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### AuthorTopic: VCAA 2010 question 14 Electric power  (Read 472 times) Tweet Share

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#### you0006

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##### VCAA 2010 question 14 Electric power
« on: November 10, 2011, 09:03:48 pm »
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For the power loss, because there are two transmission lines I was wondering why we didn't need to multiply the power loss by two?

#### Shark 774

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##### Re: VCAA 2010 question 14 Electric power
« Reply #1 on: November 10, 2011, 09:27:01 pm »
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For the power loss, because there are two transmission lines I was wondering why we didn't need to multiply the power loss by two?

I assumed you mean question 16, where you need to calculate the power loss.

The fact that there are two transmission lines DOES have an effect on the power loss, but you don't multiply it by two. You still use the formula:
Power Loss = (Current ^ 2)(Resistance),
but the resistance is the resistance of both the lines, hence:
Power Loss =  [(P/V)^2] x (Resistance of one line x 2)
=  [(4/2)^2] x (2x2)
= 4 x 4 = 16W

I hope this helps.

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##### Re: VCAA 2010 question 14 Electric power
« Reply #2 on: November 10, 2011, 10:27:53 pm »
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For the power loss, because there are two transmission lines I was wondering why we didn't need to multiply the power loss by two?
If you really do want to multiply the power loss by 2 in that VCAA 2010 question (or other questions when you're given only the resistance  and not the total resistance), you can calculate the power loss for one of the lines and then times it by 2. (e.g. power loss of one line * 2).

Or you can do what Shark774 said and take the total resistance and just use a single formula.

It's up to you, it's good to know as many of the different ways so you can double check.

#### you0006

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##### Re: VCAA 2010 question 14 Electric power
« Reply #3 on: November 11, 2011, 09:10:29 am »
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thanx guys!