In a series LCR circuit, a resistor of 300 Ω, a capacitor of 25 nF and an inductor of 100 mH are used. For maximum current in the circuit, the angular frequency of the ac source is _________ × 10⁴ radians s⁻¹

Solution

Physics sir · 3 steps

Maximum current means resonance, so the resistor plays no part and only L and C fix ω.

1. Condition for maximum. The current is largest when the impedance is smallest. That happens at resonance, where X_L = X_C and the impedance falls to just R.

   ω₀ L = 1/(ω₀ C) ⇒ ω₀ = 1/√LC

2. Put in numbers. With L = 100 mH = 0.1 H and C = 25 nF = 25 × 10⁻⁹ F, LC = 2.5 × 10⁻⁹ s². So √LC = 5 × 10⁻⁵ s.
3. So. ω₀ = 1/(5 × 10⁻⁵) = 2 × 10⁴ rad s⁻¹, so the blank is 2. The 300 Ω only sets how large the peak current is, not where the peak occurs.

The common slip: 0.32. Using the frequency formula f₀ = 1/2π√LC gives about 3.2 × 10³ Hz, which is 0.32 in these units. The question asks for angular frequency, so do not divide by 2π.

The idea to keep. In a series LCR circuit, maximum current always means resonance, and ω₀ = 1/√LC depends only on L and C. Convert mH and nF to SI units before taking the root.

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