2. Solved Papers
  3. NEET

NEET AIPMT SOLVED PAPER SCREENING 2007

  • 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                 \[\phi ={{\phi }_{0}}+4\,t\]                 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            \[{{V}_{s}}={{V}_{p}}\frac{{{N}_{s}}}{{{N}_{p}}}\]                 \[=4\left( \frac{1500}{50} \right)\]                 = 120 V                 Note:    As in case of given transformer, voltage if secondary is increased, hence it is a step-up transformer.


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