Question

An electric bulb is rated at 220 V, 100 W. What is its resistance ?

• A. $2.2\Omega$
• B. $480\Omega$
• C. $484\Omega$
• D. $490\Omega$

Answer 1

The correct option is C $484\Omega$

In any electrical appliance their is a simple relationship between the power output of a component, the potential difference across that component, and the current running through it. This is because charge times potential difference is energy, so (charge per unit time) times (potential difference) gives (energy per unit time). Thus, we have $P=VI.$

Substituting $I=V/R$ from ohm's law, the power relationship can be rewritten as $P=\frac{V^2}{R}$. From this formula, the resistance can be calculated as follows.

$P=\frac{V^2}{R}$ that is, $R=\frac{V^2}{P}$. The power and voltage are given as$100W$ and $220V.$

Therefore, $R=\frac{{220}^2}{100}=\frac{48400}{100}=484\Omega$.

Hence, the resistance of the bulb is $484\Omega .$