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NEET 2018
Exam
NEET
Year
2018
Subject
Physics
Chapter
Alternating Current

NEET 2018 · Physics · Alternating Current

An inductor 20 mH, capacitor 100μ F and a resistor 50Ω are...

An inductor 20mH,20 mH, capacitor 100μF100\mu F and a resistor 50Ω50Ω are connected in series across a source of emf, V=10sin⁡(314t)V = 10\sin ( 314t ). The power loss in the circuit is

Options

  1. A2.74W2.74\text{W}
  2. B0.43W0.43\text{W}
  3. C0.79W0.79\text{W}
  4. D1.13W1.13\text{W}

Correct answer

C. 0.79W0.79\text{W}

Step-by-step solution

V0=10V,ω=314rad/s{\text{V}}_{0} = 10V, \omega = 314rad/s

P=VrmsIrmscosϕP = {V}_{rms}{I}_{rms}cos ϕ

P=VrmsVrmsZRZP = {V}_{rms} \dfrac{{V}_{rms}}{Z} \dfrac{R}{Z}

=(Vrms)2RZ2= \dfrac{{( {V}_{rms} )}^{2} R}{{Z}^{2}}

XL=\omegaL=(314)(20×10−3)=6.280{X}_{L} = \omegaL = \left(314\right)\left(20 \times {10}^{- 3}\right) = 6.280

XC=1\omegaC=1314×100×10−6=31.84Ω{X}_{C} = \dfrac{1}{\omegaC} = \dfrac{1}{314 \times 100 \times {10}^{- 6}} = 31.84Ω

R=50ΩR = 50Ω

Z=(XC−XL)2+R2Z = \sqrt{{( {X}_{C} - {X}_{L} )}^{2} + {R}^{2}}

=(31.84−6.28)2+(50)2=56Ω= \sqrt{{\left(31.84 - 6.28\right)}^{2} + {\left(50\right)}^{2}} = 56Ω

⇒P=(102)2×50(56)2=0.79W⇒ P = \dfrac{{( \dfrac{10}{\sqrt{2}} )}^{2} \times 50}{{( 56 )}^{2}} = 0.79\text{W}
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