Find the peak current and resonant frequency of the following circuit (as shown in figure).

Solution

Physics sir · 4 steps

Use the source's ω = 100 for the current, but resonance is fixed by L and C alone.

1. Find the reactances. From V = 30 sin 100t, the peak voltage is V₀ = 30 V and ω = 100 rad s⁻¹. So X_L = ω L = 100 × 0.1 = 10 Ω and X_C = 1/(ω C) = 1/(100 × 100 × 10⁻⁶) = 100 Ω.

   Z = √(R² + (X_C - X_L)²) = √(120² + 90²) = 150 Ω

2. Peak current. I₀ = V₀/Z = 30/150 = 0.2 A.
3. Resonant frequency. At resonance X_L = X_C, so ω₀ = 1/√LC = 1/√(0.1 × 10⁻⁴) = 1/√10⁻⁵ ≈ 316 rad s⁻¹. Then f₀ = ω₀/2π = 316/6.28 ≈ 50 Hz.
4. So. The peak current is 0.2 A and the resonant frequency is about 50 Hz, which is option (d).

The idea to keep. The current you get depends on the driving ω, but resonance is fixed by L and C alone: f₀ = 1/2π√LC. And Z can never be smaller than R.

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