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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Alternating Current

BITSAT 2015 · Physics · Alternating Current

In the circuit shown below, the ac source has voltage V=20 (ω t)...

In the circuit shown below, the ac source has voltage V=20cos⁡(ωt)V=20 \cos (\omega t) volt with ω=2000rad/s\omega=2000 \mathrm{rad} / \mathrm{s} The amplitude of the current will be nearest to In the circuit shown below, the ac source has voltage V=20 (ω t) volt with ω=2000 rad / s The amplitude of the current

Options

  1. A2A2 \mathrm{ A}
  2. B3.3A3.3 \mathrm{ A}
  3. C2/5A2 / \sqrt{5} \mathrm{ A}
  4. D5A\sqrt{5} \mathrm{ A}

Correct answer

A. 2A2 \mathrm{ A}

Step-by-step solution

Z=R2+(ωL−1ωC)2=102+(2000×5×103−12000×50×10−6)2=10Ωi=V0Z=2010=2A.\begin{aligned} Z= \sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}\\ &=\sqrt{10^{2}+\left(2000 \times 5 \times 10^{3}-\frac{1}{2000 \times 50 \times 10^{-6}}\right)^{2}}\\ =& 10 \Omega \\ \quad i=\frac{V_{0}}{Z}=\frac{20}{10}=2 A. \end{aligned}
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