A potential difference V is applied to a wire of cross-section area A, length l,

Question

A potential difference V is applied to a wire of cross-section area A, length l, and conductivity sigma. You want to change the applied potential difference and add length to the wire so that the power dissipated is increased by a factor of 30 and the current is increased by a factor of 4. Let the wire have the same cross-sectional area before and after it is lengthened. What is the ratio of the new length to the original length? What is the ratio of the new applied voltage to the original voltage?

Explanation / Answer

from conservation of momentum

initial momentum of the system = final momentum of the system

initial momentum = 0

final momentum = (68 + 120) * v + (15 * 4.8)

(68 + 120) * v + (15 * 4.8) = 0

v = 0.38 m/s

opposite to the direction of package

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we know that

i = V / R

power dissipation

P = i^2 * R

we have the relation that

R = rho * l / A

here rho and A are constant only l is variable

let initial length be l1 and final be l2

initial potential v1 and final v2

now i1 = v1 / l1

and i2 = v2 / l2

i2 = 4 * i1

therefore v2 / v1 = 4 * l2/l1 ------------------(1)

power is

P1 = i1^2 * l1

P2 = i2^2 * l2

here P2 = 30 P1

and i2 = 4 i1

therefore l2 /l1 = 1.875----------------Answer--------------

from eq (1)

v2 / v1 = 4 * 1.875

v2 / v1 = 7.5-----------Answer-------------

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