watts and power, converting dw/dt to an integral, and changes in integral interval units

[MonDie]

I am reading Schaum's Outlines, Electric Circuits Fifth Edition having taken both precalculus courses but not calculus.

Firstly, I am confused by the definition of power as the time derivative of watts (p = dw/dt) even though watts is a unit of power. More importantly, I am confused about how this derivative is reversed into an integral of power as below:

 

[math]w = \int_0^t p dt[/math]

 

I recognize that both have an undefined variable t, but I am baffled at why the below should be true regardless of the value of t:

 

[math](\int_0^t p dt) \div w = \frac{dw}{dt} \div p = 1[/math]

 

I get even more confused by some equations given to describe the behavior of an inductor, which stores energy in its magnetic field, charging when current (i) rises and discharging when current drops. I think I understand Ohm's law for an inductor, stated below with the constant of inductance L.

 

[math]v = L \frac{di}{dt}[/math]

 

[math]i = \frac{1}{L} \int_?^? v dt + k_1[/math]

 

I get confused at the inductor's equations for power and for the amount of energy stored in its magnetic field. The energy in its magnetic field is confusingly expressed in watts (w subscript L) rather than joules. Furthermore, I am confused by the change of interval from t1-t2 to i1-i2 followed by the change to a bracketed [i-squared-1 minus i-squared-2].

 

[math]p = vi = L \frac{di}{dt} i = \frac{d}{dt}[ \frac{1}{2} Li^2][/math]

 

[math]w_L = \int_t^t p dt = \int_i^i Li di = \frac{1}{2} L [i^2_2 - i^2_1][/math]

The book then concludes that "the energy stored in the magnetic field of an inductance is W_L = (1/2)Li^2."

 

Obviously v is converted into the form of inductance times the time derivative of current. Then di/dt is multiplied by dt to get di, and this somehow changes the integral's interval (or the x-axis of the integral's function) from units of time to units of current. I am not sure exactly why this happens even though I see it happening a lot in the chapter. I presume that the function is now being plotted against the backdrop of changing current rather than changing time, but that is all I can say. Lastly, I have no clue what to make of the empty time derivative d/dt, nor the bracketed squares of i.

 

Thank you for any help you can give!

Edited April 25, 2017 by MonDie

[Sensei]

I am reading Schaum's Outlines, Electric Circuits Fifth Edition having taken both precalculus courses but not calculus.

Firstly, I am confused by the definition of power as the time derivative of watts (p = dw/dt) even though watts is a unit of power. More importantly, I am confused about how this derivative is reversed into an integral of power as below:

 

[math]w = \int_0^t p dt[/math]

In this context, w is shortcut from "work" rather than "watt".

Work is energy with Joules as unit. Watt = Joules / second.

https://en.wikipedia.org/wiki/Work_(physics)

 

There is very similar equation in Wikipedia article

[latex] w = \int_0^t p(t) dt [/latex]

https://en.wikipedia.org/wiki/Power_(physics)

in "Peak power and duty cycle" section.

It could be used to calculate energy of AC pulses.

Edited April 25, 2017 by Sensei

[MonDie]

That is very, very helpful! However there is still a lot unanswered with regards to the time derivative with the empty numerator, the changing integral interval units, and the mysterious use of square brackets.

[MonDie]

It would seem that part of what I want is a good explanation of the fundamental theorem(s) of calculus. I found something on berkeley dot edu, but it is actually a proof and not intuitive at all. https://math.berkeley.edu/~ogus/Math_1A/lectures/fundamental.pdf

I am confused by page two by the primes inside the functions: f(x') instead of f'(x).

[imatfaal]

It would seem that part of what I want is a good explanation of the fundamental theorem(s) of calculus. I found something on berkeley dot edu, but it is actually a proof and not intuitive at all. https://math.berkeley.edu/~ogus/Math_1A/lectures/fundamental.pdf

I am confused by page two by the primes inside the functions: f(x') instead of f'(x).

