Current resistor and time

In summary: That's right, in part A you used the first equation for electric potential. In part B, you used the second equation for capacitance. In part C, you used the third equation for current. And in part D, you used the fourth equation for power.
  • #1
scholio
160
0

Homework Statement



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


Homework Equations



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

capacitance C = Q/V where Q is charge ---> not sure about this one

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

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...
 
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  • #2
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
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!
 
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  • #4
For the number of electrons per coulumb... h t t p : / / en.wikipedia.org/wiki/Electric_current

That should help.
 
  • #5
scholio,
scholio said:
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!

how would i got about handling part b using I = nAqv_d, do i hold A area, and drift velocity v_d constant?

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
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
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
scholio said:
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

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).
 
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  • #9
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
No, it's more like changing units. There is one electron in [itex]1.6\times 10^{-19}[/itex] coulombs. How many electrons are in 1200 coulombs?
 
  • #11
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
That's right. Did you get part d?
 
  • #13
no not yet, still need to finish up part c, any tips before i start part d?
 
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  • #14
I'll let you work on it first. You have everything you need (just don't use the equations with capacitance).
 
  • #15
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 joulesdid i do it correctly? i got the answers i was looking for.
 
  • #16
scholio said:
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

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
 

What is a current resistor and how does it work?

A current resistor is an electrical component that is used to limit the flow of electric current in a circuit. It works by introducing resistance, which is measured in ohms, to the flow of electrons. This reduces the current and prevents damage to other components in the circuit.

What factors affect the resistance of a current resistor?

The resistance of a current resistor can be affected by several factors, including the type of material it is made of, the length and thickness of the resistor, and the temperature. Generally, materials with higher resistivity, longer and thinner resistors, and higher temperatures will result in higher resistance.

How is current resistor measured?

Current resistor is typically measured using a multimeter, which can measure the resistance in ohms. The multimeter is connected to the resistor and will display the resistance value on a digital screen. It is important to ensure that the resistor is not connected to a power source while measuring its resistance.

What is the relationship between current, voltage, and resistance in a circuit?

According to Ohm's Law, there is a direct relationship between current, voltage, and resistance in a circuit. This means that as the voltage increases, the current will also increase, but the resistance will decrease. Similarly, if the resistance increases, the current will decrease, while the voltage will remain constant.

What are the different types of current resistors?

There are several types of current resistors, including fixed resistors, variable resistors, and thermistors. Fixed resistors have a set resistance value and cannot be changed, while variable resistors (also known as potentiometers) can be adjusted to change the resistance. Thermistors are resistors that change their resistance based on temperature.

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