JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank Graphical Questions

  • question_answer
               When an ac source of e.m.f. \[e={{E}_{0}}\]\[\sin (100\,t)\] is connected across a circuit, the phase difference between the e.m.f. e and the current i in the circuit is observed to be \[\pi /4\], as shown in the diagram. If the circuit consists possibly only of RC or LC in series, find the relationship between the two elements                         [IIT-JEE (Screening) 2003]

    A)            \[R=1k\Omega ,\,C=10\mu F\]                                        

    B)            \[R=1k\Omega ,\,C=1\mu F\]

    C)            \[R=1k\Omega ,\,L=10H\]      

    D)            \[R=1k\Omega ,\,L=1H\]

    Correct Answer: A

    Solution :

                       As the current i leads the voltage by \[\frac{\pi }{4},\] it is an RC circuit, hence \[\tan \varphi =\frac{{{X}_{C}}}{R}\]Þ \[\tan \frac{\pi }{4}=\frac{1}{\omega \,CR}\]                    Þ \[\omega \,CR=1\] as w = 100 rad/sec                    Þ \[CR=\frac{1}{100}se{{c}^{-1}}\].            From all the given options only option (a) is correct.


You need to login to perform this action.
You will be redirected in 3 sec spinner