In an LCR series circuit, an inductor 30 mH and a resistor 1 Ω are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which, the current leads the voltage by 45° is 1/x × 10⁻³ F. Then the value of x is ____________.

Solution

Physics sir · 4 steps

Current leading means the capacitor wins: at 45° the extra capacitive reactance must equal R exactly.

1. Inductive reactance. X_L = ω L = 300 × 30 × 10⁻³ = 9 Ω.
2. Use the phase angle. Current leads voltage, so the circuit is capacitive and X_C > X_L. For a phase difference φ between current and voltage, the reactances satisfy the relation below. With φ = 45°, tan φ = 1.

   tan φ = (X_C - X_L)/R = 1 ⇒ X_C = X_L + R = 9 + 1 = 10 Ω

3. Find C. C = 1/(ω X_C) = 1/(300 × 10) = 1/3000 F = 1/3 × 10⁻³ F.
4. So. Comparing with 1/x × 10⁻³ F gives x = 3.

The common slip: 2.4. Taking X_L - X_C = R, as if voltage led, gives X_C = 8 Ω and x = 2.4, not 3.

The idea to keep. In a series RLC circuit, tan φ = (X_L - X_C)/R. If current leads, X_C is bigger; at 45° the net reactance equals R.

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