Calculate number of turns for an electric heater

  • Thread starter Thread starter MisterP
  • Start date Start date
  • Tags Tags
    Electric Heater
AI Thread Summary
To calculate the number of turns needed for a nichrome wire heater with a resistance of 40 Ω, the area of the wire is determined to be 0.785 mm² using its diameter. The resistance formula R = ρ*l/S is applied, where ρ is the specific resistance of nichrome, estimated at 1.1 Ω⋅mm²/m. The calculation indicates that to achieve the desired resistance, approximately 200 turns of wire are required. However, confusion arises regarding the length of wire needed for these turns, with calculations suggesting a total length of 3140 cm, which seems inconsistent. Clarification on the specific resistance value and algebraic simplification shows that the correct approach leads to the expected 200 turns.
MisterP
Messages
47
Reaction score
3

Homework Statement


Need to calculate, how many turns of nichrome wire do I need on a cylinder to make a heater with resistance of 40 Ω.
Wire diameter = 1mm
Cylinder radius = 2.5cm
R = 40 Ω
Wire diameter = 1mm
Correct answer - 200 turns

Homework Equations


R = ρ*l/S
R = resistance
ρ = specific resistance (I guess) and it is 1.1 Ω⋅mm2/m
l = length of wire
S = area of wire "cross section"

The Attempt at a Solution


Area of wire = Π*r2 = 3.14 * 0.52 = 0.785mm2
So I used this formula, but get nowhere close to 200 turns. 40 = 1,1l/0,785...
 
Physics news on Phys.org
MisterP said:
40 = 1,1l/0,785...
Solving that will give you a length. What is it in turns?
 
haruspex said:
Solving that will give you a length. What is it in turns?
Yes, I know, but it does not add up to near "200" or 0,2 or something like that..
To calculate turns I did this: 3.14 * 5cm (2*r) = 15.7cm is for one turn.
So, 200 turns would be 3140cm.. where did I go wrong?

If I continue with "40 = 1,1l/0,785..." I get 28.57.. let`s just change it to 2857cm.. too short for answer..
 
Last edited:
MisterP said:
Yes, I know, but it does not add up to near "200" or 0,2 or something like that..
To calculate turns I did this: 3.14 * 5cm (2*r) = 15.7cm is for one turn.
So, 200 turns would be 3140cm.. where did I go wrong?

If I continue with "40 = 1,1l/0,785..." I get 28.57.. let`s just change it to 2857cm.. too short for answer..
The difference is about 10%, right?
Note that if you solve it algebraically first the πs cancel and you end up with 200/1.1. So to get 200, you have to take ρ as 1mm2/m.
Looking it up, I see it quoted as 1.1 to 1.5.
 
Last edited:
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top