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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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