Electromagnetic Induction and Alternating Current 4 Question 5

5. A series AC circuit containing an inductor (20mH), a capacitor (120μF) and a resistor (60Ω) is driven by an AC source of 24V/50Hz. The energy dissipated in the circuit in 60s is

(Main 2019, 9 Jan II)

(a) 3.39×103J

(b) 5.65×102J

(c) 2.26×103J

(d) 5.17×102J

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Answer:

Correct Answer: 5. (d)

Solution:

  1. The given series RLC circuit is shown in the figure below.

Here,

VR= potential across resistance (R)

VL= potential across inductor (L) and

VC= potential across capacitor (C)

Impedance of this series circuit is

Z=R2+(XLXC)2XL=ωL=(2πf)(L)=2π×50×20×103ΩXL=6.28ΩXC=1ωC=12πfC=12π×50×120×106=2503πΩ and XLXC=6.282503π=20.23Ω

RMS value of current in circuit is

Irms=VrmsZ=24R2+(XLXC)2Irms=24602+(20.23)2=2463.18Irms=0.379A

Therefore, energy dissipated is

=Irms2×R×tE=(0.379)2×60×60=517.10=5.17×102J



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