WAEC SSCE - Physics - 2001 (Objective)

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\(\Omega\) Bonne réponse

Détails de la réponse

The power of an electrical appliance can be calculated by the formula P = V\(^2\)/R, where P is the power in watts, V is the voltage in volts, and R is the resistance in ohms. In this case, the bulb is rated 60W and 220V. So, we can rearrange the formula to find the resistance: R = V\(^2\)/P. Substituting the values gives us: R = 220\(^2\)/60. Evaluating this expression gives us: R = 806.7\(\Omega\). Therefore, the resistance of the filament when the bulb is operating normally is 806.7\(\Omega\). Note: It is important to remember that the actual resistance of the filament may vary due to factors such as temperature, age, and manufacturing tolerances.






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