 

Primed outside the brackets refers to the differential . The x being primed inside the brackets is saying this is a value of the variable which is different from x; in this case it is a difference less than delta which gives rise to a difference in the functions of less than epsilon

 

Look at the wiki on the definition of a limit of a function

https://en.wikipedia.org/wiki/Limit_(mathematics)#Limit_of_a_function

https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit

 

Although I have just noticed wiki use different notation - typical.

[MonDie]

I did not use a particular source to understand how the integral is the antiderivative. I guess I just had to get my mind on math once again. Distance traveled versus speed is more intuitive than work versus power. For motion along a straight path, speed (s) is the time derivative of distance traveled (d), which we might express as s = dd/dt, because speed is the rate at which position is changing. Why should this imply that you can calculate distance traveled from the integral of speed? Assuming that the integral of speed has time spent traveling as its x-axis, then the integral of speed for an interval of time is the average speed for that interval multiplied by the length of that interval, or the average speed multiplied by the time spent traveling, which will give you distance traveled. Thus s = dd/dt and d = integral(s dt) as long as our x-axis remains as time, thus the dt. Note that you can put an undefined k into the integral since the starting position does not matter if we are trying to calculate total distance traveled rather than position traveled to.

[studiot]

I am suprised at Schaum.

 

They are usually pretty clear as they are designed for exam cramming.

 

Difficult to comment on one I don't have.

 

Do your square brackets look something like this?

 

[math]\int\limits_0^1 {xdx} = \left[ {\frac{{{x^2}}}{2}} \right]_0^1 = \left[ {\frac{{{1^2}}}{2} - \frac{{{0^2}}}{2}} \right] = \frac{1}{2}[/math]

[MonDie]

If they are designed for exam cramming, then that might be the problem. Anyway, I posted the squared brackets in the opening post.

 

I get confused at the inductor's equations for power and for the amount of energy stored in its magnetic field. The energy in its magnetic field is confusingly expressed in watts (w subscript L) rather than joules. Furthermore, I am confused by the change of interval from t1-t2 to i1-i2 followed by the change to a bracketed [i-squared-1 minus i-squared-2].

 

[math]p = vi = L \frac{di}{dt} i = \frac{d}{dt}[ \frac{1}{2} Li^2][/math]

 

[math]w_L = \int_t^t p dt = \int_i^i Li di = \frac{1}{2} L [i^2_2 - i^2_1][/math]

The book then concludes that "the energy stored in the magnetic field of an inductance is W_L = (1/2)Li^2."

 

Obviously v is converted into the form of inductance times the time derivative of current. Then di/dt is multiplied by dt to get di, and this somehow changes the integral's interval (or the x-axis of the integral's function) from units of time to units of current. I am not sure exactly why this happens even though I see it happening a lot in the chapter. I presume that the function is now being plotted against the backdrop of changing current rather than changing time, but that is all I can say. Lastly, I have no clue what to make of the empty time derivative d/dt, nor the bracketed squares of i.

I do not think it matters that the original content has brackets that are taller than the fraction inside the brackets.

Anyway, it looks as though you found the derivative of x^2/2, which is x, and then set it equal to the integral of its derivative, as if the antiderivative of a derivative is equal to the original, undifferentiated equation. Is that right?

 

Regardless that is only the first step, whereas I am still baffled at how the second step mysteriously removes the variable from the equation.

Edited April 26, 2017 by MonDie

[imatfaal]

I am going to have a go at this - even though I hope StudioT will provide a better answer and tell me where I have gone wrong

 

First line of equations

1. power equals voltage times current

2. as voltage equals Inductance times change in current over time (L di/dt) substitute this in for V. Giving power equals L di/dt i

3. rearrange to make the whole expression a change over time by changing i outside the influence of the di/dt to i^2/2 inside the influence

 

Second line of equations

 

1. Statement that Work is the integral of power over the time taken

2. Using the logic and expressions in the first line of equations you can rework this as integral of (L di/dt i) dt (not shown)

3. This collapse to integral with respect to di of L i. Remember i is a function of t

4. This integral equals 1/2L i^2

[studiot]

If they are designed for exam cramming, then that might be the problem. Anyway, I posted the squared brackets in the opening post.

