A 100Ω resistance, a 0.1 μF capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.

Solution

Physics sir · 3 steps

At resonance X_L = X_C, so only f and C fix L. The resistance and the 250 V play no part.

1. Resonance condition. In a series LCR circuit, resonance happens when ω L = 1/(ω C). Rearranging gives L in terms of the frequency and the capacitance alone.

   L = 1/(ω² C) = 1/(4π² f² C)

2. Put in numbers. C = 0.1 μF = 1 × 10⁻⁷ F and f = 60 Hz. Then 4π² f² = 39.48 × 3600 = 1.421 × 10⁵ s⁻², and 4π² f² C = 1.421 × 10⁻².
3. So. L = 1/(1.421 × 10⁻²) ≈ 70.4 H, which matches option (d), 70.3 H.

The idea to keep. For resonance problems, reach for ω₀ = 1/√LC first. R only decides how sharp the peak is, not where it sits. Convert μF to farads before you calculate.

Answer from the NTA final answer key. Solution by PrepWiser.