JEE Main 2020 — Alternating Current Question with Solution
From: JEE Main 2020 (Online) 3rd September Morning Slot
Question
A 750 Hz, 20 V (rms) source is connected to a
resistance of 100 , an inductance of 0.1803 H
and a capacitance of 10 F all in series. The
time in which the resistance (heat capacity
2 J/oC) will get heated by 10oC. (assume no loss
of heat to the surroudnings) is close to :
Choose an option
Show full solutionCorrect option: A
Correct answer
A348 s
Step-by-step explanation
f = 750 Hz, Vrms = 20 V,
R = 100 , L = 0.1803 H,
C = 10 F, S = 2 J/°C
|Z| =
=
=
=
= 834
In AC, power (P) = irmsVrms cos
and irms =
Power factor (cos ) =
P =
=
=
= 0.0575 J/S
Also, H = Pt = S
t = = 348 sec
R = 100 , L = 0.1803 H,
C = 10 F, S = 2 J/°C
|Z| =
=
=
=
= 834
In AC, power (P) = irmsVrms cos
and irms =
Power factor (cos ) =
P =
=
=
= 0.0575 J/S
Also, H = Pt = S
t = = 348 sec
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This is a previous-year question from JEE Main 2020, 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.