Question

An electric heating element to dissipate 400 W on 220 V mains is to be made from nichrome ribbon 1 mm wide and 0.05 mm thick. Calculate the length of the ribbon required if the resistivity of nichrome is 1.1 × 10−6 Ω m.

Answer 1

Answer:

Power P=400 W

Voltage=220 V

Since P=VI

I=P/V=400/220 A

I=1.8A

Now , R=V/I= 220V/1.82A

R=122 ohm

Area of the nichrome ribbon

(A)= (0.05×10^-3)(1×10^-3)

A=5×10^-8 meter square

Now

R=pL/A

L=R/pA

L=122/(1.1×10^-6)(5×10^-8)

L=22.18×10^14 metre

Explanation:

Answer 2

Answer:

The length of the ribbon required is 5.5m

Explanation:

Given the power dissipation, P = 400 W

Voltage supply, V = 220 V

Width of nichrome ribbon, $b = 1 mm = 10^{-3} m$

Thickness of nichrome ribbon, $l = 0.05 mm = 5\times 10^{-5} m$

Resistivity of ribbon,  $\rho= 1.1 \times 10^{-6}\Omega m$

The power dissipation through an element is given by

$P=\dfrac{V^{2} }{R}$

Thus, the resistance through nichrome is

$R=\dfrac{V^{2} }{P} =\dfrac{(220)^{2} }{400}$

  $=\dfrac{220\times 220 }{400} =11\times 11 =121\Omega$

The resistance of wire is directly proportional to its length and inversely proportional to its cross-sectional area. i.e.,

$R=\rho \dfrac{l}{A} =\rho \dfrac{l}{b\times t}$

where $\rho$ is the resistivity of the material.

Substituting the values in the above formula,

$121=1.1 \times 10^{-6} \times\dfrac{l}{10^{-3} \times 5\times 10^{-5}}$

$\implies121=11 \times 2 \times l$

$\implies l =\dfrac{121}{11 \times 2} =5.5m$

Therefore, the length of the ribbon required is 5.5m.

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