Exponential Curve

[Eddy Current]

In the problem it takes 100 milliseconds to reach 1.5 amps, and i know that each time constant is equal to 63.2% times some value. And i know that there is 5 times constants so 100/5 is 20. So they get 1.5 x .632 =.948 from 0-20. But, what i don't understand is how they get the other figures from there, the other times constants after that are 1.297, 1.425, 1.472, and 1.489.

 

[GoldDigger]

In the problem it takes 100 milliseconds to reach 1.5 amps, and i know that each time constant is equal to 63.2% times some value. And i know that there is 5 times constants so 100/5 is 20. So they get 1.5 x .632 =.948 from 0-20. But, what i don't understand is how they get the other figures from there, the other times constants after that are 1.297, 1.425, 1.472, and 1.489.

I think we need a lot more context to be able to comment on your question.

Exponential increases in current, for example, are very different from exponential decay of current when the drive is removed from an inductive circuit. And both of those are different from exponential approach to a steady state value when closing a circuit into an inductor.

My general feeling is that you do not understand the math behind exponential functions and so the explanation will have to start at a deeper level than your question.

But in case you are able to make use of the information, what is changing exponentially is not the current itself, but rather the difference between that current and 1.5A.
So, the numbers you need to explain are really .203, .075, .028, and .011. Notice that each one of those numbers is ~.37 times the previous number. And your first number, .948 is (1.5 - .552) and .552 times approximately .37 is .203.

There is a little problem with significant digits and rounding in the answers to the problem.

 

[Eddy Current]

I think we need a lot more context to be able to comment on your question.

Exponential increases in current, for example, are very different from exponential decay of current when the drive is removed from an inductive circuit. And both of those are different from exponential approach to a steady state value when closing a circuit into an inductor.

My general feeling is that you do not understand the math behind exponential functions and so the explanation will have to start at a deeper level than your question.

But in case you are able to make use of the information, what is changing exponentially is not the current itself, but rather the difference between that current and 1.5A.
So, the numbers you need to explain are really .203, .075, .028, and .011. Notice that each one of those numbers is ~.37 times the previous number. And your first number, .948 is (1.5 - .552) and .552 times approximately .37 is .203.

There is a little problem with significant digits and rounding in the answers to the problem.

There really isn't much more to the example other than the graph from the Delmars Electrical book which shows the amps(0-1.5) on one side and the milliseconds(0-100) on the other. Something that i might not have made clear is that .948 is 0-20, 1.297 20-40, 1.425 40-60, 1.427 60-80, and 1.489 80-100.

I thought the current was changing in the graph it goes from 0-1.5 in 100 milliseconds.

 

[gar]

130915-2356 EDT

Eddy Current:

Your problem is still that you don't start with basics.

The equations for the type of exponential curves you are looking at are:

x1 = K1*e^(-t/K2)
and
x2 = (1-K1*e^(-t/K2) )

where K1 and K2 are constants, and t is the independent variable time. e is a constant approximately equal to 2.718281828 . This should be available in most scientific calculators.

In finite time x1 will never equal K1, but it will get very close.
In finite time x2 will never equal 0, but it will get very close.

x1 is said to asymptotically approach K1 as time increases.
See http://en.wikipedia.org/wiki/Asymptote

In electrical circuits K2 is called the time constant. When t=K2 the value of e^(-1) = 0.367879441 (about).

When t = 5 time constants e^(-5) = 0.00674 (about) and at 10 time constants about 0.0000454 .

You have not adequately described your problem, but suppose it is a series LR circuit that ultimately after a long time (much longer than your 100 mS) reaches 1.5 A, and using your values and assuming 5 time constants at 100 mS, then at:

1 time constant- i = 1.5*0.632120559 = 0.948181 your value was 0.948 .
2 time constants i = 1.5*0.864664717 = 1.296997 your value was 1.297 .
3 time constants i = 1.5*0.950212932 = 1.425319 your value was 1.425 .
Jump to
5 time constants i = 1.5*0.993262953 = 1.489893 your value was 1.489 .

Your values are good based on the assumptions I made. And if the values match the curve, then 5 time constants at 100 mS is a good assumption.

.

 

[Eddy Current]

130915-2356 EDT

Eddy Current:

Your problem is still that you don't start with basics.

The equations for the type of exponential curves you are looking at are:

x1 = K1*e^(-t/K2)
and
x2 = (1-K1*e^(-t/K2) )

where K1 and K2 are constants, and t is the independent variable time. e is a constant approximately equal to 2.718281828 . This should be available in most scientific calculators.

In finite time x1 will never equal K1, but it will get very close.
In finite time x2 will never equal 0, but it will get very close.

x1 is said to asymptotically approach K1 as time increases.
See http://en.wikipedia.org/wiki/Asymptote

In electrical circuits K2 is called the time constant. When t=K2 the value of e^(-1) = 0.367879441 (about).

