JEE Main 2019 — Alternating Current Question with Solution
From: JEE Main 2019 (Online) 9th January Evening Slot
Question
A series AC circuit containing an inductor (20 mH), a capacitor (120 F) and a resistor (60 ) is driven by an AC source of 24V/50 Hz. The energy dissipated in the circuit in 60 s is :
Choose an option
Show full solutionCorrect option: C
Correct answer
C5.17 102 J
Step-by-step explanation
Energy dissipated in 60 Sec
= (Pavg) 60
= Vrms Irms cos 60
= Vrms cos 60
XL = L = 2FL
= 2(50) 20 103
= 2
XC =
=
=
= 26.52
XC XL = 20.24 20
Z =
=
= 20
Also cos = =
Energy dissipated in 60 sec
=
= 5.17 102 J
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