WAEC - Physics (2001 - No. 37)
An electric bulb is rated 60W, 220V. Calculate the resistance of its filament when it is operating normally
296.7\(\Omega\)
400.0\(\Omega\)
512.2\(\Omega\)
806.7\(\Omega\)
Explanation
R = \(\frac{v^2}{P} = \frac{220^2}{100} = 806.7\Omega\)
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