A car headlight bulb working on a $12 \mathrm{~V}$ car battery draws a current of $0.5 \mathrm{~A}$. The resistance of the light bulb is :(a) $0.5 \Omega$(b) $6 \Omega$(c) $12 \Omega$(d) $24 \Omega$

Given:

Potential difference, V = 12 volt

Current, I = 0.5 ohms

To find:

Resistance, R

Solution:

According to the ohm's law:

$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$

$\mathrm{R}=\frac{\mathrm{12}}{\mathrm{0.5}}$

$\mathrm{R} = 24 \Omega $

Answer is (d)

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Updated on: 10-Oct-2022

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