 

I do not think it matters that the original content has brackets that are taller than the fraction inside the brackets.

Anyway, it looks as though you found the derivative of x^2/2, which is x, and then set it equal to the integral of its derivative, as if the antiderivative of a derivative is equal to the original, undifferentiated equation. Is that right?

 

Regardless that is only the first step, whereas I am still baffled at how the second step mysteriously removes the variable from the equation.

 

Apologies mondie, I only looked at the worry over square brackets so I didn't notice you had displayed some.

 

If your entire post #1 is confusing you, it is certainly comfusing me to unravel.

 

I'm not sure why you are reading an electrical engineering book.

Are you now studying electrical engineering?

 

There was a long discussion on an electrical engineering forum where many engineers observed that although they learned calculus in college, they never used it in their working life.

But they had to know calculus to pass their exams to become electrical engineers.

 

Further complications for here are added because those studying electrical engineering at college (you will not meet those equations in school) are expected to have some basic physics so they understand that power is the rate of doing work and that energy is work times time.

 

Further you have mentioned Ohms law for an inductor incorrectly.

 

Your equation is Faraday's law of induction except that you have missed out the all important minus sign which shows that the induced voltage opposes the change in current.

 

You also stated that you haven't taken a calculus course so not suprisingly you are struggling with the equations incolving calculus.

 

 

 

 

I am now going to offer you an alternative life view called black box theory.

 

Black box theory originated (I think) with electrical engineers, but it has spread to many disciplines.

It is heavily used in electrical engineering and has the advantage that it can use both digital and analog techniques.

 

Essentially you have a black box with an input and an output.

The output is a well defined function of the input.

But you don't worry how it works inside the box.

 

The black box may be a physical device such as a breadmaker where the inputs are ingredients and the output is a loaf of bread;

or it may be theoretical where the input is a number and the output is the square of that number.

 

In the first example I have shown the basic box.

I have added one extra line called control. This control allows me to vary the output that is produced from a given input.

 

The second example is a nice double your money machine or scheme.

 

The third example is the Las Vegas version.

 

The fourth example takes a mathematical function and outputs the derivative of that function.

 

The fifth example does the same for an integral, but also uses the control input to apply the boundary condions or the limits of integration.

 

The final example is known as a two port network and is a fundamental building block in electrical circuit theory.

 

So help us to help you by saying where you are coming from and where you want to go to.

 

Are you interested in the Physics, the Electrical Engineering or the Mathematics, because, without black box methods, you need to get all of these right to succeed.

 

 

[EudecioGabriel]

Hi!

 

You should read about The Fundamental Theorem of Calculus, it's really easy believe me, you'll understand why the range of the Integral changed!

It's not a eletrical understanding problem is just a few calculus basis . Good Look.

 

And search for questions!! do some questions until you find one you have to make changes in the range of the integral!

 

I lost one question about this last test , so I took 9 from 10 ( In Brazil we score in numbers, not in letters)

Edited April 29, 2017 by EudecioGabriel

[MonDie]

The calculus may be overkill, but I still want to get a grasp of this math since I already started and since it will help me comprehend the book.

Anyway, the odd thing about the brackets is how they have an interval like the integral has an interval. I have now seen two examples that are clearer than the example in Schaum's. These examples clearly show that the content in the brackets is the derivative of the function in the integral across the equal sign. It is puzzling why the brackets also have an interval after them. For the integral, the interval is static but the variable in the function is dynamic. The integral essential says to take the average output of that function for the given interval—once x is given a value—and then multiply that by the width of the interval. In contrast, the derivative says to determine the slope of the line at a particular value of x—once x is given a value. This bracketed content appears to be some sort of inbetween, between the integral and the derivative, even though I do not know what it would want me to do other than solve for y at x as if it were an ordinary function f(x).

