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JEE Main 20203 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
3 September
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2020 · 3 September · Morning Shift · Physics · Alternating Current

A 750Hz, 20 V(rms) source is connected to a resistance of 100Ω, an...

A 750Hz,20750Hz, 20V\left(rms\right)sourceisconnectedtoaresistanceofsource is connected to a resistance of100Ω,aninductanceofan inductance of0 . 1803Handacapacitanceofand a capacitance of10\muFallinseries.Thetimeinwhichtheresistance(heatcapacityall in series. The time in which the resistance (heat capacity2J / ° C)willgetheatedby) will get heated by10 ° C . (assume no loss of heat to the surroundings) is close to :A 750Hz, 20 V(rms) source is connected to a resistance of 100Ω, an inductance of 0 . 1803H and a capacitance of 10 all

Options

  1. A418s418s
  2. B245s245s
  3. C365s365s
  4. D348s348s

Correct answer

D. 348s348s

Step-by-step solution

The AC circuit is shown below.

A 750Hz, 20 V(rms) source is connected to a resistance of 100Ω, an inductance of 0 . 1803H and a capacitance of 10 all

Here,R=100,XL=Lω=0.1803×750×2π=850ΩR = 100, {X}_{L} = L\omega = 0 . 1803 \times 750 \times 2\pi = 850Ω

The capacitive reactance Xc=1Cω=110−5×2π×750=21.23Ω{X}_{c} = \dfrac{1}{C \omega} = \dfrac{1}{{10}^{- 5} \times 2\pi \times 750} = 21 . 23Ω

Total impedance Z=R2+(XL−XC)2Z = \sqrt{{R}^{2} + {\left({X}_{L} - {X}_{C}\right)}^{2}}

=1002+(850−21.23)2=834.77≈835= \sqrt{{100}^{2} + {\left(850 - 21 . 23\right)}^{2}} = 834 . 77 \approx 835

The heat loss H=irms2Rt=(ms)∆tH = {i}_{rms}^{2}Rt = \left(ms\right) ∆ t

20835×20835×100t=(2)×10\dfrac{20}{835} \times \dfrac{20}{835} \times 100t = \left(2\right) \times 10

t=348.61sect = 348 . 61sec

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