An ideal transformer has a primary power input of 10 kW. The secondary current when the…

Physics · NEET · NTA ExamsElectromagnetic Induction and Alternating Currents

An ideal transformer has a primary power input of 10 kW. The secondary current when the transformer is on load is 25 A. If the primary to secondary turns ratio is 8 : 1, then the potential difference applied in the primary coil will be
  1. \(\frac{(10)^{4}}{25(8)^{2}} V\)
  2. \(\frac{10^{4}}{(25)(8)} \mathrm{V}\)
  3. \(\frac{(10)^{4} 8}{25} v\)
  4. \(\frac{(10)^{4}(8)^{2}}{25} \mathrm{v}\)

Answer

(C) (10)^ 4 8 25 v

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