J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    The number of turns of the primary and the secondary coils of a transformer are 10 and 100 respectively. The primary voltage and the current are given as  \[2\text{ }V\]and\[1\text{ }A\]. Assuming the efficiency of the transformer as \[90%,\] the secondary voltage and the current respectively are

    A)   \[20\text{ }V\]and\[0.1\text{ }A\]  

    B)   \[0.2\text{ }V\]and \[1\text{ }A\]

    C)  \[20\,\,V\]and \[0.09\text{ }A\]

    D)   \[0.2\,\,V\]and \[0.9\text{ }A\]

    Correct Answer: A

    Solution :

    Given, \[{{N}_{P}}=10\] \[{{N}_{S}}=100,\,\,\,\,\,\,\,\,\,\eta =90%\] \[{{N}_{P}}=2V,\,\,\,\,\,\,\,\,\,{{V}_{S}}=?\] \[{{i}_{P}}=1\,A,\,\,\,\,\,\,\,\,\,{{i}_{S}}=?\] For a transformer \[\frac{{{N}_{S}}}{{{N}_{P}}}=\frac{{{V}_{S}}}{{{V}_{P}}}=\frac{{{i}_{P}}}{{{i}_{S}}}\] \[\frac{100}{10}=\frac{{{V}_{S}}}{2}\] \[\Rightarrow \] \[{{V}_{S}}=20V\] Similarly,    \[\frac{100}{10}=\frac{1}{{{i}_{S}}}\] \[{{i}_{S}}=0.1A\]


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