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A lightbulb with an intrinsic resistance of 270 \OmegaΩ is hooked up to a 12-volt battery. How much power is output by the lightbulb? Give your answer out to the thousandths place in units of watts (W).

Respuesta :

Answer:

P = 0.533 W

Explanation:

given,

Resistance of the bulb, R = 270 Ω

Potential of the battery, V=  12 V

Power output of the bulb = ?

we know,

P = I² R

also, V = IR

[tex]P = \dfrac{V^2}{R}[/tex]

[tex]P =\dfrac{12^2}{270}[/tex]

[tex]P =\dfrac{144}{270}[/tex]

     P = 0.533 W

Hence, the Power delivered by the bulb is equal to 0.533 W.

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