Current Flowing through High Voltage AC Transmission Lines

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Homework Help Overview

The discussion revolves around a problem related to current flow in high voltage AC transmission lines, specifically focusing on the calculations involving transformer turns, voltage, resistance, and efficiency of the wires. The original poster attempts to calculate the current flowing through the wires after a transformer steps up the voltage for transmission.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants question the validity of the original calculation, particularly regarding the implications of connecting wires directly at both ends and the concept of efficiency in relation to current flow.
  • There is discussion about the definition of wire efficiency and its impact on current calculations, as well as the importance of understanding circuit conditions for current flow.
  • Some participants suggest considering power equations and the role of load impedance in the context of the problem.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some have offered insights into the implications of wire resistance and efficiency, while others are questioning the assumptions made in the original calculations. There is no explicit consensus yet, but the conversation is delving into important aspects of the problem.

Contextual Notes

Participants note that the problem does not specify the load connected to the transmission lines, which is crucial for understanding the current flow. There is also mention of the need for a clear definition of efficiency as it pertains to the wires, which remains a point of contention.

Dave Johnson
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Homework Statement



Q1: A power plant produces energy at a voltage of Vi = 18667 V. Before being sent along long distance power lines this electricity is sent through a transformer with 143 turns in the primary coil and 8971 turns in the secondary coil.

Voltage is calculated to be 1171060.538V

If the wires have an efficiency of 99.60% and a resistance of 1096 Ω, what current flows through these wires?

Homework Equations



V = IR

The Attempt at a Solution



I = VR = 1171060.538 / 1096 = 1068A
1068A x 0.996 = 1064A

However, answer is wrong. Can anyone point out my mistake?
 
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Dave Johnson said:
I = VR = 1171060.538 / 1096 = 1068A
That calculation tells you the current that would flow if the wires were tied together at the other end, i.e. no other load in the circuit. The consequences would be most unfortunate.
Dave Johnson said:
1068A x 0.996 = 1064A
That makes even less sense. Inefficiency in the wires does not mean electrons disappear along the way. Current in = current out.

Any other ideas?
 
What is this thing called the efficiency of the wires?
 
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haruspex said:
That calculation tells you the current that would flow if the wires were tied together at the other end, i.e. no other load in the circuit. The consequences would be most unfortunate.

That makes even less sense. Inefficiency in the wires does not mean electrons disappear along the way. Current in = current out.

Any other ideas?

what do you mean current when both ends are connected? what are they asking if not maximum current flow?

##spoiler
I thought the reason OP got his answer wrong was because he didn't take into account impedence scaling by transformer.
ie : R1/R2 = square of turns ratio.
 
Vriska said:
what do you mean current when both ends are connected? what are they asking if not maximum current flow?
They did not mention maximum current flow, nor the maximum power transfer theorem. This problem hinges on an understanding of what is meant by "efficiency of the wires". The OP needs to root out a definition from his text or course notes.
 
Vriska said:
what do you mean current when both ends are connected? what are they asking if not maximum current flow?
No current flows unless there is a circuit. Nobody in their right mind connects the line directly to the neutral without a fat resistor in between. Somewhere out there in the grid there is a real load, with a big voltage drop across it.
The power lines have resistance too. With the efficiency as given, how much of the total voltage drop is caused by the wires?
 
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haruspex said:
No current flows unless there is a circuit. Nobody in their right mind connects the line directly to the neutral without a fat resistor in between. Somewhere out there in the grid there is a real load, with a big voltage drop across it.
The power lines have resistance too. With the efficiency as given, how much of the total voltage drop is caused by the wires?

woh thanks, I didn't see the response but I was thinking of this problem again so I can back to see if OP got answered.

so they mean something like (voltage on thing being driven) /(voltage that should actually be there (eg 220)? ,if that were the case you could find the load?

What about generator impedence multiplication though?
 
Vriska said:
so they mean something like (voltage on thing being driven) /(voltage that should actually be there
yes, but not 220V. It's the voltage you calculated.
Vriska said:
What about generator impedence multiplication though?
I don't think you have enough information to take impedance into account. But it would be more a question of impedance of the load, no?
 
Perhaps the OP could consider using the power equations instead of just Ohm's Law. Also, a sanity check would be that the current flowing through wires is really small, so as to minimise power losses such as by Joule Heating.
 

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