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  • Solve this problem - Electromagnetic Induction and alternative Currents - NEET

 The instantaneous values of alternating current and voltages in a circuit are given as

i = \frac{1} {{\sqrt 2 }}\sin \left( {100\,\pi \text{t}} \right)\text{ampere}

e = \frac{1} {{\sqrt 2 }}\sin \left( {100\,\pi \text{t} + {\pi \mathord{\left/ {\vphantom {\pi 3}} \right. \kern-\nulldelimiterspace} 3}} \right)\text{Volt}

The average power in Watts consumed in the circuit is :

  • Option 1)

    {\raise0.7ex\hbox{{\sqrt 3 }} \!\mathord{\left/ {\vphantom {{\sqrt 3 } 4}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{4}}

  • Option 2)

    \frac{1} {2}

  • Option 3)

    \frac{1} {8}

  • Option 4)

    \frac{1} {4}

 

Answers (1)

best_answer

As discussed in 

 

 

Given, i=\frac{1}{\sqrt 2}\sin(100\pi t)\\

Compare    i=U_{o}\sin wt\:\:\:\:\Rightarrow i_{o}=\frac{1}{\sqrt 2}A

e=\frac{1}{\sqrt 2}\sin (100 \pi t + \frac{\pi}{3})

e_{o}=\frac{1}{\sqrt 2}V, \Phi =\frac{\pi}{3}

i_{rms}=\frac{i_{o}}{\sqrt 2}=\frac{\frac{1}{\sqrt 2}}{\frac{1}{\sqrt 2}}A=\frac{1}{2}A

e_{rms}=\frac{e_{o}}{\sqrt 2}=\frac{\frac{1}{\sqrt 2}}{{\sqrt 2}}=\frac{1}{2}V

Average power    P=i_{rms}.e_{rms}\: \cos \phi

 

p=\frac{1}{8}

 


Option 1)

{\raise0.7ex\hbox{${\sqrt 3 }$} \!\mathord{\left/ {\vphantom {{\sqrt 3 } 4}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$4$}}

This option is incorrect.

Option 2)

\frac{1} {2}

This option is incorrect.

Option 3)

\frac{1} {8}

This option is correct.

Option 4)

\frac{1} {4}

This option is incorrect.

Posted by

divya.saini

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