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Unit WorkBook 4 â Level 4 ENGâ U2 Engineering Maths © 2018 UniCourse Ltd. All Rights Reserved.
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Pearson BTEC Higher Nationals in Electrical and Electronic Engineering (RQF)
Unit 2: Engineering Maths (core)
Unit Workbook 4 in a series of 4 for this unit
Learning Outcome 4
Calculus
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Unit WorkBook 4 â Level 4 ENGâ U2 Engineering Maths © 2018 UniCourse Ltd. All Rights Reserved.
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3.1 Calculus There are a couple of great webpages for checking your answers to Calculus problemsâŠ
Check Differentiation answers
Check Integration answers
3.1.1 The Concept of the Limit, Continuity and the Derivative Calculus deals with functions which continually vary and is based upon the concept of a limit and continuity. Letâs refer to a diagram to understand the concept of a limitâŠ
Here we have indicated a point P on part of a function (the red curve) with Cartesian co-ordinates (ð¥ð¥1,ðŠðŠ1). We require another point Q on the function to be a small increment away from P and will designate a small increment with the symbol ð¿ð¿ (delta).
What we then have isâŠ
ð·ð· = (ðððð,ðððð)
ðžðž = (ðððð + ð¹ð¹ðð, ðððð + ð¹ð¹ðð)
Look now at the chord PQ (drawn as the straight line in blue). If we can determine the slope of this chord and then make it infinitesimally short we will end up with a tangent to the function. It is the slope of this tangent which forms the basis of Differential Calculus (normally called differentiation).
By inspection, we see that the slope of the chord is given byâŠ
ðððððððððð ðððð ðððððððððð ð·ð·ðžðž =ð¹ð¹ððð¹ð¹ðð
If we deliberately make ð¿ð¿ð¥ð¥ approach zero (i.e. make it as short as possible) then we shall reach a limit, which may expressed mathematically asâŠ
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Worked Example 1
ðððððððð
= ð¹ð¹ððððððððððððï¿œâ¯â¯ï¿œ ðð
ð¹ð¹ððð¹ð¹ðð
The term ðððŠðŠ ððð¥ð¥â is written in Leibnitz notation and indicates the slope of the chord when the chord only touches the function at one single point. This is achieved by continual reduction of ð¿ð¿ð¥ð¥.
What we can now say is that we are able to find the slope of any function by adopting this process. Hence, given a function ðð(ð¥ð¥) we are able to differentiate that function, meaning find its slope at all points. We can writeâŠ
The slope at any point of a function ðð(ð¥ð¥) is given by ðð(ðð(ðð))ðððð
This process is called finding the derivative of a function.
3.1.2 Derivatives of Standard Functions What we donât want to be doing is to spend too much time drawing graphs of functions just to work out the derivative. Fortunately there are standard ways to determine the derivative of functions and some of the frequent ones which engineers meet are given in the table below.
Function Derivative ðŽðŽð¥ð¥ðð ðððŽðŽð¥ð¥ððâ1
A sin (ð¥ð¥) A cos (ð¥ð¥)
A cos (ð¥ð¥) âðŽðŽ sin (ð¥ð¥)
ðŽðŽ ðððððð ðððŽðŽðððððð
ðŽðŽ ðððððððð(ð¥ð¥) ðŽðŽ ð¥ð¥ï¿œ
A sinh (ð¥ð¥) A cosh (ð¥ð¥)
A cosh (ð¥ð¥) A sinh (ð¥ð¥)
Letâs look at some examples of using these standard derivativesâŠ
Differentiate the function ðð = ðððððð with respect to ðð.
We see that this function is similar to that in row 1 of our table. We can see that ðŽðŽ = 3 and ðð = 4. We may then writeâŠ
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Unit WorkBook 4 â Level 4 ENGâ U2 Engineering Maths © 2018 UniCourse Ltd. All Rights Reserved.
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Worked Example 2
Worked Example 3
Worked Example 4
ðððŠðŠððð¥ð¥
= ðððŽðŽð¥ð¥ððâ1 = (4)(3)ð¥ð¥4â1 = 12ð¥ð¥3
Differentiate the function ðð = ððððâðð with respect to ðð.
