A light bulb is rated at 100 W for a 220 V supply. Find the resistance of the bulb.

Question

Accepted Answer

The correct option is (2):

To find the resistance of the light bulb, we can use the relationship between power ($P$), voltage ($V$), and resistance ($R$). The formula for electric power in terms of voltage and resistance is given by:
$P = \frac{V^2}{R}$

We are given the following values:
Power ($P$) = 100 W
Voltage ($V$) = 220 V

We need to find the resistance ($R$). We can rearrange the formula to solve for $R$:
$R = \frac{V^2}{P}$

Now, substitute the given values into the rearranged formula:
$R = \frac{(220,\mathrm{V})^2}{100,\mathrm{W}}$
$R = \frac{48400,\mathrm{V^2}}{100,\mathrm{W}}$
$R = 484,\Omega$

Therefore, the resistance of the bulb is $484,\Omega$. This calculation directly applies the power formula, which is a fundamental concept in electrical circuits.

The incorrect option is (1):
848 $\Omega$. This value would be obtained if there was a calculation error, such as multiplying 220 by 4 instead of squaring it, or other incorrect numerical operations.

The incorrect option is (3):
242 $\Omega$. This value might result from dividing $V^2$ by $2P$ or other erroneous mathematical operations, not adhering to the correct power formula.

The incorrect option is (4):
424 $\Omega$. This value does not correspond to any standard derivation from the given power and voltage, suggesting a computational mistake in applying the formula.