A 2.0 kWheater element from a dryer has a length of 80 cm . If a 10 cm section is removed, what power is used by the now shortened element at 120 V?

Power of the heater element, P = 2.0 kW

Original length of the dryer, ${\mathrm{L}}_{\mathrm{original}}=80\mathrm{cm}$

Removed length of the dryer, ${\mathrm{L}}_{\mathrm{removed}}=10\mathrm{cm}$

Potential difference, V = 120 V

You can use the equation for power in terms of potential and resistance. Resistance can be written as resistivity, length, and cross-sectional area. Using the entered data in the task, you can compare the original length and the new length. Also find that resistance is inversely proportional to length. According to the problem, the initial length is 80 cm. If 10 cm is removed, the new length is equal to 7/8 of the original value. So you can calculate the performance for truncated elements.

Formulas:

The electric power generated due to the potential difference,

$\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}$ ….. (1)

Here, P is the electric power, V is the potential, and R is the resistance.

The resistance value of a material is,

$\mathrm{R}=\frac{\mathrm{pL}}{\mathrm{A}}$ ….. (2)

Here, $p$is the resistivity, L is the length, and A is the area.

As known that, the resistance is directly proportional to length from equation (2) and power is inversely proportional to resistance from equation (1); thus, you can conclude that power is inversely proportional to length.

Thus, the new power after the removed length is given using the length relation as follows:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{new}}=\frac{\mathrm{original}\mathrm{length}\times {\mathrm{P}}_{\mathrm{original}}}{\left(\mathrm{original}\mathrm{length}-\mathrm{removed}\mathrm{length}\right)} \\ =\frac{80\mathrm{cm}}{\left(80-10\right)\mathrm{cm}}\times 2.0\mathrm{kW} \\ =\frac{8}{7}(2.0\mathrm{kW}) \\ =2.4\mathrm{kW} \end{gathered}$$

Hence, the value of the power is 2.4 kW .