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BHU PMT BHU PMT (Screening) Solved Paper-2010

  • question_answer
    A 50 Hz AC current of crest value 1 A flows through the primary coil of a transformer. If mutual induction between primary and secondary is 1.5 H, the crest voltage induced in the secondary coil is

    A)  272V                                    

    B)  320V

    C)  415V                                    

    D)  471V

    Correct Answer: D

    Solution :

                     \[i={{i}_{0}}\sin \omega t={{i}_{0}}\sin 2\,\pi nt\] \[{{E}_{\max }}=M\frac{di}{dt}=N.\frac{d}{dt}({{i}_{0}}.\sin 2\pi nt)\]                 \[=M{{i}_{0}}.2\pi n\,\,\cos 2\pi \,nt\] \[\therefore \]  \[{{E}_{\max }}=M{{i}_{0}}.2\pi n\]                 \[=1.5\times 1\times 2\pi \times 50\]                 \[=150\pi V=471\,V\]


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