JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Application of EMI (Motor, Dynamo, Transformer)

  • question_answer
    A loss free transformer has 500 turns on its primary winding and 2500 in secondary. The meters of the secondary indicate 200 volts at 8 amperes under these conditions. The voltage and current in the primary is              [MP PMT 1996]

    A)            100 V, 16 A                            

    B)             40 V, 40 A

    C)            160 V, 10 A                            

    D)             80 V, 20 A

    Correct Answer: B

    Solution :

                       \[\frac{{{V}_{p}}}{{{V}_{s}}}=\frac{{{N}_{p}}}{{{N}_{s}}}=\frac{500}{2500}=\frac{1}{5}\Rightarrow {{V}_{p}}=\frac{200}{5}=40\ V\]            Also \[{{i}_{p}}{{V}_{p}}={{i}_{s}}{{V}_{s}}\Rightarrow {{i}_{p}}={{i}_{s}}\frac{{{V}_{s}}}{{{V}_{p}}}=8\times 5=40\ A\]


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