Punjab Medical Punjab - MET Solved Paper-2003

  • question_answer
    A lamp in which \[10\,\,amp\] current can flow at\[15\,\,V\], is connected with an alternating source of potential\[220\,\,V\]. The frequency of source is\[50Hz\]. The impedance of choke coil required to light the bulb is:

    A) \[0.07\,\,H\]                      

    B) \[0.14\,\,H\]

    C) \[0.28\,\,H\]                      

    D) \[1.07\,\,H\]

    Correct Answer: A

    Solution :

    The resistance of lamp\[R=\frac{V}{I}=\frac{15}{10}=1.5\Omega \] (Given: \[V=15\,\,volt\], \[I=10\,\,amp\], A.C. voltage \[=220\,\,V\], \[f=50Hz\]) The impedance of circuit\[Z=\frac{V}{I}=\frac{220}{10}=22\,\,ohm\] Again impedance\[Z=\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}\] or            \[22=\sqrt{{{(1.5)}^{2}}+{{(2\times \pi \times 50)}^{2}}{{L}^{2}}}\] or            \[{{(22)}^{2}}={{(1.5)}^{2}}+4{{\pi }^{2}}\times {{50}^{2}}\times {{L}^{2}}\] or            \[L=0.07\,\,henry\]


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