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  • Need explanation for: - Electromagnetic Induction and alternative Currents - NEET

An inductor 20 mH, a capacitor 50 \mu F and a resistor 40 \Omega are connected in series across a source of emf V = 10 sin 340 t. The power loss in A.C circuit :

  • Option 1)

    0.51 W

  • Option 2)

    0.67 W

  • Option 3)

    0.76 W

  • Option 4)

    0.89 W

 

Answers (1)

best_answer

 

Impedence -

Z= sqrt{R^{2}+left ( X_{L}-X_{c} ight )^{2}}

-

 

 

Average power (True power) -

P_{av}= V_{rms}: : i_{rms}cos phi

-

 

 and

 

Power factor -

cos phi = frac{R}{Z}
 

- wherein

R ightarrow resistance

Z ightarrow impedence

 

 

X_{c}=\frac{1}{wc}=\frac{1}{340 \times 50 \times 10 ^{-6}}= \frac{10^{4}}{34 \times 5}=58.82 \Omega

X_{L}= W_{L} = 340 \times 20 \times 10 ^{-3}= 6.8 \Omega

Z= \sqrt{R^{2}+(X_{L}-X_{c})^{2}} = \sqrt{(40)^{2}+(58.82- 6.8)^{2}

Z= \sqrt{(40)^{2}+ (52.02)^{2}} = 65.62 \Omega

Peak Current - I_{o}=\frac{V_{o}}{z}=\frac{10}{65.62}A

\cos \phi = \frac{R}{Z}=\frac{40}{65.62}

 Power loss in A.C circuit = P = Vrms I rms \cos \phi = \frac{1}{2}V^{o}I^{o} \cos \phi

P = \frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62} =0.51 W

 

 


Option 1)

0.51 W

Correct option

Option 2)

0.67 W

incorrect option

Option 3)

0.76 W

incorrect option

Option 4)

0.89 W

incorrect option

Posted by

prateek

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