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


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