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Alternating Current question

2018 · Q126
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Alternating Current question

2018 · Q126

NEETPhysicsAlternating CurrentMCQ+4 / −1
An inductor 20 mH, a capacitor 100 μ\muμF and a resistor 50 Ω\OmegaΩ are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is
  1. A
    0.79 W
  2. B
    0.43 W
  3. C
    2.74 W
  4. D
    1.13 W
View written solutionFree

Correct answer: A

Average power in impedance

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

where XC = capacitive reactance and XL = inductive reactance.

Also XC = 1ωC{1 \over {\omega C}}ωC1​ and XL = ωL\omega LωL

∴\therefore∴ Z = (50)2+(1314×100×10−6−314×20×10−3)2\sqrt {{{\left( {50} \right)}^2} + {{\left( {{1 \over {314 \times 100 \times {{10}^{ - 6}}}} - 314 \times 20 \times {{10}^{ - 3}}} \right)}^2}} (50)2+(314×100×10−61​−314×20×10−3)2​

⇒\Rightarrow⇒ Z = 56.15 Ω\Omega Ω

Irms = VrmsZ{{{V_{rms}}} \over Z}ZVrms​​ = Vm2×Z=102×56{{{V_m}} \over {\sqrt 2 \times Z}} = {{10} \over {\sqrt 2 \times 56}}2​×ZVm​​=2​×5610​

Hence power loss in the circuit

= (102×56)2×50{\left( {{{10} \over {\sqrt 2 \times 56}}} \right)^2} \times 50(2​×5610​)2×50 = 0.79 W

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