# how to make (i) the subject



## flyinghigher2011

1.4	State the formula for calculating the resistance of a conductor given its length (l), its cross-sectional area (a) and its resistivity (_p_).

R=pl/a

1.5	Transpose the formula in 1.4 to make l the subject.

Am I right with the following formula

I=R*A


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## Zog

Where do you get "I" from, they want you to solve for l (Length), unless there is more to the quoestion you didn't tell us.


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## flyinghigher2011

Zog said:


> Where do you get "I" from, they want you to solve for l (Length), unless there is more to the quoestion you didn't tell us.


that is all the question, judging by your answer i must have written the formula down wrong then?

should it be 

R= I/A


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## Zog

There is no "I" in the question. You just have to solve for "l" (Length)

l=(a*r)/p


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## flyinghigher2011

Zog said:


> There is no "I" in the question. You just have to solve for "l" (Length)
> 
> l=(a*r)/p


so what about the first part of the question did i get that right

or should the answer to the first part of the question should be the following

R=p*l/a


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## Zog

flyinghigher2011 said:


> so what about the first part of the question did i get that right


Yes


flyinghigher2011 said:


> or should the answer to the first part of the question should be the following
> 
> R=p*l/a


 That is the same thing that you wrote, just a different way of showing it.


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## flyinghigher2011

Zog said:


> Yes
> That is the same thing that you wrote, just a different way of showing it.


there was two question in my first post though!!


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## Going_Commando

So if your formula has R for the subject, and you want to swap it around to make l (length) the subject, then you just have to do some simple multiplication and division. Step by step would be:

R=(p*l)/a
(R*a)=(p*l)
(R*a)/p=l or
l=(r*a)/p

It can get difficult to transcribe fractions into text, so if the formula does start out as R= p*l
----
a

Then you can use parentheses to show the order of operations a bit more clearly. Hope this helps.


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## etb

It sounds like the problem you're having is that you don't know what the formula means, because you were even more confused in your last post about it.

R is the resistance of a conductor of a particular length and gauge (cross-sectional area)

L is the length

The Greek letter RHO (looks like a slanted P) is the resistivity of the material. It does not depend on the size or length of the wire. RHO for copper is different than for aluminum, etc

A is the cross-sectional area

Learn to use dimensional analysis. For example, in the formula
D = R * T (distance = rate * time)

D [km] = R [km/hr] * [hr]

The hr in the bottom cancels with the hr in the top and you're left with km. That should make very good sense both in your head and on paper.

Now try R = P * L / A

R has units [ohm], P is [ohm*m], L is [m], A is [m^2] or [m*m]

So [ohm*m] * [m] / [m*m] works out to [ohm], which is the correct unit for resistance.

Of course, your resistivity is probably specified in micro-ohm-centimeters and area is likely mm^2, so just do the metric conversions. You're from the UK so that should be something you learned at a young age school.

Think about the formula. The smaller the cross sectional area, the more resistance that wire should have. The longer the wire, the more resistance it should have. The higher the resistivity of the material, the higher the resistance of the wire for any gauge or length. The formula backs up all those common-sense mental checks.

Your posts are very confusing.

Yes, the resistance of a conductor is given by the first formula you posted.

No, your derivation,I=R*A, is not correct. It should be

L = R * A / P(rho)

To get that, start with

R = P*L/A

and multiply each side by A/P, that leaves

L = R*A/P


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## flyinghigher2011

etb said:


> It sounds like the problem you're having is that you don't know what the formula means, because you were even more confused in your last post about it.
> 
> R is the resistance of a conductor of a particular length and gauge (cross-sectional area)
> 
> L is the length
> 
> The Greek letter RHO (looks like a slanted P) is the resistivity of the material. It does not depend on the size or length of the wire. RHO for copper is different than for aluminum, etc
> 
> A is the cross-sectional area
> 
> Learn to use dimensional analysis. For example, in the formula
> D = R * T (distance = rate * time)
> 
> D [km] = R [km/hr] * [hr]
> 
> The hr in the bottom cancels with the hr in the top and you're left with km. That should make very good sense both in your head and on paper.
> 
> Now try R = P * L / A
> 
> R has units [ohm], P is [ohm*m], L is [m], A is [m^2] or [m*m]
> 
> So [ohm*m] * [m] / [m*m] works out to [ohm], which is the correct unit for resistance.
> 
> Of course, your resistivity is probably specified in micro-ohm-centimeters and area is likely mm^2, so just do the metric conversions. You're from the UK so that should be something you learned at a young age school.
> 
> Think about the formula. The smaller the cross sectional area, the more resistance that wire should have. The longer the wire, the more resistance it should have. The higher the resistivity of the material, the higher the resistance of the wire for any gauge or length. The formula backs up all those common-sense mental checks.
> 
> Your posts are very confusing.
> 
> Yes, the resistance of a conductor is given by the first formula you posted.
> 
> No, your derivation,I=R*A, is not correct. It should be
> 
> L = R * A / P(rho)
> 
> To get that, start with
> 
> R = P*L/A
> 
> and multiply each side by A/P, that leaves
> 
> L = R*A/P


Thank you for your post, its just that im learning Electrical Installations at collage and did not do to well at school.


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## etb

flyinghigher2011 said:


> Thank you for your post, its just that im learning Electrical Installations at collage and did not do to well at school.


In in the US but some books on UK installation I've seen are
Basic Electrical Installation by Linsley
Advanced Electrical Installation by Linsley
Electrical installation work by scaddan

Some of them you can browse portions of for free on the web.

Best of luck!


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## flyinghigher2011

etb said:


> In in the US but some books on UK installation I've seen are
> Basic Electrical Installation by Linsley
> Advanced Electrical Installation by Linsley
> Electrical installation work by scaddan
> 
> Some of them you can browse portions of for free on the web.
> 
> Best of luck!


this is the one im trying to figure out at the moment

Basic Electrical Installation by Linsley


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