A 750 Hz,20V( rms) source is connected to a resistance of 100Ω , an inductance of 0.1803H and a capacitance of 10μ F all in series. The time in which…
JEE Main 2020 — Physics Electromagnetism
2020mcqhard
A 750Hz,20V(rms) source is connected to a resistance of 100Ω, an inductance of 0.1803H and a capacitance of 10μF all in series. The time in which the resistance (heat capacity 2J/∘C ) will get heated by 10∘C. (assume no loss of heat to the surroundings) is close to :
Official previous-year question
Held on 3 Sept 2020 · Verified 6 Jul 2026.
Options
A
418s
B
245s
C
365s
D
348s
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Solution
The AC circuit is shown below.
Here,R=100,XL=Lω=0.1803×750×2π=850Ω
The capacitive reactance Xc=Cω1=10−5×2π×7501=21.23Ω
Total impedance Z=R2+(XL−XC)2
=1002+(850−21.23)2=834.77≈835
The heat loss H=irms2Rt=(ms)Δt
83520×83520×100t=(2)×10
t=348.61sec
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