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?

A. P = 18.75 watts

B. P = 75 watts

Explanation:

V = 120 Volts

P = VI

I = P/V = 75/120 = 0.625 Amps

V = IR

R = V/I

R = 120/0.625 = 192 Ω

So the resistance of the bulb is 192 Ω and it does not change as it is given in the question.  

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?

As P = VI and I = V/R

P = V*(V/R)

P = V²/R

P = (60)/192

P = 18.75 watts

As expected, it will dissipate less power (18.75 watts) than rated power due to not having rated voltage of 120 Volts.

I = V/R = 60/192 = 0.3125 Amps

or I = P/V = 18.75/60 = 0.3125 Amps

Since the resistance is being held constant, decreasing voltage will also decrease current as V = IR voltage is directly proportional to the current.

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?

P = V*(V/R)

P = V²/R

P = 120/192 = 75 watts

I = P/V = 75/120 = 0.625 Amps

As expected, it will dissipate rated power of 75 watts at rated voltage of 120 Volts.

b.a device that inputs heat into a cool object