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JCECE Engineering JCECE Engineering Solved Paper-2005

  • question_answer
    If a transformer of an audio amplifier has output impedance \[8000\,\,\Omega \] and the speaker has input impedance of \[8\,\,\Omega \], the primary and secondary turns of this transformer connected between the output of amplifier and to loud speaker should have the ratio:

    A) \[1000:1\]                           

    B) \[100:1\]

    C) \[1:32\]                               

    D) \[32:1\]

    Correct Answer: A

    Solution :

    From Faraday's law, the induced emf across primary and secondary is                 \[{{e}_{p}}=-{{N}_{p}}\frac{\Delta \phi }{\Delta t}\]                 \[{{e}_{s}}=-{{N}_{s}}\frac{\Delta \phi }{\Delta t}\] Also,      \[e=iR\] \[\therefore \]  \[\frac{{{R}_{p}}}{{{R}_{s}}}=\frac{{{N}_{p}}}{{{N}_{s}}}\] Given,   \[{{R}_{s}}=8000\Omega ,\,\,{{R}_{p}}=8\Omega \] \[\therefore \]  \[\frac{{{N}_{s}}}{{{N}_{p}}}=\frac{{{R}_{s}}}{{{R}_{p}}}=\frac{8000}{8}=\frac{1000}{1}\]


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