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The current through an inductor of impedance 10 Ω lags behind the voltage by a phase of 60 ° when just the inductor is connected to the ac source. Now the inductor is connected to a 5 Ω resistance in series, then the net impedance of the circuit is

Options

  1. A15 Ω
  2. B12 Ω
  3. C13.2 Ω
  4. D18 Ω

Correct answer

C. 13.2 Ω

Step-by-step solution

10 = r 2 + X L 2 and X L r = tan ⁡ 60 ° 10 = r 2 + r 3 2 or r = 5   Ω ,   X L = 5 3   Ω Z = 5 + 5 2 + 5 3 2 = 175 = 13.2   Ω

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