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JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank Self Evaluation Test - Alternating Current

  • question_answer
    The primary and secondary coil 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 webers, t is time in seconds and \[{{\phi }_{0}}\] is a constant, the output voltage across the secondary con is 

    A) 120 volts    

    B)        220 volts

    C) 30 volts          

    D)        90 volts

    Correct Answer: A

    Solution :

    [a] Since \[\frac{{{V}_{s}}}{{{V}_{p}}}=\frac{{{N}_{s}}}{{{N}_{p}}}\] Where \[{{N}_{s}}\] = No. of turns across primary coil = 50 \[{{N}_{p}}\] = No. of turns across secondary coil = 1500 and \[{{V}_{p}}=\frac{d\phi }{dt}=\frac{d}{dt}\,({{\phi }_{0}}+4t)=\,4\Rightarrow \,V=\frac{1500}{50}\times 4\] =120 V   


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