- #1

- 1

- 0

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))