What is the charge of the 10 billion electrons?

[Marcin H]

Homework Statement

:[/B]
I have a battery and lightbulb circuit (in series). 12V battery rated at 35Ah.

Ten billion electrons pass a planar cross section of the wire. What is the charge of the 10 billion electrons?

Homework Equations

Q=Ne
I=dq/dt
1A=1C/s

The Attempt at a Solution

Is this as simple as using Q=Ne or not? I may be overthinking this, but this doesn't seem right.

Q = Ne
Q=(#electrons)(electron charge)
Q= (1E10)(1.6E-19 C)
Q = 1.6 E -9 C

 

[Mark44]

Not an advanced physics problem, so moved to Intro Physics

 

[Doc Al]

Is this as simple as using Q=Ne or not? I may be overthinking this, but this doesn't seem right.

Assuming you have the question right, that's all there is to it. (Get the sign right.)

 

[Marcin H]

Assuming you have the question right, that's all there is to it. (Get the sign right.)

Ok. Follow up question. How many Joules of energy does the battery I mention store? I would use .5CV^2 or .5QV, but this is what I learned in E&M physics, so I think there is another way. This is EGR 110 intro to electrical and computer engineering, so there are no pre-recs. Is there a way to do this using unit conversions?

 

[Ray Vickson]

Homework Statement

:[/B]
I have a battery and lightbulb circuit (in series). 12V battery rated at 35Ah.

Ten billion electrons pass a planar cross section of the wire. What is the charge of the 10 billion electrons?

Homework Equations

Q=Ne
I=dq/dt
1A=1C/s

The Attempt at a Solution

Is this as simple as using Q=Ne or not? I may be overthinking this, but this doesn't seem right.

Q = Ne
Q=(#electrons)(electron charge)
Q= (1E10)(1.6E-19)
Q = 1.6 E -9

Always state your units; chances are the question would be marked wrong without them. Otherwise, it looks OK.

 

[Doc Al]

How many Joules of energy does the battery I mention store?

Hint: If the battery delivered a current I, what power would it be delivering?

 

[Marcin H]

Hint: If the battery delivered a current I, what power would it be delivering?

Yeah I was thinking about using P=IV, but I'm not sure we can. Can we still use that if we are given current in Amp hours?
12V battery rated at 35Ah.

 

[Doc Al]

Yeah I was thinking about using P=IV, but I'm not sure we can. Can we still use that if we are given current in Amp hours?
12V battery rated at 35Ah.

Using P = VI is the first step. Now how do you go from power to energy? (Don't plug in numbers yet, just play with the equations symbolically.)

 

[Marcin H]

Using P = VI is the first step. Now how do you go from power to energy? (Don't plug in numbers yet, just play with the equations symbolically.)

Oh I see. P=IV will give you units of WattHours and then you just do unit conversions and get it to Joules. Got it! Thanks!

 

[Doc Al]

P=IV will give you units of WattHours

Power will be in units of Watts; if you multiply by the time, then you'll have energy.

 

[Marcin H]

Power will be in units of Watts; if you multiply by the time, then you'll have energy.

Yeah, that's what I did. P= (12V)(35Ah) =420WattHours*(3600)

 

[Doc Al]

Yeah, that's what I did. P= (12V)(35Ah) =420WattHours*(3600)

Good.

 

[Marcin H]

Good.

Ran into another problem. Assume the battery is supplying .5A of current to a light bulb. How long will the battery last? (Battery is 12V rated at 35Ah) I am supposed to use the 35Ah to find the time or do I have to somehow use the .5A. I don' t think there is a way to find how long it will last with just Amps. You need Amp hours to do this right?

 

[Doc Al]

I am supposed to use the 35Ah to find the time or do I have to somehow use the .5A.

You need to use both Amp-Hours and Amps to solve for the time.

 

[Marcin H]

You need to use both Amp-Hours and Amps to solve for the time.

Ok so, I used P=IV to find that the battery supplies 6W. I can do the same thing to find that it provides 420Wh. So can I just use unit conversions to find the hours? 6W/420Wh. Watts cancel and you are left with 1/70h. So 70 hours? Is this correct?

 

[Doc Al]

Ok so, I used P=IV to find that the battery supplies 6W. I can do the same thing to find that it provides 420Wh. So can I just use unit conversions to find the hours? 6W/420Wh. Watts cancel and you are left with 1/70h. So 70 hours? Is this correct?

Sure.

All you have to do is realize that the ampere-hours are given so: 35 A-hr = I*T = .5A*T. Solve for T and get 70 hours.

 

[Marcin H]

Sure.

All you have to do is realize that the ampere-hours are given so: 35 A-hr = I*T = .5A*T. Solve for T and get 70 hours.

Can you explain that a bit more? Where did you get the "35 A-hr = I*T = .5A*T." part from

 

[Doc Al]

The battery capacity, expressed in terms of Amp-Hours, equals Current*Time. Draw a lower current and the battery can run for a longer time.

 

[Doc Al]

35 A-hr = (.5 A)*(70 hr) = (1 A)*(35 hr) = ... etc.

 

[Marcin H]

Ok, that makes sense. More sense than what I did. Is the way I did it a bad way to go about this? I don't know if I would have noticed to use that equation BC=AT on a test or quiz. I was just looking at units and trying to get hours through unit conversion.

 

[Doc Al]

Is the way I did it a bad way to go about this?

No, just a bit more work than needed. Don't think of it as "unit conversion" though.