Energy efficiency, Transformer?

[Daisy]

Homework Statement

A power plant produces energy at a voltage of Vi = 12693 V. Before being sent along long distance power lines this electricity is sent through a transformer with 166 turns in the primary coil and 6917 turns in the secondary coil. The voltage of the electricity sent through the long distance wires is 528900V.
If the wires have an efficiency of 99.30% and a resistance of 1089 Ω, what current flows through these wires?

Homework Equations

I have used V=IR. Of which I got V to be 485.67 and I mil

The Attempt at a Solution

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I got I = V/R to be 485.67 and I multiplied it by 0.993 in order to determine the current flowing through these wires. But I got it wrong. I don't seem to understand what is wrong. Did I use the correct equation? Or was my computation of efficiency incorrect?

 

[gneill]

R is the resistance of the transmission wires and doesn't include any load resistance at the receiving end. That means your current calculation is not correct.

Instead, start by writing an expression for the voltage received at the far end assuming some (as yet unknown) current $I$ is flowing.

Next consider the definition of Efficiency in this context (power transmission).

 

[Daisy]

I am really confused. What do you mean by voltage received at the far end?

 

[gneill]

I am really confused. What do you mean by voltage received at the far end?

Power station at one end of the transmission line, some load at the far end. The transmission line has resistance so there will be a voltage drop between the sending end and the receiving end. Thus the voltage received at the far end is not the same as the voltage at the sending end.

 

[HalfLostAndSearching]

I would like to clarify that energy efficiency refers to the amount of energy that is converted into useful work compared to the total amount of energy consumed. In the context of a transformer, energy efficiency refers to the ratio of output power to input power. In this case, the efficiency of the transformer would be calculated as the ratio of the output voltage (528900V) to the input voltage (12693V). This gives an efficiency of 41.7%.

Regarding the calculation for the current flowing through the wires, using Ohm's Law (V=IR) is the correct approach. However, there may have been an error in the calculation of the resistance (1089 Ω). It would be helpful to double check the values and ensure they are in the correct units (ohms in this case) before calculating the current. Additionally, since the efficiency of the wires is given as 99.30%, the current should be multiplied by this efficiency to get the actual current flowing through the wires. This would give a final current of approximately 483.2 A.