NEET Sample Paper NEET Sample Test Paper-58

  • question_answer
    An ideal transformer has a primary power input of 10kW. The secondary current when the transformer is on load is 25A. If the primary, secondary turns ratio is 8 : 1, then the potential difference applied to the primary coil is:

    A) \[\frac{{{10}^{4}}}{25\,{{(8)}^{2}}}\]                      

    B) \[\frac{{{10}^{4}}}{25\,(8)}\]

    C) \[\frac{{{(10)}^{4}}8}{25\,}\]              

    D) \[\frac{{{(10)}^{4}}\,{{(8)}^{2}}}{25\,}\]

    Correct Answer: C

    Solution :

    [c] \[\eta =\frac{powe{{r}_{\,(output)}}}{powe{{r}_{input}}}=\frac{{{V}_{s}}{{I}_{s}}}{{{V}_{p}}{{I}_{p}}}\] \[\frac{{{I}_{s}}}{{{I}_{p}}}=\frac{{{V}_{p}}}{{{V}_{s}}}=\frac{{{n}_{p}}}{{{n}_{s}}}=\frac{8}{1}\] \[\frac{25}{{{I}_{p}}}=\frac{8}{1}\] \[{{I}_{p}}=\frac{25}{8}A\] \[{{P}_{input}}=10KW={{V}_{p}}{{T}_{p}}\] \[10\times {{10}^{3}}={{V}_{p}}\times \frac{25}{8}\] \[{{V}_{p}}\times \frac{8\times {{10}^{4}}}{25}\,volt\]


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