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JEE Main 202431 JanuaryEvening Shift

Exam: JEE Main
Year: 2024
Date: 31 January
Shift: Evening Shift
Subject: Physics
Chapter: Alternating Current

JEE Main 2024 · 31 January · Evening Shift · Physics · Alternating Current

An AC voltage V = 20 200 is applied to a series LCR circuit which...

An AC voltage

V = 20\sin 200\pit

is applied to a series LCR circuit which drives a current

I = 10\sin \left(200\pit + \dfrac{\pi}{3}\right)

. The average power dissipated is:

Options

A $21 . 6W$
B $200W$
C $173 . 2W$
D $50W$

Correct answer

D. $50W$

Step-by-step solution

The average power dissipated is, $< P > = IVcosϕ$

$= \dfrac{10}{\sqrt{2}} \times \dfrac{20}{\sqrt{2}} \times \cos 60 °$

$= 50W$

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