Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A series AC circuit containing an inductor 20 mH, a capacitor 120 μF and a resistor 60 Ω is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is:
PhysicsAlternating CurrentJEE MainJEE Main 2019 (09 Jan Shift 2)
Options:
  • A 5.17×102 J
  • B 3.39×103 J
  • C 2.26×103 J
  • D 5.65×102 J
Solution:
2705 Upvotes Verified Answer
The correct answer is: 5.17×102 J

(Power factor), Pf=cosϕ=RZ

Where R, Z are the resistance and impedance respectively.
(Energy dissipated) E=Vrms2Z×cosϕ×t
E=Vrms2Z×RZ×t

E=Vrms2Z2×R×t

Vrms, t are the RMS voltage and time respectively.

We know that the Impedance of a LCR circuit is given by the formula,
     Z2=R2+ωL-1ωC2Z2=602+2π×50×20×10-3-12π×50×120×10-62

Where ω, L, C are the angular frequency, inductance and capacitance in the circuit respectively.
Z2=602+2π-100012π2

Z2=4009.7

E=2424009.7×60×60 J 

= 517.14 J

=5.17×102 J

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.