An ideal transformer has a primary power input of 10 kW. The secondary current when the…
Physics · NEET · NTA Exams — Electromagnetic 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
- \(\frac{(10)^{4}}{25(8)^{2}} V\)
- \(\frac{10^{4}}{(25)(8)} \mathrm{V}\)
- \(\frac{(10)^{4} 8}{25} v\)
- \(\frac{(10)^{4}(8)^{2}}{25} \mathrm{v}\)
Answer
(C) (10)^ 4 8 25 v
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