When t = 5 time constants e^(-5) = 0.00674 (about) and at 10 time constants about 0.0000454 .

You have not adequately described your problem, but suppose it is a series LR circuit that ultimately after a long time (much longer than your 100 mS) reaches 1.5 A, and using your values and assuming 5 time constants at 100 mS, then at:

1 time constant- i = 1.5*0.632120559 = 0.948181 your value was 0.948 .
2 time constants i = 1.5*0.864664717 = 1.296997 your value was 1.297 .
3 time constants i = 1.5*0.950212932 = 1.425319 your value was 1.425 .
Jump to
5 time constants i = 1.5*0.993262953 = 1.489893 your value was 1.489 .

Your values are good based on the assumptions I made. And if the values match the curve, then 5 time constants at 100 mS is a good assumption.

.

This is all the information they give in the book, its an example with a graph.

According to the book the time constants are 20, 100/5 time constants =20, that is why they have it broke down 0-20, 20-40 and so on until 100 milliseconds right?


Im still not sure i understand how you get 0.864664717, 0.864664717, 0.950212932, 0.993262953

 

[Smart $]

Hello, Eddy.

I believe the subject is electrical time constant of an RC circuit. Simply put, the time constant is:

= RC

where...

time constant

is in seconds,
resistance R is in ohms, and
capacitance C is in farads.

http://en.wikipedia.org/wiki/Time_constant#Time_constants_in_electrical_circuits

PS: Gar is correct in that with time constants 100% is never reached, but is considered [generally] to be so after 5

.

 

[gar]

130916-1254 EDT

Eddy Current:

One time constant is a unit of time that has a value in conventional units like seconds. I have identified this as K2 in the equations.

In the original question presented to you it appears that the 1.5000000 A is the asymptotic value for the circuit in question at a very large value of t (time). I believe you were told that 5 time constants equaled 100 mS, but this is not clear. Possibly you made the assumption. The current values on the curve at 1, 2, 3, 4, and 5 time constants would be useful to know.

From your first post

But, what i don't understand is how they get the other figures from there, the other times constants after that are 1.297, 1.425, 1.472, and 1.489.

These numbers are not time constants, they are current values.


How did I get the numbers 0.632120559, 0.864664717, 0.950212932, ......, 0.993262953, by solving the equation

f = ( 1-e^(-n) ) where n = 1, 2, 3, 4, 5
except I did not do the calculation for 4.


.

 

[Eddy Current]

The current values on the curve at 1, 2, 3, 4, and 5 time constants would be useful to know.

I provided the current values in the post above.

.948A is 0-20(Time), 1.297A 20-40(Time), 1.425A 40-60(Time), 1.427A 60-80(Time), and 1.489A 80-100(Time).

According to the book it takes 100 milliseconds to go from 0-1.5A.

 

[gar]

130916-1412 EDT

Eddy Current:

If the current actually is 1.500000 at 5 time constants, which means it becomes 1.5 * 1.0 / 0.993262953 = 1.5 * 1.006782743 = 1.510174114 A at very many time constants, then the current at 1 time constant (20 mS) is 1.5 * 0.632120559 / 0.993262953 = 1.5 * 0.636540807 = 0.9546 A and not 0.948 A.

The statement of the problem should have been ---
The asymptotic value of the current is 1.5 A, and 100 mS equals 5 time constants. What are the current values at 1, 2, 3, 4, and 5 time constants? The answers to this question are the values you provided in post #1.

.

 

[Eddy Current]

130916-1412 EDT

Eddy Current:

If the current actually is 1.500000 at 5 time constants, which means it becomes 1.5 * 1.0 / 0.993262953 = 1.5 * 1.006782743 = 1.510174114 A at very many time constants, then the current at 1 time constant (20 mS) is 1.5 * 0.632120559 / 0.993262953 = 1.5 * 0.636540807 = 0.9546 A and not 0.948 A.

The statement of the problem should have been ---
The asymptotic value of the current is 1.5 A, and 100 mS equals 5 time constants. What are the current values at 1, 2, 3, 4, and 5 time constants? The answers to this question are the values you provided in post #1.

.

The problem is i'm not understanding how they are getting the values i posted in post #1.


I provided some photos.

 

[mivey]

The problem is I'm not understanding how they are getting the values I posted in post #1.

1) Change = 1.5 * 63.2% = 0.948. Remaining = 1.5 - 0.948 = 0.552. Position = 0 + 0.948 = 0.948

2) Change = 0.552 * 63.2% = 0.349. Remaining = 0.552 - 0.349 = 0.203. Position = 0.948 + 0.349 = 1.297

3) Change = 0.203 * 63.2% = 0.128. Remaining = 0.203 - 0.128 = 0.075. Position = 1.297 + 0.128 = 1.425

4) Change = 0.075 * 63.2% = 0.047. Remaining = 0.075 - 0.047 = 0.028. Position = 1.425 + 0.047 = 1.472

5) Change = 0.028 * 63.2% = 0.017. Remaining = 0.028 - 0.017 = 0.011. Position = 1.472 + 0.017 = 1.489

Which is not exact but has some rounding errors which they did not seem to think was important.

