Time constant

[sunny1982]

Hi folks I was wondering if someone could help me understand the questions below and advise me on the formulas to use. Many thanks in advance.

A 100 µF capacitor is connected in series with an 8000Ω resistor. Determine the time constant of the circuit.
If the circuit is suddenly connected to a 100 dc supply find,

I. The initial rise of P.D across the capacitor, (rate of voltage rise per volt to the first time constant)

II. The initial charging current, (at zero time)

 

[MikeMl]

What is P.D?

 

[BioniC187]

I'm guessing it means potential difference?

 

[Grossel]

If you can find the answer on those questions, you should have potential to solve the rest yourself:

1
How fast (V/sec) would that cap's voltage rise if the current was constant?

2
How many volts is there over the resistor at t0?

 

[WTP Pepper]

You need to do some reading on the subject.

 

[MrAl]

Hi folks I was wondering if someone could help me understand the questions below and advise me on the formulas to use. Many thanks in advance.

A 100 µF capacitor is connected in series with an 8000Ω resistor. Determine the time constant of the circuit.
If the circuit is suddenly connected to a 100 dc supply find,

I. The initial rise of P.D across the capacitor, (rate of voltage rise per volt to the first time constant)

II. The initial charging current, (at zero time)

Hello,


To get you started, the initial current is just the voltage divided by the resistance because the capacitor is considered a short circuit at t=0. If there is any initial cap voltage though you have to consider that too, but for this problem there is none.

 

[sunny1982]

Hi can anyone tell me if this is correct please? thanks

I. The initial rise of P.D across the capacitor, (rate of voltage rise per volt to the first time constant)

T= CR= 100 * 10-^6 *8000= 8seconds

 

[MrAl]

Hello,

The voltage rises to 63.2 percent of the applied voltage step in the first time constant. So if your time constant is 8 seconds and the applied step is 100 volts then it takes 8 seconds to reach 63.2 volts, and that's the voltage after first time constant. Since it went from 0 to 63.2 in 8 seconds, the rate of rise of voltage was:
dv/dt=63.2/8=7.9 volts per second.

This is of course an approximation anyway as the voltage really rises exponentially as:
Vc=100*(1-e^(-t/RC))

 

[sunny1982]

Vc=100*(1-e^(-t/RC))[/QUOTE]

Thankyou so much for your help I understand it abit more now, so how do i calculate the formula above?

so far i understand that the VC = 100v, I don't understand how to calculate this into the formula 1-e^, -t=1 as we are looking for initial voltage rise, RC= 8 seconds which is the time constant of the circuit. 1-e^ this is the bit i'm having trouble to understand.

 

[MrAl]

Hello again,

You dont have to worry too much about that unless you want to understand that formula too. If you just need to know the voltage after 8 seconds then you can use the fact that the voltage rises to 63.2 percent of the applied step voltage after 8 seconds.

That formula tells you the voltage at any instant of time, such as 1 second, 2 seconds, 3, 4, etc., or even 5.345 seconds for example. Just plug the time into the variable 't' and use RC=R*C for RC. e^(-t/RC) means compute t/RC first, make it negative, then take the number 'e' up to that power. 'e' is the base of the natural log system of numbers, and is approximately equal to:
2.7182818284590452353602874713527

 

[sunny1982]

Hello again,

You dont have to worry too much about that unless you want to understand that formula too. If you just need to know the voltage after 8 seconds then you can use the fact that the voltage rises to 63.2 percent of the applied step voltage after 8 seconds.

That formula tells you the voltage at any instant of time, such as 1 second, 2 seconds, 3, 4, etc., or even 5.345 seconds for example. Just plug the time into the variable 't' and use RC=R*C for RC. e^(-t/RC) means compute t/RC first, make it negative, then take the number 'e' up to that power. 'e' is the base of the natural log system of numbers, and is approximately equal to:
2.7182818284590452353602874713527

Mr Al thankyou very much that is fantastic great help.

 

[sunny1982]

Mr Al thankyou very much that is fantastic great help.

Another thing Mr Al, in my case the variable t will be 1?

Anyway seperate to that, what will the initial charging current be at zero time?

 

[MrAl]

Hello again,



Well, if you want to know the voltage after one time constant and one time constant is 8 seconds, then in your case the variable t should be 8 because 't' is "time in seconds" and you want to know the voltage of the capacitor after 8 seconds.

The initial charging current is easy to calculate because the voltage across the capacitor at t=0 (no time passed yet) is zero volts. So the current is just the applied step voltage divided by the resistance:
i=100/R

so with R=8000 we get:
i=100/8000

which is easy to calculate

 

[sunny1982]

Its fantastic the way you explain things its really interesting once you know what you are doing wish you was my college teacher lol. I really appreciate your help, just the last thing i need to understand is how to calculaet the ultimate charge in the capicitor and the ultimate energy stored in the capicitor.