We see that this function is similar to that in row 1 of our table. We can see that ðŽðŽ = 6 and ðð = â5. We may then writeâŠ
ðððŠðŠððð¥ð¥
= ðððŽðŽð¥ð¥ððâ1 = (â5)(6)ð¥ð¥â5â1 = â30ð¥ð¥â6
Differentiate the function ðð = ðððð ðððððð(ðð) with respect to ðð.
Donât worry that ðŠðŠ and ð¥ð¥ have disappeared here. We can use any letters we like. The letter ð£ð£ means voltage and ð¡ð¡ is time. All we want to do here then is to find ððð£ð£ ððð¡ð¡ï¿œ .
We see that the function is similar to that in row 2 of our table. We may writeâŠ
ððð£ð£ððð¡ð¡
= ðŽðŽ cos(ð¡ð¡) = 12 cos (ð¡ð¡)
Differentiate the function ðð = ðð ðððððð(ðð) with respect to ðð.
The letter ðð means current and ð¡ð¡ is time. All we want to do here then is to find ðððð ððð¡ð¡ï¿œ .
We see that the function is similar to that in row 3 of our table. We may writeâŠ
ððððððð¡ð¡
= âðŽðŽ sin(ð¡ð¡) = â5 sin (ð¡ð¡)
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Unit WorkBook 4 â Level 4 ENGâ U2 Engineering Maths © 2018 UniCourse Ltd. All Rights Reserved.
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Worked Example 5
Worked Example 6
Worked Example 7
Worked Example 8
Questions
Differentiate the function ðð = ðððððððð with respect to ðð.
From row 4 of our table we can writeâŠ
ðððŠðŠððð¥ð¥
= ðððŽðŽðððððð = (2)(4)ðð2ðð = 8ðð2ðð
Differentiate the function ðð = ðððð ðððððððð(ðð) with respect to ðð.
We refer to row 5 of our table and writeâŠ
ðððŠðŠððð¥ð¥
=ðŽðŽð¥ð¥
=16ð¥ð¥
Differentiate the function ðð = ðð ðððððððð(ðð) with respect to ðð.
We refer to row 6 of our table and writeâŠ
ðððŠðŠððð¥ð¥
= ðŽðŽ cosh(ð¥ð¥) = 9 cosh (ð¥ð¥)
Differentiate the function ðð = âðððð.ðð ðððððððð(ðð) with respect to ðð.
We refer to row 7 of our table and writeâŠ
ðððŠðŠððð¥ð¥
= ðŽðŽ sinh(ð¥ð¥) = â14.5 sinh (ð¥ð¥)
Q3.1 Differentiate the function ðð = ðððððð with respect to ðð.
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Video
Q3.2 Differentiate the function ðð = ððððâðð with respect to ðð.
Q3.3 Differentiate the function ðð = ðððð ðððððð(ðð) with respect to ðð.
Q3.4 Differentiate the function ðð = ðððð ðððððð(ðð) with respect to ðð.
Q3.5 Differentiate the function ðð = ðððððððð with respect to ðð.
Q3.6 Differentiate the function ðð = ðððð ðððððððð(ðð) with respect to ðð.
Q3.7 Differentiate the function ðð = ðððð ðððððððð(ðð) with respect to ðð.
Q3.8 Differentiate the function ðð = âðððð.ðððð ðððððððð(ðð) with respect to ðð.
ANSWERS
These videos will boost your knowledge of differentiation
3.1.3 Notion of the Derivative and Rates of Change The notion of taking the derivative of a function and examining rate of change can be readily explained with reference to the diagram belowâŠ
Q3.1 40ð¥ð¥4
Q3.2 â28ð¥ð¥â5
Q3.3 18 cos(ð¡ð¡)
Q3.4 â11 sin(ð¡ð¡)
Q3.5 18ðð3ðð
Q3.6 20 ð¥ð¥â
Q3.7 13 cosh(ð¥ð¥)
Q3.8 â19.62 sinh(ð¥ð¥) Sample
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Challenge
Here we have plotted a section of a sine wave in red. Notice that we have used pink arrows to indicate the slope at various points on the sine wave. For example, at the start the sine wave is rising to a positive more rapidly than at any other point. The pink arrow indicates this positive growth by pointing upwards in the direction of the sine wave at that beginning point. The second arrow shows the slope of the sine wave at ðð 2â radians (or 900 if you like). Here it is at a plateau and has zero slope (meaning that it just moves horizontally, with no vertical movement at all). The third arrow indicates a negative slope (its moving down very steeply). The picture continues in a similar way for the remaining arrows.
