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JIPMER 2013
Exam
JIPMER
Year
2013
Subject
Physics
Chapter
Alternating Current

JIPMER 2013 · Physics · Alternating Current

220 ~V, 50 ~Hz, AC source is connected to an inductance of 0.2 H...

220 V,50 Hz,AC220 \mathrm{~V}, 50 \mathrm{~Hz}, \mathrm{AC} source is connected to an inductance of 0.2 H and a resistance of 20Ω20 \Omega in series. What is the current in the circuit?

Options

  1. A3.33 A
  2. B33.3 A
  3. C5 A
  4. D10 A

Correct answer

A. 3.33 A

Step-by-step solution

From the formula I=VR2+(ωL)2I=220(20)2+(2π×50×0.2)2=22066=3.33 A\begin{aligned} I & =\frac{V}{\sqrt{R^2+(\omega L)^2}} \\ I & =\frac{220}{\sqrt{(20)^2+(2 \pi \times 50 \times 0.2)^2}} \\ & =\frac{220}{66}=3.33 \mathrm{~A}\end{aligned}
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