 

[sunny1982]

The ultimate charge in the capacitor

Q = C E = 100µF * 10¯⁶ * 100v = 0.1 coulomb

The ultimate energy stored in the capacitor

E = V² * C/2 = 100v (100µF)/2 = 500 joules

I've had a go myself Mr Al just wondering if this is correct -

 

[MrAl]

Hi again,


The energy stored in a capacitor is:
E=(C*v^2)/2

[LATEX]E=(1/2)*C*v^2[/LATEX]

and the charge transferred between times t1 and t2 is:
q=integral(t1 to t2) i dt

[LATEX]q=\int _{t1}^{t2} {i}\ dt [/LATEX]

so the charge is basically the integral of the current over the time period the charge is being transferred. If we happen to have constant current then the charge is just the current times the time period:
q=i*t

 

[sunny1982]

Hi again,


The energy stored in a capacitor is:
E=(C*v^2)/2

[LATEX]E=(1/2)*C/v^2[/LATEX]

and the charge transferred between times t1 and t2 is:
q=integral(t1 to t2) i dt

[LATEX]q=\int _{t1}^{t2} {i}\ dt [/LATEX]

so the charge is basically the integral of the current over the time period the charge is being transferred. If we happen to have constant current then the charge is just the current times the time period:
q=i*t

I'm lost again lol, How do i determine t1 and t2 E=(1/2)* C/v^2, what do you mean by (1/2), and I don't understand the ultimate charge bit

 

[sunny1982]

W = 1/2 C V² = (100µF * 10^-6 * 100v²)/2 = 0.5 joules I've understood this bit about the energy stored, but I still can't understand the ultimate charge in the capacitor.

 

[sunny1982]

W = 1/2 C V² = (100µF * 10^-6 * 100v²)/2 = 0.5 joules I've understood this bit about the energy stored, but I still can't understand the ultimate charge in the capacitor.

 

[MrAl]

Hello again,


I'm very sorry as i typed the Latex version of the equation improperly. I'm correcting that first then we'll get to the final question about the charge.

The energy stored in a capacitor is:
E=(C*v^2)/2

[LATEX]E=(1/2)*C/v^2[/LATEX]

and the charge transferred between times t1 and t2 is:
q=integral(t1 to t2) i dt

[LATEX]q=\int _{t1}^{t2} {i}\ dt [/LATEX]

so the charge is basically the integral of the current over the time period the charge is being transferred. If we happen to have constant current then the charge is just the current times the time period:
q=i*t

CORRECTION:


The energy stored in a capacitor is:
E=(C*v^2)/2

[LATEX]E=(1/2)*C*v^2[/LATEX]

Luckily you didnt have a problem with that part anyway.

The charge stored is:
[LATEX]q=\int _{t1}^{t2} {i}\ dt [/LATEX]

and all that means is that we sum all the infinitesimal contributions of the current 'i' over time, and the sum is the total charge transferred. It's simpler to look at a constant current and also when the current changes in discrete steps. Lets do that.

Do you understand how to integrate?

Say we have a current that is 1 amp for 2 seconds (starting from t=0) and then the current changes to 2 amps for 3 seconds, then goes to zero. To find the total charge, we multiply the current times the time for each interval and then sum all the results.
The first interval has 1 amp for 2 seconds, so that comes out to:
q1=1*2=2

The second interval has 2 amps for 3 seconds, so that comes out to:
q2=2*3=6

Now we sum the results:
q=q1+q2=2+6=8

So the total charge transferred is 8 Coulombs.

Make sense so far?

Now lets say those intervals were shorter, 0.2 seconds and 0.3 seconds respectively.

For interval 1 we have:
q1=1*0.2=0.2

and for interval 2 we have:
q2=2*0.3=0.6

and the total now is:
q=0.2+0.6=0.8 Coulombs.

It follows that if we let the intervals get even smaller, the total would get smaller too, so for intervals of 0.002 and 0.003 seconds we would have a total of:
q=0.002+0.006=0.008 Coulombs.

It just so happens that a continuous time signal could be broken up into small time units (say 1ms each) and we could approximate by summing all the contributions over the entire wave for time increments of 1ms. We might then have a reasonable approximation. But for any wave if we let the time increments get extremely small like 1e-12 then we get a more accurate result, and if we let them get even smaller like 1e-24 then we'd get an even more accurate result than before. Ultimately when we let the intervals tend to zero we get an exact value, and that is equal to the integral above.

Does this make sense?