Why does a re-formed copper rod have a resistance of 4R instead of 2R?

AI Thread Summary
When a cylindrical copper rod is re-formed to twice its original length while maintaining the same volume, its cross-sectional area decreases to half of the original size. The resistance of a conductor is determined by the formula R = ρ(L/A), where ρ is resistivity, L is length, and A is cross-sectional area. With the new length being 2L and the new area being A/2, the resistance becomes R' = ρ(2L/(A/2)), simplifying to R' = 4R. Thus, the resistance increases to four times the original value due to the combined effect of increased length and decreased area. Understanding this relationship is crucial for solving similar problems in physics.
juggalomike
Messages
49
Reaction score
0

Homework Statement


A cylindrical copper rod of length L and cross-sectional area A is re-formed to
twice its original length with no change in volume. If the resistance between its
ends was originally R, what is it now?


Homework Equations


R=p(L/A)


The Attempt at a Solution


Using the above equation the answer would be 2R, however the correct answer is 4R. Why is this? I am studying for a test tomorrow and need to understand why this is so.
 
Physics news on Phys.org
If it has the same volume but twice the length, what will happen to the cross-sectional area?
 
alex3 said:
If it has the same volume but twice the length, what will happen to the cross-sectional area?

ahhh, it will be 1/2 of the original, thanks.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Electric current and resistance question
Replies
2
Views
2K
Why is resistance of similar cases calculated differently?
Replies
4
Views
2K
Calculating Resistance of Coaxial Cable with Isolator
Replies
5
Views
2K
Platinum resistance thermometer and Kirchoff's Laws
Replies
1
Views
3K
Resistance, current and current density
Replies
8
Views
2K
Which equation is used to define resistance?
Replies
10
Views
2K
The resistance of a wire (conductor) in cylindrical form is:
Replies
5
Views
2K
Finding the drift speed of a conduction electrons
Replies
13
Views
3K
Conductor with shell problem getting zero as answer?
Replies
2
Views
807
Calculating Copper Wire Gauge for Resistance Thermometer Bridge Circuit
Replies
6
Views
2K

Hot Threads

Recent Insights

Back
Top