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Physics · Alternating Current

JEE Main 2024 — 31 January, Shift 2 — Question 33

An AC voltage V=20sin⁡200πt\mathrm{V}=20 \sin 200 \pi \mathrm{t} is applied to a series LCR circuit which drives a current

I=10sin⁡(200πt+π3).I=10 \sin \left(200 \pi t+\frac{\pi}{3}\right) . \quad The average power dissipated is:

  1. Option A:

    21.6 W

  2. Option B:

    200 W

  3. Option C:

    173.2 W

  4. Option D:

    50 W

    Correct

Answer: D

Step-by-step solution

⟨P⟩=IVcos⁡ϕ\langle\mathrm{P}\rangle=\mathrm{IV} \cos \phi =202×102×cos⁡60∘=\frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times \cos 60^{\circ}

=50 W=50 \mathrm{~W}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Physics
Chapter
Alternating Current
Topic
Series LCR Circuit and Power Factor
An AC voltage V =20 sin 200 π t is applied to a series LCR circuit… | JEE Main 2024 PYQ with Solution · DhiX AI