AMU Medical AMU Solved Paper-2012

  • question_answer
    When an AC source of emf E = eq sin (100 O is connected across a circuit, the phase difference between the emf E and the current I is observed to be \[\pi /4\], as shown in the figure. If the circuit consists possibly only of R -C or R - L in series, which of the following combinations is possible?

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

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

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

    D)  \[R=1\,k\Omega ,\,\,C=1\,H\]

    Correct Answer: C

    Solution :

                     Emf of AC source \[E={{E}_{0}}\sin \,(100\,t)\] We have co = 100 and \[\phi =\frac{\pi }{4}={{45}^{o}}\]                 \[\tan \phi =\frac{\omega L}{R}\]                 \[1=\frac{100\times L}{R}\]                 \[R=100\,L\] If L = 10 H then \[R=1\,k\Omega \]


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