Confused about the answer given by vcaa on electric power

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Homework Help Overview

The discussion revolves around a question from the 2006 physics exam regarding electric power and current calculation at a specific point in a circuit. Participants are examining the use of a voltage value in the context of potential voltage loss due to resistance in the power line.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the rationale behind using a specific voltage value without accounting for potential voltage loss. There is discussion about estimating the current and the significance of voltage drop due to wire resistance.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the problem. Some have suggested methods to check if the voltage loss is significant and how to approach the calculations. There is no explicit consensus on the necessity of deducting voltage loss before calculating current.

Contextual Notes

Participants note the ambiguity in the exam question regarding whether to consider voltage loss and how to justify assumptions made during calculations. The resistance of the wire and its impact on current calculations are also under discussion.

torud
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hello!
I was going over some questions on the vcaa site and there was this question in the year 2006 physics exam two. It was question ten.

Here's the question (page 7 question 10)
http://www.vcaa.vic.edu.au/Documents/exams/physics/2006physics2-w.pdf

The answer said that in order to find the current at point R, I=100KW/22000V. I don't get why they used 22000V. Shouldn't they deduct the voltage loss from that value?

here's their answer! (page 3 question 10)
http://www.vcaa.vic.edu.au/Documents/exams/physics/physics2_assessrep_06.pdf

please help!
 
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torud said:
hello!
I was going over some questions on the vcaa site and there was this question in the year 2006 physics exam two. It was question ten.

Here's the question (page 7 question 10)
http://www.vcaa.vic.edu.au/Documents/exams/physics/2006physics2-w.pdf

The answer said that in order to find the current at point R, I=100KW/22000V. I don't get why they used 22000V. Shouldn't they deduct the voltage loss from that value?

here's their answer! (page 3 question 10)
http://www.vcaa.vic.edu.au/Documents/exams/physics/physics2_assessrep_06.pdf

please help!
What voltage loss ?
 
torud said:
hello!
I was going over some questions on the vcaa site and there was this question in the year 2006 physics exam two. It was question ten.

Here's the question (page 7 question 10)
http://www.vcaa.vic.edu.au/Documents/exams/physics/2006physics2-w.pdf

The answer said that in order to find the current at point R, I=100KW/22000V. I don't get why they used 22000V. Shouldn't they deduct the voltage loss from that value?

here's their answer! (page 3 question 10)
http://www.vcaa.vic.edu.au/Documents/exams/physics/physics2_assessrep_06.pdf

please help!

According to the question they are looking for an approximate value for the current. Take the approximate value and estimate the potential drop that results due to the power line resistance. If you then deduct that drop from the supply voltage and recalculate the current using this reduced voltage, what is the "new" value for the current? Has it changed significantly?
 
sammyS- the voltage loss due to the resistance in the wire (2 ohms)
gneil- ah, i see.. the answer for the current after voltage loss deduction is 4.55A also. But in an exam how would you know for certain that you don't need to deduct the voltage loss first? would you have to write ' the voltage loss is negligible' The question is a bit vague.. right?
 
torud said:
sammyS- the voltage loss due to the resistance in the wire (2 ohms)
gneil- ah, i see.. the answer for the current after voltage loss deduction is 4.55A also. But in an exam how would you know for certain that you don't need to deduct the voltage loss first? would you have to write ' the voltage loss is negligible' The question is a bit vague.. right?

It's simple enough to check to see if the calculated current will cause a significant potential drop. So write a statement like, "Check to see if power line potential drop is significant", then show the calculation: plug in the change and see if it steps on the significant figures of the result.
 
okay, i see. thanks! it'll be a pain if the loss was significant.. :P
 
torud said:
okay, i see. thanks! it'll be a pain if the loss was significant.. :P

Not too painful. If 100kW is being drawn then E*I = 100kW at point R (where E is the potential there). So E = 100kW/I . But the potential there is also the the source potential less the resistive line loss, or E = 22000V - I*2Ω. Equate the E's and solve for I. Yes, it'll be a quadratic and you'll have to sort out which root yields the reasonable result.
 
mm.. yes.. i see. By using the quadratic formula, i achieved two answers - the desired 4.55A and 10995.5A. To eliminate the second answer, can you say that it is not possible for 10995.5A being larger than 400A (which is the current of the secondary coil of the step down transformer) since the transformer is a step down transformer?
 
Why not check to see what the line voltage drop would be at 10995.5A?
 
  • #10
lol... thanks so much for your help! ^^
 

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