JEE Main 2021 · 27 July · morning · Single correct · +4 / −1

A 0.07 H inductor and a 12Ω resistor are connected in series to a 220V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take π as 22/7]

  1. (a) 8.8 A and tan ⁻¹( 11/6 )
  2. (b) 88 A and tan ⁻¹( 11/6 )
  3. (c) 0.88 A and tan ⁻¹( 11/6 )
  4. (d) 8.8 A and tan ⁻¹( 6/11 )
Answer and solution

Answer (a) 8.8 A and tan ⁻¹( 11/6 )

Solution

Physics sir · 4 steps

Find X_L first; then the impedance gives the current and X_L/R gives the phase angle.

  1. Inductive reactance. X_L = 2π f L = 2 × 22/7 × 50 × 0.07 = 22 Ω.
  2. Impedance. In a series LR circuit, Z = √(R² + X_L²) = √(144 + 484) = √628 ≈ 25.06 Ω.
  3. Current. I = V/Z = 220/25.06 ≈ 8.78 A, which is about 8.8 A.

    tan φ = X_L/R = 22/12 = 11/6

  4. So. The current is about 8.8 A and it lags the voltage by tan⁻¹(11/6), which is option (a).

The idea to keep. In an LR circuit the phase angle comes from tan φ = X_L/R, reactance on top. The current needs the full impedance √(R² + X_L²), not just R or X_L.

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

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Asked about 140 times in JEE Main since 2020. This question is from 2021.