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Asked in NEET 2026 · Resonance condition and tuning
At resonance X_L=X_C, which gives f₀=1/(2π√LC).
The resistance plays no part in setting the resonant frequency; it only sets how sharp the peak is.
LC=(1×10⁻³)(0.1×10⁻⁶)=10⁻¹⁰ s², so
√LC=10⁻⁵ s.
f₀=1/(2π×10⁻⁵)=1.59×10⁴ Hz=15.9 kHz.
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