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Potential Difference and Power Dissipation Problem

  • #1
A potential difference V is applied to a wire of cross-section area of 1 unit, length 1 unit,
and conductivity σ. You want to change the applied potential difference and draw out the
wire so the power dissipated is increased by a factor of 30 and the current is increased by a
factor of 4 .

a) What should be the new values of the length?

b) What is the new cross sectional area?

What i've done so far is:
Linitial = 1 unit, Ainitial = 1 unit
Initial power Pi = I2 * R, R = ρ * Linitial / Ainitial = 1/σ (as ρ = 1/σ, )
Pi = I2 / σ
Pf = 30 * Pi
If2 * Rf = 30 * I2 / σ

Substituting If = 4 * I
16 * I2 * L / σ * A = 30 * I2 / σ
16 * L / A = 30
16 * L = 30 * A

L * A = volume of wire = Linitial * Ainitial (as volume does not change)
=> L * A = 1 * 1
A = 1 / L
Substituting
16 * L = 30 * (1/L)
L2 = 30 / 16
final length L = √1.875 = 1.369 units

But it's wrong. (Its electronic and I only have 5 guess left (I've added 1 to it too for 2.369, didn't work either))
 

Answers and Replies

  • #2
collinsmark
Homework Helper
Gold Member
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final length L = √1.875 = 1.369 units

But it's wrong. (Its electronic and I only have 5 guess left (I've added 1 to it too for 2.369, didn't work either))
Well, if it's any consolation, I got the same answer you did. 1.369 units. :confused:
 

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