# Current resistor and time

1. Jun 11, 2008

### scholio

1. The problem statement, all variables and given/known data

a current of 5 ampere exists in a 10 ohm resistor for 4 minutes

a) how many coulombs pass through the resistor in this time? should be 1200 coulombs
b) how many electrons pass through the resistor in this time? should be 7.5*10^21
c) what potential drop exists across the resistor? should be 50 volts
d) how much energy is dissipated in the resistor during this time? should be 60000joules

2. Relevant equations

resistance R = V/I where V is electric potential, I is current

current I = nAqv_d where n is number of electrons, A is area, q is charge, v_d is drift velocity

power P = IV where I is current, V is electric potential.

work W = 1/2(CV^2) where C is capacitance

3. The attempt at a solution

somewhat started part a:

use first eq to solve for electric potential V --> V = IR = 5(10) = 50 volts

tried using second eq, C = Q/V but did not have capacitance, i don't even think you can use capacitance in this problem

i need help with an equation containing time, because t = 4mins = 240seconds. other wise i could use the fourth eq, and let P = W/t where W is work, t is time and solve for t. the work work formula requires capacitance, which i don't have

as for part b and finding the number of elections i think i use the third eq, hold A as constant but how do i figure out the drift velocity v_d, or do i hold that constant too?

i haven't tried the other parts yet, looking for some insight...

2. Jun 11, 2008

### alphysicist

Hi Scholio,

You're right that you cannot use C=Q/V here. Instead, since you know the current, what is the definition of current?

3. Jun 11, 2008

### scholio

current I is the change in Q over the change in time --> I = dQ/dt

oh i figured it out, i didn't even think of that relationship:

since V = IR = 50 volts

V = (dQ/dt)R
50 = (dQ/240)10
dQ = 12000/10
Q = 1200 coulombs

that was straightforward, thanks!

Last edited: Jun 11, 2008
4. Jun 11, 2008

### lukas86

For the number of electrons per coulumb... h t t p : / / en.wikipedia.org/wiki/Electric_current

That should help.

5. Jun 11, 2008

### alphysicist

scholio,
That's right. Another way to think about it, is that a current of 5A means 5 coulombs per second. Since you know how many seconds the current is running and it's constant, you can just multiply current times time to get coulombs.

6. Jun 11, 2008

### lukas86

The way I did part A which works without needing the voltage is...

I = Q/t...
Q = I*t
= 5A * 4min*(60s/1min)
= 1200 Coul

7. Jun 11, 2008

### scholio

thanks, i got part a now.

how would i got about handling part b using I = nAqv_d, do i hold A area, and drift velocity v_d constant? the q in this equation is not the 1200 i found in part a is it?

anyways i tried it and it didn't work out: i took A and v_d constant

5 = n(1200)
n = 4.16*10^-3 ---> not right, need 7.5*10^21 electrons

8. Jun 11, 2008

### alphysicist

I don't think you need to use that equation. What's the charge on an electron? Be sure to include the units.

(You can argue that A and v_d are constant, but they still have to be in the equation!)

Edit: Also, the n in that equation is not the number, it is the number density (particles / volume).

Last edited: Jun 11, 2008
9. Jun 11, 2008

### scholio

oh okay, the charge of an electron is 1.6*10^-19 coulombs, is that what i substitute in for q in the equation?

there is another equation involving n, q and v_d. its for current density J.

J = nqv_d where J = sigma(E) where E is electric field, and sigma is conductivity. the material is not specified though.

electric field E = F/q where F is force and q is charge

is this along the right lines now?

10. Jun 11, 2008

### alphysicist

No, it's more like changing units. There is one electron in $1.6\times 10^{-19}$ coulombs. How many electrons are in 1200 coulombs?

11. Jun 11, 2008

### scholio

oh, again i overanalyzed the problem. 1200coulombs/(1.6*10^-19 coulombs) = 7.5*10^21 electrons

thanks again

as for part c, finding the potential drop that exists along the resistor, did i already solve for that in part a? am i correct to say that the V in V = IR is potential difference?

then isn't V = 50 volts then?

12. Jun 11, 2008

### alphysicist

That's right. Did you get part d?

13. Jun 11, 2008

### scholio

no not yet, still need to finish up part c, any tips before i start part d?

Last edited: Jun 11, 2008
14. Jun 11, 2008

### alphysicist

I'll let you work on it first. You have everything you need (just don't use the equations with capacitance).

15. Jun 11, 2008

### scholio

i got part c and d.

part c:
what potential drop exists across the resistor?

QV/t = I^2R where Q is 1200 coulombs, I is 5 amp, R is 10 ohms, t = 240 sec
V = I^2RT/Q = 50 votls

part d:
how much energy is dissipated in the resistor during this time?

energy lost/t = I^2R where t = 240 sec, I = 5amp, R = 10 ohms

energy lost = I^2Rt = 60000 joules

did i do it correctly? i got the answers i was looking for.

16. Jun 11, 2008

### alphysicist

It looks right, but I don't think you need to do quite all that. Your original approach V=IR in the first post is exactly what's needed.

(In the equation you have above, notice that Q/t =I, when the current is constant, so:

QV/t = I^2R ----> I V = I^2 R ------> V = I R