Hereâs the really interesting bit. When we record all of the slopes on our sine wave we get the dotted waveform in blue. What does that look like? Of course, it is a cosine wave. So we have just proved graphically that finding the slope (differentiating) at all points on a sine wave leads to a cosine wave.
⎠ðð{ð¬ð¬ð¬ð¬ð¬ð¬ (ðð)}
ðððð= ððððð¬ð¬ (ðð)
⊠as given in our table of standard derivatives in section 3.1.2.
If we had loads of time to kill we could prove all of the other derivatives graphically.
Determine a graphical proof (as above) forâŠ
ðð{ððððð¬ð¬ (ðð)}ðððð
= âð¬ð¬ð¬ð¬ð¬ð¬ (ðð)
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3.2.3 Differentiation of Inverse Trigonometric Functions Suppose we have the equationâŠ
ðŠðŠ = ðŽðŽððððâ1(ð¥ð¥)
How do we then go about finding ðððŠðŠ ððð¥ð¥â ? Maybe we should write this in a different formâŠ
ð¥ð¥ = sin(ðŠðŠ)
Now can differentiateâŠ
ððð¥ð¥ðððŠðŠ
= cos(ðŠðŠ) ⎠ðððððððð ð ð ððð¡ð¡â ðŽðŽðððððððŽðŽ: ðððŠðŠððð¥ð¥
=1
cos (ðŠðŠ)
Next we shall remember one of our handy trig. IdentitiesâŠ
ðððððŽðŽ2(ðŠðŠ) + ðŽðŽðððð2(ðŠðŠ) = 1 ⎠ðððððŽðŽ2(ðŠðŠ) = 1 â ðŽðŽðððð2(ðŠðŠ) ⎠cos(ðŠðŠ) = ï¿œ1 â ðŽðŽðððð2(ðŠðŠ)
Since we know ð¥ð¥ = sin(ðŠðŠ) then if we square both sides we shall have ð¥ð¥2 = ðŽðŽðððð2(ðŠðŠ)
So we may re-write our transposed trig. identity asâŠ
cos(ðŠðŠ) = ï¿œ1 â ð¥ð¥2
Now we can put this latest result back into our differentialâŠ
ðððððððð
=ðð
ððððð¬ð¬ (ðð)=
ððâðð â ðððð
=ðð
(ðð â ðððð)ðð.ðð = (ðð â ðððð)âðð.ðð
That was a bit tricky, but we managed to find ðŠðŠâ².
Letâs go on and see if we can find ðŠðŠâ²â²âŠ
ðŠðŠâ² = (1 â ð¥ð¥2)â0.5
This is where that shorthand notation for derivatives comes in really handyâŠ
ð¿ð¿ððð¡ð¡ ð¢ð¢ = 1 â ð¥ð¥2 ⎠ððð¢ð¢ððð¥ð¥
= â2ð¥ð¥
⎠ðŠðŠâ² = ð¢ð¢â0.5
⎠ðððŠðŠâ²
ððð¢ð¢= â0.5ð¢ð¢â1.5
⎠ðŠðŠâ²â² =ððð¢ð¢ððð¥ð¥
â ðððŠðŠâ²
ððð¢ð¢= â2ð¥ð¥ à â0.5ð¢ð¢â1.5
⎠ððâ²â² = ððð ð âðð.ðð = ðð(ðð â ðððð)âðð.ðð
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Question
Question
Q3.23 Find ððâ²â² for ðð = ððððððâðð(ðððð)
ANSWER
3.2.4 Differentiation of Inverse Hyperbolic Functions Consider the functionâŠ
ðŠðŠ = ðŽðŽððððââ1(ð¥ð¥)
Our task here is to find ðŠðŠâ²â². We proceed in a similar manner to the previous problemâŠ
ð¥ð¥ = sinh (ðŠðŠ)
⎠ððð¥ð¥ðððŠðŠ
= cosh(ðŠðŠ) ⎠ðððŠðŠððð¥ð¥
