Question

The heater element of a particular 120-V toaster is a 8.9-m length of nichrome wire, whose diameter is 0.86 mm. The resistivity of nichrome at the operating temperature of the toaster is 1.3 × 10-6 Ω ∙ m. If the toaster is operated at a voltage of 120 V, how much power does it draw

Answer 1

Power, P = 722.96 watts

Step-by-step explanation:

It is given that,

Voltage, V = 120 V

Length of nichrome wire, l = 8.9 m

Diameter of wire, d = 0.86 mm

Radius of wire, r = 0.43 mm = 0.00043 m

Resistivity of wire,
`\rho=1.3* 10^(-6)\ \Omega-m`

We need to find the power drawn by this heater. Power is given by :

`P=(V^2)/(R)`

And,
`R=\rho(l)/(A)`

`P=(V^2* A)/(\rho* l)`

`P=(120^2* \pi (0.00043)^2)/(1.3* 10^(-6)* 8.9)`

P = 722.96 watts

So, the power drawn by this heater element is 722.96 watts. Hence, this is the required solution.