Question

What is the voltage drop across a $1\mathrm{kW}$ electric heater element whose resistance, when hot, is $40\mathrm{\Omega }$?

Answer 1

Step 1: Given data

Power of the electric heater element, $P=1\mathrm{kW}=1\times {10}^3\mathrm{W}$

Resistance of the element, $R=40\Omega$

Step 2: Formula used

Electric power is defined as the time rate at which work is done or energy is transferred.

$P=\frac{workdone}{time}=\frac{W}{t}$

The power of an electric element in terms of voltage is given by

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

where $V$= Voltage drop across the heating element, $R$= Resistance and $P$= Electric Power

Step 3: Calculating the voltage drop across the electric heater element

We have

$$\begin{gathered} \Rightarrow P=\frac{V^2}{R} \\ \Rightarrow V^2=PR \\ \Rightarrow ={10}^3\mathrm{W}\times 40\mathrm{\Omega } \\ \Rightarrow =40000\mathrm{V} \\ \Rightarrow V=\sqrt{40000}=200\mathrm{V} \end{gathered}$$

Hence, the voltage drop across the electric heater element is $200\mathrm{V}$.