=1
cosh (ðŠðŠ)
A useful identity isâŠ
ðððððŽðŽâ2(ðŠðŠ)â ðŽðŽððððâ2(ðŠðŠ) = 1 ⎠ðððððŽðŽâ2(ðŠðŠ) = 1 + ðŽðŽððððâ2(ðŠðŠ)
⎠ðððððŽðŽâ2(ðŠðŠ) = 1 + ð¥ð¥2 ⎠cosh(ðŠðŠ) = ï¿œ1 + ð¥ð¥2
⎠ðððððððð
= ððâ² =ðð
âðð + ðððð= (ðð + ðððð)âðð.ðð
Now to find that second derivativeâŠ
ðŠðŠâ² = (1 + ð¥ð¥2)â0.5
ð¿ð¿ððð¡ð¡ ð¢ð¢ = 1 + ð¥ð¥2 ⎠ððð¢ð¢ððð¥ð¥
= 2ð¥ð¥
⎠ðŠðŠâ² = ð¢ð¢â0.5 ⎠ðððŠðŠâ²
ððð¢ð¢= â0.5ð¢ð¢â1.5
⎠ðŠðŠâ²â² =ððð¢ð¢ððð¥ð¥
â ðððŠðŠâ²
ððð¢ð¢= 2ð¥ð¥ à â0.5ð¢ð¢â1.5 = âð¥ð¥(1 + ð¥ð¥2)â1.5
Q3.24 Find ððâ²â² for ðð = ððððððððâðð(ðððð)
ANSWER
Q3.23 â8ðð(1â4ðð2)1.5
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Worked Example 22
3.3 Further Integration
3.3.1 Integration by Parts This technique provides us with a way to integrate the product of two simple functions. We shall develop a formula to use as a framework for the technique.
The starting point for our formula lies with the Product Rule, which you mastered in Section 3.1.5âŠ
ððððð¥ð¥
(ð¢ð¢ð£ð£) = ð¢ð¢ððð£ð£ððð¥ð¥
+ ð£ð£ððð¢ð¢ððð¥ð¥
Letâs rearrange this formulaâŠ
ð¢ð¢ððð£ð£ððð¥ð¥
=ððððð¥ð¥
(ð¢ð¢ð£ð£) â ð£ð£ððð¢ð¢ððð¥ð¥
If we now integrate both sides with respect to ð¥ð¥ then we need to place a â« sign before each term and a ððð¥ð¥ at the end of each term, like soâŠ
ï¿œð¢ð¢ððð£ð£ððð¥ð¥
ððð¥ð¥ = ï¿œððððð¥ð¥
(ð¢ð¢ð£ð£)ððð¥ð¥ âï¿œð£ð£ððð¢ð¢ððð¥ð¥
ððð¥ð¥
Some simplifications can be noticed here:
In the first integral the two ððð¥ð¥ terms cancel each other out In the second integral we are taking the âintegral of the differentialâ of ð¢ð¢ð£ð£. Since integration is the
reverse of differentiation then the âintegral of a differentialâ drops away, just leaving ð¢ð¢ð£ð£ In the third integral the two ððð¥ð¥ terms cancel each other out
Performing these simplifications givesâŠ
ï¿œð ð ðððð = ð ð ðð âï¿œðð ððð ð
That is our formula for integration by parts.
What we do with the formula is to look at the left hand side and see the integral of a product. That product is composed of ð¢ð¢ and ððð£ð£. When using integration by parts you have a choice of which part of the product to call ð¢ð¢ and which to call ððð£ð£. Fortunately, your choice is guidedâŠ
The ð¢ð¢ part will become a constant after taking multiple derivatives The ððð£ð£ part is readily integrated by using standard integrals.
