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`        A heater coil is to be constructed with a nichrome wire (ρ = 1.0 × 10^-6 Ω-m) which can operate at 500 W when connected to a 250 V supply. (a) What would be the wire is 0.5 mm^2, what length of the wire will be needed? (c) If the radius of each turn is 40 mm, how many turns will be there in the coil?`
5 years ago Navjyot Kalra
654 Points
```							Sol.  f = 1 * 10^-6 Ωm P =  500 W E = 250 v
(a) R = V^2/P = 250 * 250/500 = 125 Ω
(b) A = 0.5 mm^2 = 0.5 * 10^-6 m^2 = 5 * 10^-7 m^2
R = fl/A = l = RA/f = 125 * 5 * 10^-7/1 * 10^-6 = 625 * 10^-1 = 62.5 m
(c) 62.5 = 2πr * n, 62.5 = 3 * 3.14 * 4 * 10^-3 * n
⇒ n = 62.5/2 * 3.14 *4 *10^3 ⇒ n = 62.5 * 10^-3/8 * 3.14 = 2500 turns

```
5 years ago
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