An inductor of 0.5 mH, a capacitor of 200 μF and a resistor of 2 Ω are connected in series with a 220 V ac source. If the current is in phase with the emf, the frequency of ac source will be ____________ × 10² Hz.

Solution

Physics sir · 3 steps

Current in phase with emf means resonance, so set X_L = X_C and find the resonant frequency.

1. Spot resonance. In a series RLC circuit the current is in phase with the emf only when X_L = X_C. The two reactances then cancel and the circuit behaves like a pure resistor.

   ω L = 1/(ω C) ⇒ f = 1/2π√LC

2. Put in numbers. LC = (0.5 × 10⁻³) × (200 × 10⁻⁶) = 10⁻⁷ s², so √LC = 3.16 × 10⁻⁴ s. Then f = 1/(2π × 3.16 × 10⁻⁴) ≈ 503 Hz.
3. So. f ≈ 5.03 × 10² Hz, so the blank is 5. The 2 Ω resistor and the 220 V supply do not affect the resonant frequency.

The common slip: 31.6. Stopping at ω = 1/√LC = 3162 rad s⁻¹ gives 31.6, not 5. Divide by 2π to turn angular frequency into frequency.

The idea to keep. In phase means resonance, and the resonant frequency depends only on L and C. Ignore R and the source voltage, and remember to divide ω by 2π.

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