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What happens with the resistance of a wire with twice the diameter 
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#1
Nov1610, 12:52 PM

P: 40

hi everybody
Question: a length of uniform wire has a R of 2 ohms. calculate the R of a wire of the same metal and original length but twice the diameter. I thought as the wire gets twice as thick the R must halve to 1ohms. But the solution is 0.5 ohms. what did i do wrong? thanks in advance 


#2
Nov1610, 12:56 PM

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PF Gold
P: 5,197

Can you list for me the parameters that resistance depends upon? (Hint: it does not depend *directly* on the diameter.) Which one of these parameters is relevant here?



#3
Nov1610, 01:00 PM

P: 40

is it that the R decreases when the A increases?



#4
Nov1610, 01:03 PM

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PF Gold
P: 5,197

What happens with the resistance of a wire with twice the diameter
Yes, crosssectional area is one of the parameters on which resistance depends. (I had asked you to list all of the parameters, but this one is the relevant one here). So, what does changing the diameter do to the area? What does that resulting change in area do to the resistance?



#5
Nov1610, 01:06 PM

P: 40

is it that when you double the diameter you double the A?



#6
Nov1610, 01:08 PM

P: 40

as the area increases the R decrease



#7
Nov1610, 01:09 PM

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PF Gold
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#8
Nov1610, 01:11 PM

P: 40

r ^2 times pi



#9
Nov1610, 01:17 PM

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PF Gold
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#10
Nov1610, 01:20 PM

P: 40

sorry i don't get what you mean with the A depends on the square of diameter.
can you show me an example with numbers please so that i can visualise this problem? 


#11
Nov1610, 01:33 PM

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PF Gold
P: 5,197

Alright look. Let's call the original diameter d_{1}. Now:
[tex] A_1 = \pi r_1^2 [/tex] But the radius is just half the diameter, so: [tex] r_1 = d_1 / 2 [/tex] Agreed? Therefore: [tex] A_1 = \pi \left(\frac{d_1}{2}\right)^2 = \pi \frac{d_1^2}{4} [/tex] This is how the area of a circle depends upon its diameter. As you can see, area is equal to a constant times the diameter squared. Now, what happens if we change the diameter by doubling it? Let's call the new diameter d_{2} so that: [tex] d_2 = 2d_1 [/tex] We can plug this new diameter into the formula for the area in order to find the new area: [tex] A_2 = \pi \frac{d_2^2}{4} = \pi \frac{(2d_1)^2}{4} = 4 \pi \frac{d_1^2}{4} = 4A_1 [/tex] So we have the result that A_{2} = 4A_{1}. After doubling the diameter, the new area is equal to FOUR times the original area. This is because the area of a circle is proportional to its diameter SQUARED. So if you double the diameter, you quadruple the area. If you triple the diameter, you increase the area by a factor of nine. If you quadruple the diameter, the area increases by a factor of 16. Now do you understand? 


#12
Nov1610, 01:42 PM

P: 40

wow I'm impressed with that.
Yes I understand now. Thanks a lot for your answer :) 


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