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
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top