Question

How much power is used by an ultrasonic jewelry cleaner that draws 0.042 A
of current from a 120-V line?


Celeste’s air conditioner uses 2160 W of power as a current of 9.0 A passes
through it. a) What is the voltage drop when the air conditioner is running?
b) How does this compare to the usual household voltage? c) What would
happen if Celeste tried connecting her air conditioner to a usual 120-V line?


Which has more resistance when plugged into a 120.-V line, a 1400.-W
microwave oven or a 150.-W electric can opener?

Answer 1

Electrical power is defined as\[P = VI\]We know voltage and the current.

We can use the same equation to determine the voltage drop across the air-conditioner. It will be much higher than 120. The air-conditioner will draw a much higher amperage if we try to connect it to 120-V power source. The increased amperage will blow the circuit breaker, or worse, catch the house on fire.

We can use Ohm's law to determine the resistance of the different devices. \[V = IR, ~I = {V \over R}\]Substituting into the power equation\[P = VI = {V^2 \over R}\]We know the voltage and the power, we can find the resistance.

Answer 2

Is power in volts or watts..or neither? I have 8.4 as the answer for the first question. I don't know if that's right though. I'm working on the other two, but wanna make sure I'm on the right track.

Answer 3

Power is in watts which 1 watt = 1 joule per second. Not sure how you came up with 8.4 though. Instantaneous power for the first question would be 120V*.042A = 5.04W

Answer 4

Oops my mistake I accidentally plugged the wrong numbers into my calculator. So for the second problem I have P=IV, so 2160 x 9. So a) is 19440? That seems a little high though.

Answer 5

For the second problem you're given power (P) and current (I). From Ohm's law you have:
P = IV and V = IR

You know P = 2160W and I = 9A. Solving for V you get 240V.

Answer 6

Sorry physics doesn't usually click the first time around for me. I have a B in that class right now. For the last problem for the microwave, I have \[1400=(120^{2})/R\] and for the can opener I have \[150=(120^{2})/R\] After that I just solve them and the one with the highest answer is correct?

Answer 7

This site is SO slow.
Anyway...that's right but you don't even need to perform any calculations to answer that question. Look at the proportions...if R goes up, power must go down.

Answer 8

Ah-hah! Got it. I'll have to do some more practice problems before my test to make sure I understand it completely. Thank you. :)

Answer 9

no problem, good luck :)