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When an AC source of emf e=E_0 \sin \sin (100 t) is connected across a circuit, the phase difference between the emfe and the current i in the circuit is observed to be \frac{\pi}{4}, as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C in series, find the relationship between the two elements

Option: 1

R=1 k \Omega, C=10 \mu F


Option: 2

R=1 k \Omega, C=1 \mu F


Option: 3

R=1 \mathrm{k} \Omega, L=10 \mathrm{H}


Option: 4

R=1 k \Omega, L=1 \mathrm{H}


Answers (1)

best_answer

As the current i leads the emf e by \frac{\pi}{4} , it is an  R-C \text { circuit }

\tan \tan \phi=\frac{x_c}{R}

or   \tan \tan \frac{\pi}{4}=\frac{\frac{1}{\omega C}}{R}

\therefore \quad \omega C R=1
As \quad \omega=100 \, \mathrm{rads}^{-1}
The product of C-R should be \frac{1}{100} s^{-1}.

Posted by

shivangi.bhatnagar

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