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Worked Example 23
Determine: â«ðð ð¬ð¬ð¬ð¬ð¬ð¬(ðð)ðððð
One part of our product is ð¥ð¥ and the other part is sin (ð¥ð¥). We need to make a choice as to which one to make ð¢ð¢ and which to make ððð£ð£. Our guidance tells us to make the ð¥ð¥ part equal to ð¢ð¢ since the derivative of ð¥ð¥ becomes 1 (i.e. a constant)âŠ
ð¿ð¿ððð¡ð¡ ð¢ð¢ = ð¥ð¥
ð¿ð¿ððð¡ð¡ ððð£ð£ = sin (ð¥ð¥)
We now try to form the terms in our integration by parts formulaâŠ
ðŒðŒðð ð¢ð¢ = ð¥ð¥ ⎠ððð¢ð¢ððð¥ð¥
= 1 ⎠ððð¢ð¢ = ððð¥ð¥
ðŒðŒðð ððð£ð£ = sin(ð¥ð¥) ⎠ð£ð£ = â cos (ð¥ð¥)
We can now writeâŠ
ï¿œð ð ðððð = ð ð ðð âï¿œðð ððð ð
ï¿œð¥ð¥ sin(ð¥ð¥)ððð¥ð¥ = ð¥ð¥(â cos(ð¥ð¥)) âï¿œ(â cos(ð¥ð¥))ððð¥ð¥
⎠ᅵð¥ð¥ sin(ð¥ð¥)ððð¥ð¥ = âð¥ð¥ cos(ð¥ð¥) + ï¿œ cos(ð¥ð¥)ððð¥ð¥
⎠ᅵððð¬ð¬ð¬ð¬ð¬ð¬(ðð)ðððð = âððððððð¬ð¬(ðð) + ð¬ð¬ð¬ð¬ð¬ð¬(ðð) + ðªðª
That was fairly straightforward. Always remember the constant of integration ð¶ð¶ at the end.
Letâs now try to integrate the product of an exponential and a cosine.
Determine: â«ðððððð ððððð¬ð¬(ðð)ðððð
Our two products here are ðð2ðð and cos (ð¥ð¥). Neither of them will reduce to a constant when differentiated so it doesnât matter which one we call ð¢ð¢ and which we call ððð£ð£. You will see here that we need to do integration by parts twice to arrive at our answer.
Just as in the previous worked example we need to decide on assignments for ð¢ð¢ and ððð£ð£, then find ð£ð£ and ððð¢ð¢. Letâs do thatâŠ
ð¿ð¿ððð¡ð¡ ð¢ð¢ = ðð2ðð ⎠ððð¢ð¢ððð¥ð¥
= 2ðð2ðð ⎠ððð¢ð¢ = 2ðð2ððððð¥ð¥
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Question
ð¿ð¿ððð¡ð¡ ððð£ð£ = cos(ð¥ð¥) ⎠ð£ð£ = sin (ð¥ð¥)
To aid our working here (and save some ink) we normally assign the capital letter ðŒðŒ to the integral in our question.
⎠ðŒðŒ = ï¿œðð2ðð cos(ð¥ð¥)ððð¥ð¥
⎠ðŒðŒ = ðð2ðð sin(ð¥ð¥) âï¿œ sin(ð¥ð¥) 2ðð2ððððð¥ð¥
This needs tidyingâŠ
⎠ðŒðŒ = ðð2ðð sin(ð¥ð¥) â 2ï¿œðð2ððsin(ð¥ð¥)ððð¥ð¥
We were rather hoping for an answer, but what we seem to have is another âintegration by partsâ problem embedded into our development. Donât worry, thatâs quite normal for this type of problem. All we do is attack that last part with âintegration by partsâ once more, using square brackets. Our desired result will then appearâŠ
ðŒðŒ = ðð2ðð sin(ð¥ð¥) â 2 ï¿œðð2ðð(âcos (ð¥ð¥)) âï¿œâcos (ð¥ð¥)2ðð2ððððð¥ð¥ï¿œ
Expanding these terms givesâŠ
ðŒðŒ = ðð2ðð sin(ð¥ð¥) + 2ðð2ðð cos(ð¥ð¥) â 4ï¿œðð2ðð cos(ð¥ð¥)ððð¥ð¥
That last integral is the same as ðŒðŒ, which is really handy, so we may now writeâŠ
ðŒðŒ = ðð2ðð sin(ð¥ð¥) + 2ðð2ðð cos(ð¥ð¥) â 4ðŒðŒ
⎠5ðŒðŒ = ðð2ðð sin(ð¥ð¥) + 2ðð2ððcos (ð¥ð¥)
If we divide both sides by 5 then we have our final answerâŠ
ð°ð° =ðððððð ð¬ð¬ð¬ð¬ð¬ð¬(ðð) + ððððððððððððð¬ð¬ (ðð)
ðð
Q3.25 Determine: â«ðððððððððððð(ðððð)ðððð
ANSWER
Q3.25 4ðð4ð¥ð¥ sin(2ðð)â2ðð4ð¥ð¥cos(2ðð)
20
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3.4 Solution of Engineering Problems with Calculus
3.4.1 Charging RC Circuit For the circuit below, use Calculus to find the charge stored in the capacitor 2 seconds after the switch closes. Assume zero charge on the capacitor before this event.
