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  • Help me solve this - Electromagnetic Induction and alternative Currents - NEET

An inductor 20 mH, a capacitor 100 \muF and a resistor 50 \Omega are connected in series across a source of emf V = 10 sin 314 t. The power loss in circuit is

  • Option 1)

    2.74 W

  • Option 2)

    0.43 W

  • Option 3)

    0.79 W

  • Option 4)

    1.13 W

 

Answers (1)

best_answer

As we have learned

Average power -

P_{av}= {i}'^{2}_{rms}R= \frac{V^{2}_{rms}R}{Z^{2}}

-

 

power loss= v_{rms}I_{rms} \cos \phi

\left ( \frac{v_{rms}}{Z} \right ) (V_{rms})\frac{R}{Z}

\left ( \frac{v_{rms}}{Z} \right ) ^{2}R

X_{_{l}}=WL= 314*20*10^{-3}\Omega =6.28\Omega

X_{_{c}}=1/WL= 1/314*20*10^{-3}\Omega =31.84\Omega

 Z=\sqrt{R^{2}+(X_{c}-X_{l})^{2}} = 56\Omega

p= \frac{\frac{10}{\sqrt 2}^{2}}{56^{2}}\times 50= 0.79w

 

 

 

 


Option 1)

2.74 W

This is incorrect

Option 2)

0.43 W

This is incorrect

Option 3)

0.79 W

This is correct

Option 4)

1.13 W

This is incorrect

Posted by

prateek

View full answer

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