Question

The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-V line. (a) What is the current in the heating element? Is 12-gauge wire large enough to supply this current? (b) What is the resistance of the dryer’s heating element at its operating temperature? (c) At 11 cents per kWh, how much does it cost per hour to operate the dryer?

Short answer

(a) The current in the heating element is$1=17.08\mathrm{A}$ and safe for 12-gauge wire.

(b) The resistance of the dryer’s heating element at its operating temperature is$14.1\Omega$

(c) Cost per hour to operate the dryer is 45.1 Cents.

Step-by-step solution

Definition of the heating element.

The heating element is part of an electric heating appliance in which electric energy is transformed into heat.

Heating element power $\mathrm{P}=4.1\mathrm{kW}\left(\frac{{10}^3\mathrm{W}}{1\mathrm{kW}}\right)=4.1\times {10}^3\mathrm{W}$

Connected line voltage = 240 V

Calculation of current

(a)

The current is given as,

$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}$……………….(1)

Substitute the values of P and V in equation (1) to get the current,

$$\begin{gathered} \mathrm{I}=\frac{4.1\times {10}^3\mathrm{W}}{240\mathrm{V}} \\ =17.08\mathrm{A} \\ \mathrm{Thus},\mathrm{the}\mathrm{current}\mathrm{in}\mathrm{the}\mathrm{heating}\mathrm{element}\mathrm{is}\mathrm{I}=17.08\mathrm{A}\mathrm{which}\mathrm{is}\mathrm{safe}\mathrm{for}12-\mathrm{gauge}\mathrm{wire}. \end{gathered}$$

Thus, the current in the heating element is which is safe for 12-gauge wire.

Step 3:

(b)

The power dissipated is given as,

$\mathrm{P}=\frac{{\mathrm{V}}^2\mathrm{R}}{}$ (2)

Substitute all the values in equation (2) which gives resistance,

$$\begin{gathered} \mathrm{R}=\frac{{\left(250\mathrm{V}\right)}^2}{4.1\times {10}^3\mathrm{W}} \\ =14.1\mathrm{\Omega } \end{gathered}$$

Thus, the resistance of the dryer’s heating element at its operating temperature is$14.1\mathrm{\Omega }$

Step 4:

(c)

The dryer costs 11 cents for the energy of 1 kWh.energy consumed by the dryer is,

$\mathrm{E}=\mathrm{P}\times \mathrm{t}$

E=$4.1\mathrm{kW}\times 1\mathrm{hE}=\mathrm{pt}.$pt.=(4.1kW) (1h) =4.1kWh

$\mathrm{E}=4.1\mathrm{kWh}$

The cost of operating the dryer for 1 h can be calculated as,

$$\begin{gathered} \mathrm{Operating}\mathrm{cost}=4.1\mathrm{kWh}\times 11\mathrm{cent}/\mathrm{kWh} \\ =45.1\mathrm{cent} \end{gathered}$$

Thus, the cost per hour to operate the dryer is 45.1 Cents.