 

[gar]

130917-0758 EDT

Eddy Current:

In post #4 I provided two equations. The second equation x2 provides the solution to the problem. However, I made a mistake in writing the equation.

As originally written it was
x2 = (1-K1*e^(-t/K2) )

and it should be
x2 = K1*(1-e^(-t/K2) )

Make K1 = 1.5000000, and -t/K2 = -1. -2, -3, -4, -5. This will give you the values I calculated in post #4 and #7.

The statement of the problem in the book is wrong. The exact value of 1.5 is not reached in 5 time constants. In fact it is never reached in finite time, but it gets very close.

By loosely describing the exponential curve authors create confusion that obscures a basic understanding. It is necessary that you understand that you will never reach the asymptotic value in a finite time.

A more accurate value for the change per 1 time constant period is 0.632120559 , maximum resolution of my HP 32SII calculator.

.

 

[Eddy Current]

The statement of the problem in the book is wrong. The exact value of 1.5 is not reached in 5 time constants. In fact it is never reached in finite time, but it gets very close.

Look at the second picture, the book explains that it is theoretically impossible to reach 1.5A but that it will get very close 99.3%.

 

[Eddy Current]

1) Change = 1.5 * 63.2% = 0.948. Remaining = 1.5 - 0.948 = 0.552. Position = 0 + 0.948 = 0.948

2) Change = 0.552 * 63.2% = 0.349. Remaining = 0.552 - 0.349 = 0.203. Position = 0.948 + 0.349 = 1.297

3) Change = 0.203 * 63.2% = 0.128. Remaining = 0.203 - 0.128 = 0.075. Position = 1.297 + 0.128 = 1.425

4) Change = 0.075 * 63.2% = 0.047. Remaining = 0.075 - 0.047 = 0.028. Position = 1.425 + 0.047 = 1.472

5) Change = 0.028 * 63.2% = 0.017. Remaining = 0.028 - 0.017 = 0.011. Position = 1.472 + 0.017 = 1.489

Which is not exact but has some rounding errors which they did not seem to think was important.

Ok i think im am understanding it now, thanks.

 

[gar]

130917-0943 EDT

Eddy Current:

Read the first paragraph of the statement of the problem or question in your first photograph. There it states 1.5 amperes at 100 ms. This is also what you said in your first post

In the problem it takes 100 milliseconds to reach 1.5 amps,

.

In the first photograph it states

In this example, 100 ms are required for the current to rise to its full value.

.

 

[Besoeker]

In the problem it takes 100 milliseconds to reach 1.5 amps, and i know that each time constant is equal to 63.2% times some value. And i know that there is 5 times constants so 100/5 is 20. So they get 1.5 x .632 =.948 from 0-20. But, what i don't understand is how they get the other figures from there, the other times constants after that are 1.297, 1.425, 1.472, and 1.489.

In addition to what the others have posted, here is a graph showing an exponential curve of a capacitor charging through a resistor.



Different values to your problem but the same calculation technique.

 

[Eddy Current]

130917-0943 EDT

Eddy Current:

Read the first paragraph of the statement of the problem or question in your first photograph. There it states 1.5 amperes at 100 ms. This is also what you said in your first post .

In the first photograph it states

.

You have to continue reading it on the second page where it goes on later to explain that it is theoretically impossible.

 

[gar]

130917-2353 EDT

Eddy Current:

An accurate statement of a question should precede the solution. This was not done in the first paragraph of your first photograph. This led you to present your first post without an accurate description of the question.

For a simple series RC circuit with a switch and a constant voltage source the equation can be written as follows for what happens after the switch closure assuming zero initial charge on the capacitor:

V = i*R + q/C = R*dq/dt +q/C
V is source voltage
i is the instantaneous current
q is the instantaneous charge
dq/dt is the instantaneous current (rate of change of charge),
and equals the slope of the current curve

Now you have a differential equation to solve which I don't expect you to solve. A solution for v_c is = V*(1-e^(-t/RC) ).

It turns out that after 1 time constant that v_c approximately = V*0.632120559 . After this 1 time constant the initial voltage across the resistor is approximately V-V*0.632120559 . In the above equation this new initial voltage has to be substituted for V and after one additional time constant a new initial voltage for the loop is created. This goes on forever. Each of those step ups have to be added together. But all of this is done much easier by just putting values for t into the equation.

.

 

[MichaelKelley]

I am not that much good at calculations and it is an opportunity for me to learn this here, it is great that people are giving their time for solving others issue and this is what I like mostly about the forums.

 

[Besoeker]

I am not that much good at calculations and it is an opportunity for me to learn this here, it is great that people are giving their time for solving others issue and this is what I like mostly about the forums.

Wanting to learn is a big step along the way.