Power of Electrical Circuit: Practice Problem & Solution
An electric bulb is rated $60\text{ W}$, $220\text{ V}$. The resistance of its filament is (1994)
Solution Explained:
To solve this problem, we apply the core principles of Power of Electrical Circuit. Understanding the underlying formula is key to arriving at the correct answer below:
The resistance is given by $R = \frac{V^2}{P}$. Substituting the values, $R = \frac{(220)^2}{60} = \frac{48400}{60} \approx 806.67\Omega$, which rounds to $807\Omega$.
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