For a capacitor charging via a DC source voltage, through a series resistor, the formula for instantaneous current isâŠ
ðð =ðžðžð ð â ððï¿œâð¡ð¡ ð ð ð ð ï¿œ ï¿œ ðŽðŽð ð ðððŽðŽ
Since current is the defined as the rate of change of charge (ð ð , carried by passing electrons) we may sayâŠ
ðð =ððð ð ððð¡ð¡
If we integrate both sides with respect to ð¡ð¡ we will haveâŠ
ï¿œ ðð ððð¡ð¡ = ï¿œððð ð ððð¡ð¡
ððð¡ð¡ = ï¿œððð ð = ð ð
So we may say thatâŠ
ð ð = ï¿œ ðð ððð¡ð¡
If we wish to find the amount of charge store stored within a certain time period (i.e. 2 seconds in this example) we writeâŠ
ð ð (0â2) = ï¿œ ðð ððð¡ð¡2
0
= ï¿œðžðžð ð â ððï¿œâð¡ð¡ ð ð ð ð ï¿œ ï¿œ
2
0
ððð¡ð¡
=ðžðžð ð ï¿œððï¿œâ1ð ð ð ð ï¿œð¡ð¡
â1ð ð ð¶ð¶ï¿œ
ᅵ
0
2
=âðžðžð ð ð¶ð¶ð ð
ï¿œððï¿œâð¡ð¡ ð ð ð ð ï¿œ ᅵᅵ0
2= âðžðžð¶ð¶ ï¿œððï¿œâð¡ð¡ ð ð ð ð ï¿œ ᅵᅵ
0
2
Since we know that ð ð = 1ððΩ and ð¶ð¶ = 1ððð¹ð¹ then ð ð ð¶ð¶ = 1 à 106 à 1 à 10â6 = 106â6 = 100 = 1 we may writeâŠ
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ð ð (0â2) = âðžðžð¶ð¶ ï¿œððï¿œâð¡ð¡ 1ï¿œ ᅵᅵ0
2= âðžðžð¶ð¶[ððâð¡ð¡]02 = âðžðžð¶ð¶[ððâ2 â ððâ0] = âðžðžð¶ð¶[0.135 â 1]
We also know that the DC supply voltage (ðžðž) is 10VâŠ
⎠ðð(ððâðð) = âðððð à ðð à ððððâðð[âðð.ðððððð] = ðð.ðððð à ððððâðð ðªðªððð ð ðððððððªðªðð = ðð.ðððð ðððªðª
The plot below illustrates the integration we have just performedâŠ
Current is on the vertical axis and time on the horizontal. When we multiply current by time we get charge, which is shown as the shaded area between 0 and 2 seconds. You may like to experiment with the Graph simulator to produce similar results.
3.4.2 Energising Inductor Consider the series RL circuit belowâŠ
If ð¬ð¬ = ðððð ðœðœ, ð¹ð¹ = ðððððð and ð³ð³ = ðððððððððð, use Calculus to determine the voltage across the inductor after ðð.ðððððð. Use the following formulae in your solution developmentâŠ
ðð =ð¬ð¬ð¹ð¹ï¿œðð â ððâð¹ð¹ðð ð³ð³ï¿œ ï¿œ ðððððð ððð³ð³ = ð³ð³
ðððððððð
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We start withâŠ
ð£ð£ð¿ð¿ = ð¿ð¿ððððððð¡ð¡
= ð¿ð¿ðð ï¿œðžðžð ð ï¿œ1 â ððâð ð ð¡ð¡ ð¿ð¿ï¿œ ᅵᅵ
ððð¡ð¡
=ð¿ð¿ðžðžð ð âðð ï¿œ1 â ððâð ð ð¡ð¡ ð¿ð¿ï¿œ ï¿œ
ððð¡ð¡=ð¿ð¿ðžðžð ð ï¿œð ð ð¿ð¿â ððâð ð ð¡ð¡ ð¿ð¿ï¿œ ï¿œ = ðžðžððâð ð ð¡ð¡ ð¿ð¿ï¿œ
Putting in the values for E, R, t and L givesâŠ
ððð³ð³ = ððððððï¿œâðððð
ððÃðð.ððÃððððâðððð.ðð ï¿œ
= ðð.ðððððð ðððððððððð
Sample
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