Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$220 \mathrm{~V}, 50 \mathrm{~Hz}, \mathrm{AC}$ source is connected to an inductance of 0.2 H and a resistance of $20 \Omega$ in series. What is the current in the circuit?
PhysicsAlternating CurrentJIPMERJIPMER 2013
Options:
  • A 3.33 A
  • B 33.3 A
  • C 5 A
  • D 10 A
Solution:
2064 Upvotes Verified Answer
The correct answer is: 3.33 A
From the formula
$\begin{aligned} I & =\frac{V}{\sqrt{R^2+(\omega L)^2}} \\ I & =\frac{220}{\sqrt{(20)^2+(2 \pi \times 50 \times 0.2)^2}} \\ & =\frac{220}{66}=3.33 \mathrm{~A}\end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.