Question

an electric heating element dissipate 400W on 220V mains is to be made from the wire
of area of cross section 5 x 10^-18 m^2. Calculate the length of the wire required if the resistivity of the material is 1.1 x 10^-6 ohm m

Answer 1

Answer:

$\bigstar{\bold{Length=5.5\:m}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Power (P) = 400 W
• Potential difference (V) = 220 V
• Area of cross section(A) = 5 × 10⁻¹⁸ m²
• Resistivity (ρ) = 1.1 × 10⁻⁶ Ω m

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Length of the wire (l)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ First we need to find the resistance of the conductor which is given by

  R = V²/P

→ Substituting the given datas we get,

  R = (220)²/400

  R = 121 Ω

→ Resistance of a conductor is also given by

  R = ρ l/A

→ Substituting thhhe datas, we get the value of length

 121 = 1.1 × 10⁻⁶ × l/5 × 10⁻⁸

  121 = l × 0.22 × 10²

  l = 121/22

  l = 5.5 m

$\boxed{\bold{Length=5.5\:m}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→  Resistance of a conductor is given by

     R = ρ l/A

→ Resistance is directly proportional to length and inversely proportional to area of cross section of conductor.