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Alternating Current question

2024 · 5 Apr · Shift 2 · Q68
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  5. /2024 · 5 Apr · Shift 2 · Q68

Alternating Current question

2024 · 5 Apr · Shift 2 · Q68

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
A series LCR circuit is subjected to an ac signal of 200 V,50 Hz200 \mathrm{~V}, 50 \mathrm{~Hz}200 V,50 Hz. If the voltage across the inductor (L=10 mH)(\mathrm{L}=10 \mathrm{~mH})(L=10 mH) is 31.4 V31.4 \mathrm{~V}31.4 V, then the current in this circuit is ‾\underline{\hspace{2cm}}​.
  1. A
    10 A
  2. B
    10 mA
  3. C
    68 A
  4. D
    63 A
View written solutionFree

Correct answer: A

  1. Use the inductor voltage relation in AC

For an inductor in AC, VL=IXLV_L = I X_LVL​=IXL​ where XL=ωL=2πfLX_L = \omega L = 2\pi f LXL​=ωL=2πfL

  1. Calculate inductive reactance

Given:

  • f=50 Hzf = 50\,\text{Hz}f=50Hz
  • L=10 mH=10×10−3 H=0.01 HL = 10\,\text{mH} = 10 \times 10^{-3}\,\text{H} = 0.01\,\text{H}L=10mH=10×10−3H=0.01H

So, XL=2π(50)(0.01)=π≈3.14 ΩX_L = 2\pi (50)(0.01) = \pi \approx 3.14\,\OmegaXL​=2π(50)(0.01)=π≈3.14Ω

  1. Find the current

Given voltage across inductor: VL=31.4 VV_L = 31.4\,\text{V}VL​=31.4V

Thus, I=VLXL=31.43.14=10 AI = \frac{V_L}{X_L} = \frac{31.4}{3.14} = 10\,\text{A}I=XL​VL​​=3.1431.4​=10A

  1. Match with options

The current is: 10 A\boxed{10\,\text{A}}10A​

So the correct option is A.

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