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If you triple the length of a cable and at the same time double its diameter, what will be its resistance if its original resistance was $R$ ?
$R^{\prime}=\left[\frac{3}{4} \mathrm{R}\right]$The new resistance $R^{\prime}=\frac{3}{4}$ times the original resistance.
Physics 102 Electricity and Magnetism
Chapter 19
Current, Resistance, and Direct-Current Circuit
Electric Charge and Electric Field
Gauss's Law
Electric Potential
Capacitance and Dielectrics
Current, Resistance, and Electromotive Force
Direct-Current Circuits
Electromagnetic Induction
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initially we have cable off. Resistance are let's say it has length l on diameter D Now what we have done is out increased its lend It learned its length. By threefold I've increased its diameter 24 Now we want to find out what the new resistance is. Now we know that resistance is royal by a Now let's zero is the racist ou t off this particular material. From this we can see that the new resistance we roll because it is the same major road doesn't change times the new length divided by in your area. Now we know that row times Newland is three. Hello. And we know that area is fighting square by four. So that is five Dean, You squared by four now road times three l divided by pi And we know that tin you is to me and then you squared will be to the whole square which is 40 square so that this square by four multiplied before now you can see that this is row l buy a which would be the older assistants are because our old assistances row l times the area despite the square by four and If you pull those times out that becomes our on the numbers. We have left our three by four. And when you do this, change the resistance false by 25% the new assistants will be three by fourth off. The all resistance are 75% off the older assistance.
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