Answer:
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.