How Should L and A Change to Adjust Power Dissipation and Current in a Wire?

  • Thread starter Thread starter positron96
  • Start date Start date
  • Tags Tags
    Power Resistivity
AI Thread Summary
To adjust power dissipation and current in a wire, the relationship between resistance, resistivity, length, and cross-sectional area must be considered. The resistance is defined as R = ρL/A, and power is calculated using P = IV or P = V²/R. When the wire is stretched, its volume remains constant, meaning that as the cross-sectional area A decreases, the length L increases. To achieve a power increase of 35 times and a current increase of 3 times, both L and A must be recalibrated while maintaining the wire's volume. Understanding these relationships is crucial for determining the new dimensions of the wire.
positron96
Messages
4
Reaction score
0
A potential difference V is applied to a wire of cross section A, length L, and resistivity p. You want to change the applied potential difference and stretch the wire so that the energy dissipation rate is multiplied by 35 and the current is multiplied by 3. What should be the new values of L and A [in relationship to the old values]?

I know that R = pL/A and P = IV (etc.), but it doesn't make sense that they'd want specific answers for both L and A. Can't either one change without the other and still affect the power?
 
Physics news on Phys.org
HINT: The volume of the metal does not change upon stretching.
 
So then volume V = AL. And if you're stretching it then A gets smaller and L gets bigger. But what does volume have to do with power?
 
Last edited:
Fine. It doesn't matter anymore. Thanks.
 
Resistance can be found from L, A, and rho. Power can be found with P = IV = V^2/R.

- Warren
 
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...
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...
Back
Top