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A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit.  Given that R=5 , L=25 mH and C=1000 µF.  The total impedance, and phase difference between the voltage across the source and the current will respectively be :

 

  • Option 1)

    10\Omega and\tan ^{-1}\left ( \frac{5}{3} \right )

  • Option 2)

    7\Omega and 45^{0}

     

  • Option 3)

    10\Omega \: and\: \tan ^{-1}\left ( \frac{8}{3} \right )

  • Option 4)

    7\Omega and\: \tan ^{-1}\left ( \frac{5}{3} \right )

 

Answers (1)

best_answer

As we learnt in

Impedence -

Z= \sqrt{R^{2}+\left ( \omega L -\frac{1}{\omega c}\right )^{2}}

-

 

 

V{_{0}} = 283 \ V

R = 5\Omega, L = 25 mH, C = 1000 \muF

W = 320/s

X_{L} = WL = (2\pi f) L = 320\times25\times 10^{-3}\Omega

8\Omega

X_{C} = \frac{1}{WC} = \frac{1}{320\times10^{-3}} \Omega = \frac{1}{0.32}\Omega

3.125\Omega

Total Impedance

Z = \sqrt{R^{2}+(X_{L}-X_{C})^{2}} \cong 5\sqrt{2}

Phase difference = tan^{-1} (\frac{X_{L}-X_{C}}{R})

\cong 45^{\circ}


Option 1)

10\Omega and\tan ^{-1}\left ( \frac{5}{3} \right )

This is incorrect option.

Option 2)

7\Omega and 45^{0}

 

This is correct option.

Option 3)

10\Omega \: and\: \tan ^{-1}\left ( \frac{8}{3} \right )

This is incorrect option.

Option 4)

7\Omega and\: \tan ^{-1}\left ( \frac{5}{3} \right )

This is incorrect option.

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prateek

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