For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is 2.5 nF. If resistance of 200 Ω and 100 mH inductor is being used in the given circuit. The frequency of ac source is _________ × 10³ Hz (given a²=10)

Solution

Physics sir · 4 steps

Maximum current means resonance, so ignore the 200 ohm and use only L and C.

1. Spot resonance. Current in a series LCR circuit is largest when X_L = X_C, because the impedance then drops to just R. So 2.5 nF is the resonant capacitance, and the source frequency is the resonant frequency. The resistance does not affect this frequency.

   f = 1/2π√LC = 1/2√(π² LC)

2. Put in numbers. LC = (0.1 H) × (2.5 × 10⁻⁹ F) = 2.5 × 10⁻¹⁰ s². With π² = 10, π² LC = 2.5 × 10⁻⁹ = 25 × 10⁻¹⁰, so √(π² LC) = 5 × 10⁻⁵ s.
3. Find f. f = 1/(2 × 5 × 10⁻⁵) = 1/10⁻⁴ = 10⁴ Hz.
4. So. f = 10 × 10³ Hz, so the answer is 10.

The common slip: 63. Stopping at ω = 1/√LC ≈ 6.3 × 10⁴ rad s⁻¹ and reporting it as the frequency gives about 63, not 10. Divide by 2π to get f.

The idea to keep. Maximum current in series LCR means resonance. The resonant frequency f = 1/2π√LC depends only on L and C, never on R. Put π inside the root as π² when it is given as 10.

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