An electric bulb is rated 60W and 120V. Calculate the resistance of its filament when it is operating normally
Gives
P=60W
V=120v
We know that
"P=\\frac{V^2}{R}\\\\R=\\frac{V^2}{P}\\\\put value\\\\R=\\frac{(120)^2}{60}=240\\Omega"
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