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Alternating Current question

2016 · Q127
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Alternating Current question

2016 · Q127

NEETPhysicsAlternating CurrentMCQ+4 / −1
An inductor 20 mH, a capacitor 50 μF\mu FμF and a resistor 40 Ω\OmegaΩ are connected in series across a source of emf V = 10 sin 340t. The power loss in A.C. circuit is
  1. A
    0.76 W
  2. B
    0.89 W
  3. C
    0.51 W
  4. D
    0.67 W
View written solutionFree

Correct answer: C

XL = ω\omega ωL = 340 ×\times× 20 ×\times× 10-3 = 6.8 Ω\Omega Ω

XC = 1340×50×10−6{1 \over {340 \times 50 \times {{10}^{ - 6}}}}340×50×10−61​ = 58.82 Ω\Omega Ω

Average power in impedance

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

= 402+(58.82−6.8)2\sqrt {{{40}^2} + {{\left( {58.82 - 6.8} \right)}^2}} 402+(58.82−6.8)2​ = 65.62 Ω\Omega Ω

Power loss in A.C. circuit,

= Vrms Irms cosϕ\phi ϕ

= 12{1 \over 2}21​V0 I0 cosϕ\phi ϕ

= 12×10×1065.62×4065.62{1 \over 2} \times 10 \times {{10} \over {65.62}} \times {{40} \over {65.62}}21​×10×65.6210​×65.6240​

= 0.46 W

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