Assuming you have the question right, that's all there is to it. (Get the sign right.)Marcin H said:Is this as simple as using Q=Ne or not? I may be overthinking this, but this doesn't seem 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?Doc Al said:Assuming you have the question right, that's all there is to it. (Get the sign right.)
Marcin H said: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/sThe 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
Hint: If the battery delivered a current I, what power would it be delivering?Marcin H said:How many Joules of energy does the battery I mention store?
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?Doc Al said:Hint: If the battery delivered a current I, what power would it be delivering?
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 said: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.
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 said: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.)
Power will be in units of Watts; if you multiply by the time, then you'll have energy.Marcin H said:P=IV will give you units of WattHours
Yeah, that's what I did. P= (12V)(35Ah) =420WattHours*(3600)Doc Al said:Power will be in units of Watts; if you multiply by the time, then you'll have energy.
Good.Marcin H said:Yeah, that's what I did. P= (12V)(35Ah) =420WattHours*(3600)
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 said:Good.
You need to use both Amp-Hours and Amps to solve for the time.Marcin H said:I am supposed to use the 35Ah to find the time or do I have to somehow use the .5A.
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 said:You need to use both Amp-Hours and Amps to solve for the time.
Sure.Marcin H said: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?
Can you explain that a bit more? Where did you get the "35 A-hr = I*T = .5A*T." part fromDoc Al said: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.
No, just a bit more work than needed. Don't think of it as "unit conversion" though.Marcin H said:Is the way I did it a bad way to go about this?
The basic unit of electric charge is the electron, which has a negative charge of -1.602 x 10^-19 coulombs.
A charge of 10 billion electrons is equivalent to -1.602 x 10^-8 coulombs. Therefore, it would take 10 billion electrons to make up this charge.
An electron has a negative charge of -1.602 x 10^-19 coulombs, while a proton has a positive charge of +1.602 x 10^-19 coulombs. This means that the charge of an electron is equal in magnitude but opposite in sign to that of a proton.
The charge of an electron is determined by the amount of force it experiences when placed in an electric field. This is known as the Coulomb force and is expressed by the equation F = qE, where F is the force, q is the charge, and E is the electric field strength.
The charge of an electron is a fundamental property of the particle and cannot be changed. However, the overall charge of an object can be altered by adding or removing electrons from it.