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Physics · Alternating Current

JEE Main 2026 — 24 January, Morning Shift — Question 32

For the series LCR circuit connected with 220 V , 50 Hz a.c source as shown in the figure, the power factor is α10\frac{\alpha}{10}. The value of α\alpha is ____\_\_\_\_

Question figure
  1. Option A:

    4

  2. Option B:

    10

  3. Option C:

    6

    Correct
  4. Option D:

    8

Answer: C

Step-by-step solution

Power factor =RZ=\frac{\mathrm{R}}{\mathrm{Z}}

Z=R2+(XL−XC)2=602+(150−70)2=100Ω∴ Power factor =60100=610\begin{aligned} Z & =\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} & =\sqrt{60^{2}+(150-70)^{2}}=100 \Omega \therefore & \text { Power factor }=\frac{60}{100}=\frac{6}{10} \end{aligned}

Then a=6\mathrm{a}=6

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Alternating Current
Topic
Series LCR Circuit and Power Factor