s
  2. XL=ωL=2000×0.01=20 ΩX_L = \omega L = 2000 \times 0.01 = 20\ \Omega, XC=12000×25×10−6=20 ΩX_C = \frac{1}{2000 \times 25 \times 10^{-6}} = 20\ \Omega
  3. XL=XCX_L = X_C: resonance, so Z=R=100 ΩZ = R = 100\ \Omega
  4. I0=V0R=220100=2.2I_0 = \frac{V_0}{R} = \frac{220}{100} = 2.2 A, answer (b).
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