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)
V Vakul

 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.

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