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An electric bulb is rated $$40\text{ W}$$ at $$80\text{ V}$$. If it is connected in parallel to a $$40\text{ V}$$ supply having negligible internal resistance, the current through the bulb is
The resistance of the bulb is $$R = \frac{V^2}{P} = \frac{80^2}{40} = 160\,\Omega$$. On a $$40\text{ V}$$ supply the current is $$I = \frac{40}{160} = 0.25\text{ A}$$.
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