The heating element of a toaster is a 5.4-m length of nichrome wire of diameter 0.48 mm.The resistivity of nichrome at the operating temperature of the toaster is 1.3 × 10^-6 Ω ∙ m.The toaster is designed to operate at a voltage of 120 V.How much power does it draw in normal operation? A)370 W B)360 W C)380 W D)400 W E)410 W

Question

Quizplus · Accepted Answer

The power drawn by the toaster can be calculated using the formula $P = \frac{V^2}{R}$, where $P$ is the power, $V$ is the voltage, and $R$ is the resistance of the heating element. The resistance $R$ can be calculated using the formula $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity of the material, $L$ is the length of the wire, and $A$ is the cross-sectional area of the wire. The cross-sectional area $A$ of the wire can be calculated using the formula for the area of a circle, $A = \pi r^2$, where $r$ is the radius of the wire.Given:- $\rho = 1.3 \times 10^{-6} \Omega \cdot m$- $L = 5.4 m$- Diameter of the wire $d = 0.48 mm = 0.48 \times 10^{-3} m$, so the radius $r = \frac{d}{2} = 0.24 \times 10^{-3} m$- $V = 120 V$First, calculate the cross-sectional area $A$:$$A = \pi r^2 = \pi (0.24 \times 10^{-3} m)^2$$$$A = \pi (0.24)^2 \times 10^{-6} m^2$$$$A \approx 1.805 \times 10^{-7} m^2$$Then, calculate the resistance $R$:$$R = \rho \frac{L}{A} = 1.3 \times 10^{-6} \Omega \cdot m \times \frac{5.4 m}{1.805 \times 10^{-7} m^2}$$$$R \approx 38.9 \Omega$$Finally, calculate the power $P$:$$P = \frac{V^2}{R} = \frac{120^2}{38.9} \approx 370 W$$Therefore, the toaster draws approximately 370 W in normal operation.

The Heating Element of a Toaster Is a 5