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  • I need help with - In the above circuit, C =,,and. Current in L-R1path is I1and in C-R2path it is I2. The - Electromagnetic Induction and Alternating currents - JEE Main

 In the above circuit, C = \frac{\sqrt{3}}{2} \mu F, R_{2}= 20 \Omega , L = \frac{\sqrt{3}}{10} H and R_{1} = 10 \Omega . Current in L-R1 path is I1 and in C-R2 path it is I2 . The voltage of A.C source is given by, V = 200\sqrt{2}sin\left ( 100 t \right ) volts . The phase difference between  I1 and I2 is : 

  • Option 1)

    0o

  • Option 2)

    90o

  • Option 3)

    150o

  • Option 4)

     60o

Answers (3)

best_answer

 

Phase difference -

\tan \phi = \frac{X_{L}}{R}= \tan ^{-1}\left ( \frac{X_{L}}{R} \right )

\phi = \tan ^{-1}\left ( \frac{\omega L}{R} \right )

- wherein

XL  = inductive reactance

R     = resistance

 

 

Phase difference -

\tan \phi = \frac{X_{c}}{R}

\phi = \tan ^{-1}\left ( \frac{X_{c}}{R} \right )= \tan ^{-1}\left ( \frac{1}{\omega cR} \right )

-

For current I1

\phi _{1} = \tan ^{-1}\frac{\omega L}{R_{1}} = \tan ^{-1}\left ( \frac{100\frac{\sqrt{3}}{10}}{10} \right ) = 60^{0}(lagging)

For I2 current

\phi _{2} = \tan ^{-1}\left ( \frac{1}{\omega CR_2} \right ) = \tan ^{-1}\left ( \frac{10^{6}}{100\times \frac{\sqrt{3}}{2}\times 20} \right ) \approx 90^{\circ}      (leading)

Phase difference = 90+60 = 1500

 

 


Option 1)

0o

Option 2)

90o

Option 3)

150o

Option 4)

 60o

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