An electric bulb rated 75W is operating on an 240V main supply.Determine the resistance of the heating element

April 4, 2026Physics

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Answer

\text{768 } \Omega

4 steps

Identify the given values.
The power of the electric bulb is $P = 75 \, \text{W}$ .
The operating voltage is $V = 240 \, \text{V}$ .
Use the formula relating power, voltage, and resistance.
The formula for power in terms of voltage and resistance is:
$P = \frac{V^2}{R}$
To find the resistance $R$ , we can rearrange the formula:
$R = \frac{V^2}{P}$
Substitute the given values into the formula.
$R = \frac{(240 \, \text{V})^2}{75 \, \text{W}}$
Calculate the resistance.
$R = \frac{57600 \, \text{V}^2}{75 \, \text{W}}$ $R = 768 \, \Omega$
The resistance of the heating element is $\text{768 } \Omega$ .

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