2. Solved Papers
  3. AMU Medical

AMU Medical AMU Solved Paper-2009

  • question_answer
    An LCR series circuit with \[R=100\Omega \] is connected to a 200 V, 50 Hz AC source. When only the capacitance is removed, the current leads the voltage by 60°. When only the inductance is removed, the current leads the voltage by \[60{}^\circ \]. The current in the circuit is

    A)  \[2\,A\]                                              

    B)  \[1\,A\]

    C)  \[\frac{\sqrt{3}}{2}A\]                 

    D)  \[\frac{2}{\sqrt{3}}A\]

    Correct Answer: A

    Solution :

                     If the capacitance is removed, it is an L-R circuit                 \[\tan \phi =\frac{{{X}_{L}}}{R}=\tan {{60}^{o}}=\sqrt{3}\] If inductance is removed, it is a capacitance circuit or R-C circuit \[\left| \phi  \right|\] is the same. \[\therefore \]  \[\omega L=\frac{1}{\omega C}\]. This is a resonance circuit. \[\therefore \]  \[Z=R\,;\,{{I}_{rms}}=\frac{{{E}_{rms}}}{R}\,;\,{{E}_{rms}}=200\,V\] \[\therefore \]  \[{{I}_{rms}}=\frac{200\,\,V}{100\,\,\Omega }=2\,\,A\]


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