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JCECE Engineering JCECE Engineering Solved Paper-2008

  • question_answer
    The primary and secondary coils of a transformer have \[50\] and \[1500\] turns respectively. If the magnetic flux \[\phi \] linked with the primary coil is given by \[\phi ={{\phi }_{0}}+4t\], where \[\phi \] is in weber, \[t\] is time in second and \[{{\phi }_{0}}\] is a constant, the output voltage across the secondary coil is

    A) \[90\,\,V\]                         

    B) \[120\,\,V\]

    C)  \[220\,\,V\]                      

    D)  \[30\,\,V\]

    Correct Answer: B

    Solution :

    The magnetic flux linked with the primary coil is given by So, voltage across primary                 \[{{V}_{p}}=\frac{d\phi }{dt}=\frac{d}{dt}({{\phi }_{0}}+4t)\]                      \[=4\,\,V\]     (as\[{{\phi }_{0}}=\]constant) Also, we have                 \[{{N}_{p}}=50\]and\[{{N}_{s}}=100\] From relation,                 \[\frac{{{V}_{s}}}{{{V}_{p}}}=\frac{{{N}_{s}}}{{{N}_{p}}}\] or            \[{{V}_{s}}=\frac{{{V}_{p}}{{V}_{s}}}{{{N}_{p}}}\]                      \[=4\left( \frac{1500}{50} \right)\]                     \[=120\,\,V\] Note As in case of given transformer, voltage in secondary is increased, hence it is a step-up transformer.


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