JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank Self Evaluation Test - Alternating Current

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

    A) \[\frac{{{10}^{4}}\times {{8}^{2}}}{25}V\]

    B) \[\frac{{{10}^{4}}\times 8}{25}V\]

    C) \[\frac{{{10}^{4}}}{25\times 8}V\]

    D) \[\frac{{{10}^{4}}}{25\times {{8}^{2}}}V\]

    Correct Answer: B

    Solution :

    [b] \[P={{V}_{1}}{{i}_{1}}={{V}_{2}}{{i}_{2}}\] \[{{V}_{2}}=\frac{{{10}^{4}}}{25}\] Now \[{{V}_{1}}=\frac{{{n}_{1}}}{{{n}_{2}}}\times {{V}_{2}}=\frac{8}{1}\times \frac{{{10}^{4}}}{25}\,V\]


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