Edited April 29, 2017 by MonDie

[studiot]

Hi mondie, it is kinda difficult for me to know how to help wihtout at least some reaction to my last post.

 

My next step wouold be to work through your posted example maths, line by line.

 

But I don't know if you are interested in the maths or the electrical engineering?

Edited April 29, 2017 by studiot

[MonDie]

I am interested by both, but my primary concern is electrical engineering. Regarding post #10 above, I do not know the Maxwell-Faraday equation. This equation is referred to by allaboutcircuits as "Ohm's Law for an inductor."

 

https://www.allaboutcircuits.com/textbook/direct-current/chpt-15/inductors-and-calculus/

 

The equation is also used in Schaum's to describe the behavior of an inductor. Perhaps it is over-simplified and I will see a more complex version down the road.

---

 

As I think about it more, I realize that you can graph a function or graph its integral. The interval after the brackets is useful when you are graphing the function—or its integral since they are equal and have the same graph—rather than generating an output from an input. That is to say:

 

[math]\int\limits_0^1 {xdx} = \left[ {\frac{{{x^2}}}{2}} \right][/math]

 

That would be incorrect since the integral is bound but the function is not. If I had to calculate the reverse, to calculate the integral from the function, I could not determine what interval should come after the integral except some undefined variables like t1 and t2. Thus the interval after the brackets:

 

[math]\int\limits_0^1 {xdx} = \left[ {\frac{{{x^2}}}{2}} \right]_0^1[/math]

 

Regardless, I still need to understand the final conversion in equations 5 and 6. In the case of 6, I still do not even know how to interpret the brackets in the last piece.

[studiot]

 

In the case of 6, I still do not even know how to interpret the brackets in the last piece.

 

Don't worry it will all become clear as I work through it.

 

But it will take a little time to write out all the maths.

 

So I have posted this since I see you are online right now.

 

[MonDie]

D'oh

 

I take that back....

 

The variable is in the function in the integral, not in its interval. There is nothing in the integral's function to suggest that the range of x is bound. Thus the apparent interval after the brackets is still mysterious to me.

---

 

Do not second guess myself. The integral takes the average value of x for that interval, so it only has one possible output unless the interval is left undefined too.

---

 

Uh, no. If the integral only has one possible output, then it is not equal to a function but equal to a value. A delimited function is still a function and not a value.

---

 

I did the common sense thing: I searched for more examples. It is clear that the integral is being solved by: finding the antiderivative of the function in the integral; solving the new function for the bottom and top interval values; and then subtracting the bottom's output from the top's output. I still do not know what to call that thing, but I see what is happening.

[studiot]

So here is the first session of a couple or three session crash course in calculusy things.

Many authorities advocate teaching integration first and that is probability appropriate here.

 

So what is integration?

Well integration is a way of adding things up – it is a sum. Indeed the [math]\int [/math] symbol is the Gothic letter, for the Greek letter[math]\sum [/math] , we use for summation.

 

But it is much more than just summation; it involves adding up lots of products that are the result of multiplying two quantities together.

This is very very useful to us in the real world because we often want to do just this.

Just to list a few instances.

 

Area = height x Length

Distance = velocity x time

Heat supplied = specific heat x temperature change

Potential Energy = Height x acceleration due to gravity

Electrical power = current x voltage

Metered electrical energy used = power x time

And many many more besides.

 

Integration is often introduced as ‘the area under a graph’ and indeed Area is the first on my list. But integration can provide a great deal more since calculating the left hand side of each equation involves some sort of integration of the right hand side.

