An ideal neon sign transformer provides 9430 V at 25.0 mA with an input voltage

Question

An ideal neon sign transformer provides 9430 V at 25.0 mA with an input voltage of 220 V. Calculate the transformer's input power and current. HINT (a) input power (in W) (b) input current (in A) 0/1 points | Previous Answers SerCP11 21.P045. My Notes Ask Your Teach An AC power generator produces 45 A (rms) at 3,600 V. The voltage is stepped up to 100,000 V by an ideal transformer, and the energy is transmitted through a long-distance power line that has a resistance of 116 , what percentage of the power delivered by the generator is dissipated as heat in the power line? Your response differs from the correct answer by more than 100%.96 Need Help?ReadIt

Explanation / Answer

a) P = VI = 9430 * 0.025 = 235.75 W
b) Power in equals power out
current at primary = 235.75 / 220 = 1.07 A

2) 45 A at 3600V = 162000W.
Current when transformed to 100,000V = (162000/100,000) = 1.62A.
Voltage drop along 116 ohm line = (1.62 x 116) = 187.92 V.
Dissipation in line = (187.92 x 1.62) = 304.43 W.
Percentage of loss = (304.43 /162000) x 100, = 0.188%.

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