JEE Main 2018 — Alternating Current Question with Solution
From: JEE Main 2018 (Online) 15th April Morning Slot
Question
An ideal capacitor of capacitance is charged to a potential difference of The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance The current at a time when the potential difference across the capacitor is is :
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
Capacitance, C = 0.2 F = 0.2 106 F
Inductance, L = 0.5 m H = 0.5 103 H
Let, current = I.
Using energy conservation,
UE + 0 = UE' + Ub'
cv2 + 0 = c + LI2
0.2 106 102
= 0.2 106 52 + 0.5 103 I2
By solving this,
I = 101 A
= 0.17 A.
Inductance, L = 0.5 m H = 0.5 103 H
Let, current = I.
Using energy conservation,
UE + 0 = UE' + Ub'
cv2 + 0 = c + LI2
0.2 106 102
= 0.2 106 52 + 0.5 103 I2
By solving this,
I = 101 A
= 0.17 A.
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This is a previous-year question from JEE Main 2018, covering the Alternating Current chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.