 

Important points about my list

 

1. The enormous number and variety of quantities encompassed.
   
2. Some of the quantities are variables, some are constants.
   
3. Some of the multiplied quantities have the same units, some have very different ones.

 

A consequence of (1) is that when we come to write things out as mathematical expressions we have a symbols problem.

Because we have only a few symbols to work with, mostly Roman and Greek letters, some must do double (or more) duty. Care must then be taken to avoid confusion different quantities represented by the same symbol as happened in your opening post with watts (power) and work (energy).

Further it is sometime difficult to choose which common quantity to represent by the first letter for example voltage or velocity, distance or diameter, etc.

 

Now let us look at the format of an integral

 

[math]\int\limits_{x = a}^{x = b} {\left( {{\rm{expression}}} \right)} dx[/math]

We see the product of two quantities, (expression) and dx to the right of the integral sign and two statements about different values of the variable x associated with the integral sign itself.

 

This brings us to point (2) above some quantities are variables and some are constants.

We choose one of the variables as the base variable for the integral.

This is called the running variable or the independent variable or in some cases a parameter.

The dx part is always in terms of the running variable - x in this instance.

The point is that the integral covers a range of values of this variable, not just one as is the case with a constant. So we can’t use a constant as the base.

The range of values taken on by the running variable is starting at the value at the bottom of the integral (x = a ) and extending to the value at the top (x = b) and covering all values in between.

In order of difficulty the (expression) can be

 

1. A constant
   
2. An expression in the running variable – x in this instance
   
3. An expression in another variable.

 

So what to do with this integral?

 

If we take the black box from my post#10 and define the process inside the box as

 

“Look up the integral for (expression) in a table of standard integrals

and then write the result within square brackets we have”

 

 

[math]\int\limits_{x = a}^{x = b} {\left( {{\rm{expression}}} \right)} dx = \left[ {{\rm{Integral}}\;{\rm{from}}\;{\rm{tables}}} \right] = \left[ {{\rm{new}}\;{\rm{expression}}} \right]_{x = a}^{x = b}[/math]

 

Then we write the beginning and end values of the running variable at the end of the square brackets.

I have used the full version x=a, x=b to make it clearer, but the x= part is often dropped.

However keeping the x= is not onerous and helps when we come to part 3 on the list and change of variables, which you had trouble with.

 

We evaluate this by putting the values of x at b and a into the new expression and subtracting thus.

 

[value of new expression at x=b] - [value of new expression at x=a]

 

(Note which is subtracted from which)

 

To see how this works let us look at the simplest possible examples.

 

If (expression) = 1 (that’s right one)

And a = 0 and b = 1 we have

 

 

[math]\int\limits_{x = 0}^{x = 1} 1 dx = \left[ {{\rm{Integral}}\;{\rm{from}}\;{\rm{tables}}} \right] = \left[ {{\rm{new}}\;{\rm{expression}}} \right]_{x = 0}^{x = 1}[/math]

 

If we look up the integral of 1dx in tables we find it is x so putting this in we have

 

 

[math]\int\limits_{x = 0}^{x = 1} 1 dx = \left[ {{\rm{Integral}}\;{\rm{from}}\;{\rm{tables}}} \right] = \left[ {\rm{x}} \right]_{x = 0}^{x = 1} = \left( {\left[ {x = 1} \right] - \left[ {x = 0} \right]} \right) = \left[ 1 \right] - \left[ 0 \right] = 1[/math]

 

We now have something we can start applying to your original post, so I will stop there and post this and await feedback.

 

Edited April 30, 2017 by studiot

[MonDie]

Other than introducing better terminology, your post has reviewed what I knew. Incidentally, I just applied the method above to Schaum's equations and got the same result. Furthermore, I applied that method to the integral of sin(x) from 0 to pi and got virtually the same result as I got from taking an average. I cannot understand why that simple method should output accurate results, but it does appear to do so for my test cases so far. At this point I can no longer tell you where I want to go next.