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QJEEM19Pha52a: JEE Main 2019 Physics - Class 12 JEEM19Pha52a Solution: A series AC circuit containing an inductor (20 mH), a capacitor …

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

Multiple Choice Options

  • A 5.65 ×\times 102J
  • B 2.26 ×\times 103J
  • C 5.17 ×\times 102 J
  • D 3.39 ×\times 103 J

Step-by-Step Verified Solution

JEE Main 2019 (Online) 9th January Evening Slot Physics - Alternating Current Question 143 English Explanation

Energy dissipated in 60 Sec

= (Pavg) ×\times 60

= Vrms ×\times Irms ×\times cos ϕ\phi ×\times 60

= Vrms ×\times VrmsZ×{{{V_{rms}}} \over Z} \times cos ϕ\phi ×\times 60

XL = ω\omega L = 2 π\pi FL

= 2 π\pi (50) ×\times 20 ×\times 10 −- 3

= 2 π\pi Ω\Omega

XC = 1ωC{1 \over {\omega C}}

= 12πfC{1 \over {2\pi fC}}

= 12π(50)×120×10−6{1 \over {2\pi \left( {50} \right) \times 120 \times {{10}^{ - 6}}}}

= 26.52 Ω\Omega

∴\therefore    XC −- XL = 20.24 ≃\simeq 20

∴\therefore    Z = (XC−XL)2+R2\sqrt {{{\left( {{X_C} - {X_L}} \right)}^2} + {R^2}}

= (20)2+602\sqrt {{{\left( {20} \right)}^2} + {{60}^2}}

= 20 10\sqrt {10} Ω\Omega

Also cos ϕ\phi = RZ{R \over Z} = 602010=310{{60} \over {20\sqrt {10} }} = {3 \over {\sqrt {10} }}

∴\therefore    Energy dissipated in 60 sec

= (24)22010×310×60{{{{\left( {24} \right)}^2}} \over {20\sqrt {10} }} \times {3 \over {\sqrt {10} }} \times 60

= 5.17 ×\times 102 J

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