(a) The ratio of turns in the windings of the transformer
"\\frac{N_1}{N_2}=\\frac{V_1}{V_2}=\\frac{5000}{240}=\\frac{125}{6}=20.83"(b)
"P_1=P_2=10000\\:\\rm{W}"So, the maximum current in the secondary coil
"I_2=\\frac{P_2}{V_2}=\\frac{10000\\:\\rm{W}}{240\\:\\rm{V}}=41.67\\:\\rm{A}"
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