Potential Difference and Power Dissipation Problem

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SUMMARY

The discussion centers on calculating the new length and cross-sectional area of a wire when a potential difference is applied, resulting in a power dissipation increase by a factor of 30 and a current increase by a factor of 4. The initial parameters are a wire with a length of 1 unit and a cross-sectional area of 1 unit, with conductivity σ. The calculations lead to a final length of approximately 1.369 units, but this result is questioned as potentially incorrect, indicating a need for further verification.

PREREQUISITES
  • Understanding of electrical concepts such as potential difference and conductivity.
  • Familiarity with Ohm's Law and power dissipation formulas.
  • Knowledge of geometric relationships in wire dimensions (length and cross-sectional area).
  • Basic algebraic manipulation skills to solve equations.
NEXT STEPS
  • Review the derivation of power dissipation in electrical circuits.
  • Study the relationship between resistance, length, and cross-sectional area in conductive materials.
  • Explore the implications of changing wire dimensions on electrical performance.
  • Investigate potential errors in calculations involving square roots and algebraic substitutions.
USEFUL FOR

Electrical engineers, physics students, and anyone involved in circuit design or analysis who seeks to understand the effects of wire dimensions on electrical properties.

Scrubphysicst
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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))
 
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Scrubphysicst said:
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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