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  • Try this! For the LCR circuit, shown here, the current is observed to lead the applied voltage. An additional capacitor C', when joined with the capacitor C present in the circuit, makes the power factor of the circuit unity. The capacitor C'

 For the LCR circuit, shown here, the current is observed to lead the applied voltage. An additional capacitor C', when joined with the capacitor C present in the circuit, makes the power factor of the circuit unity. The capacitor C', must have been connected in :

  • Option 1)

    series with C and has a magnitude

    \frac{1-\omega ^{2}LC}{\omega ^{2}L}

     

  • Option 2)

     series with C and has a magnitude

    \frac{C}{\left ( \omega ^{2}LC-1 \right )}

     

     

  • Option 3)

     parallel with C and has a magnitude

    \frac{C}{\left ( \omega ^{2}LC-1 \right )}

  • Option 4)

    parallel with C and has a magnitude

    \frac{1-\omega ^{2}LC}{\omega ^{2}L}

 

Answers (1)

 As we learnt in

Power factor -

Ratio of resistance and impedance(cos\phi).

-

 

 

\therefore \frac{1}{\omega C} > \omega L -------- 1

 

After inserting c^{/} both become equal since cos \phi = \frac{R}{Z} = 1

or R^{2} = R^{2} + {(X_{L} - X _{C})}^{2}

=> X_{L} = X _{C} -------- 2

This implied X_{c} must reduced

This will happen when C is increased.

\therefore C' must be inserted in parellel to C.

\therefore \omega L = \frac{1}{\omega (C+C')}

or C+C' = \frac{1}{\omega ^{2}L} or C^{/} = \frac{1}{W^{2}L} - C

or C' = \frac{1 - \omega ^{2} LC}{\omega ^{2}L}    


Option 1)

series with C and has a magnitude

\frac{1-\omega ^{2}LC}{\omega ^{2}L}

 

This is incorrect option.

Option 2)

 series with C and has a magnitude

\frac{C}{\left ( \omega ^{2}LC-1 \right )}

 

 

This is incorrect option.

Option 3)

 parallel with C and has a magnitude

\frac{C}{\left ( \omega ^{2}LC-1 \right )}

This is incorrect option.

Option 4)

parallel with C and has a magnitude

\frac{1-\omega ^{2}LC}{\omega ^{2}L}

This is correct option.

Posted by

Vakul

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