Power consumption

[tDexx]

Homework Statement

What is the power consumed when plugging an iron with a rating of 10 amps to a 120 volt outlet?

Homework Equations

(watts/volts)=amps

The Attempt at a Solution

(w/120)=10 amps
120= 10 amps (w)
120/10=12
12 watts.

 

[PeterO]

Homework Statement

What is the power consumed when plugging an iron with a rating of 10 amps to a 120 volt outlet?

Homework Equations

(watts/volts)=amps

The Attempt at a Solution

(w/120)=10 amps
120= 10 amps (w)
120/10=12
12 watts.

Is this iron designed for 120V?

I know my country uses 240V, but I could always take my iron to the US ans plug it in. I wouldn't work very well - certainly wouldn't draw 10 amps - but I could connect it without problems.

 

[cepheid]

Is this iron designed for 120V?

I know my country uses 240V, but I could always take my iron to the US ans plug it in. I wouldn't work very well - certainly wouldn't draw 10 amps - but I could connect it without problems.

Is the voltage of AC mains power [that the hypothetical iron is designed for] really relevant to the problem? Also, I know that it's dubious that an iron would draw its maximum rated current (esp. if that's 10 A!) under normal operation. But this problem really just seems to be testing whether the OP understands that power = voltage*current. What's wrong with an unrealistic problem?

Homework Statement

What is the power consumed when plugging an iron with a rating of 10 amps to a 120 volt outlet?

Homework Equations

(watts/volts)=amps

The Attempt at a Solution

(w/120)=10 amps
120= 10 amps (w)
120/10=12
12 watts.

Your algebra is way off here. If P/(120 V) = 10 A, then to solve for P (i.e. to leave it by itself on one side of the equation), you'd have to get rid of the factor of 1/(120 V) on the left-hand side. To do that, you can multiply both sides of the equation by (120 V).

 

[PeterO]

Is the voltage of AC mains power [that the hypothetical iron is designed for] really relevant to the problem? .

Certainly is.

If this iron is labelled 10A, 240V then the resistance of the heating element will be 24Ω [when hot]

If this iron is labelled 10A, 120V then the resistance of the heating element will be 12Ω [when hot]

Depending on the voltage label on the iron, the results will be quire different.

The other two questions posted by OP efer to 220V. It is common to consider what happens when a device is connected to an incorrect power supply.
For me, where we use 240V, my devices simply run slowly/inefficiently if the mains electricity is too low. If an American brings something to Australia and connects it, it will simply burn out and perhaps trip the fuse.

 

[cepheid]

Certainly is.

If this iron is labelled 10A, 240V then the resistance of the heating element will be 24Ω [when hot]

If this iron is labelled 10A, 120V then the resistance of the heating element will be 12Ω [when hot]

Depending on the voltage label on the iron, the results will be quire different.

The other two questions posted by OP efer to 220V. It is common to consider what happens when a device is connected to an incorrect power supply.

Uh huh, maybe, but it seems FAR more likely to me that the question is just asking: a device draws 10 A at 120 V, what is the power?

For me, where we use 240V, my devices simply run slowly/inefficiently if the mains electricity is too low. If an American brings something to Australia and connects it, it will simply burn out and perhaps trip the fuse.

Yeah, I know that.