Electrical power,

$\overline{){\mathbf{P}}{\mathbf{=}}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{R}}}$

We'll use the rated power and the standard household voltage to determine the resistance of the bulb

P_{rated} = 75 W

V_{standard} = 120 V

R_{bulb} = V^{2}/P = 120^{2}/75 = 192 Ω (The rated power in parts A and B is the same. R will also be the same)

A.

V_{applied} = 60 V

R_{bulb} = 192 Ω

The power rating on a light bulb indicates how much power it would dissipate when it is hooked up to the standard household voltage of 120 V (this rating does not mean that the light bulb always dissipates the same amount of power, assume that the resistance is constant in this case).

A. How much power is dissipated in a light bulb that is normally rated at 75 W, if instead we hook it up to a potential difference of 60 V?

B. How much power is dissipated in a light bulb that is normally rated at 75 W, if instead we hook it up to a potential difference of 120 V?

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