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

Answer:

Power, P = 722.96 watts

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, $\rho=1.3\times 10^{-6}\ \Omega-m$

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

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

And, $R=\rho\dfrac{l}{A}$

$P=\dfrac{V^2\times A}{\rho\times l}$

$P=\dfrac{120^2\times \pi (0.00043)^2}{1.3\times 10^{-6}\times 8.9}$

P = 722.96 watts

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