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  2. Confused! kindly explain, - Electromagnetic Induction and Alternating currents - JEE Main-4
# Engineering

A circuit connected to an ac source of emf e=e_{0} \sin(100 t) with t in seconds, gives a phase difference of \frac{\pi}{4} betweeen the emf e and current i. Which of the following circuit will exhibit this?

 

  • Option 1)

    RL circuit with R=1 \;k\Omega and  L=10 mH

  • Option 2)

    RL circuit with R=1 \;k\Omega and  L=1 mH

  • Option 3)

    RC circuit with R=1 \;k\Omega and  C=1\mu F

  • Option 4)

    RC circuit with R=1 \;k\Omega and  C=10\mu F

 

Answers (1)
S solutionqc

w=100^{-1}

for phase difference of \frac{\pi}{4}\Rightarrow \left | X_{L}-X_{C} \right |=R

if  \;X_{C}=0 \Rightarrow X_{L}=R \Rightarrow wL=R \Rightarrow L=\frac{R}{w}

if \;X_{L}=0 \Rightarrow X_{C}=R= \frac{1}{wC} \Rightarrow C=\frac{1}{wR}

for option (4)

R=1k\Omega

C=\frac{1}{wR}=\frac{1}{\frac{100}{5}\times 10^{3} \Omega }= 10^{-5}_{s/ \Omega}\Rightarrow 10 \times 0^{-6}_{s/ \Omega}\Rightarrow 10\mu F


Option 1)

RL circuit with R=1 \;k\Omega and  L=10 mH

Option 2)

RL circuit with R=1 \;k\Omega and  L=1 mH

Option 3)

RC circuit with R=1 \;k\Omega and  C=1\mu F

Option 4)

RC circuit with R=1 \;k\Omega and  C=10\mu F

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