The discussion revolves around the integration of an expression related to the charging of a capacitor in a voltage doubler circuit. Participants are attempting to clarify the correct form of the integral and its limits, as well as the appropriate variables to use in the context of the problem.
There is ongoing exploration of the problem with multiple interpretations being discussed. Some participants have provided guidance on how to approach the integration, while others are still seeking clarity on the correct expressions and assumptions. The conversation indicates a lack of consensus on the correct method to proceed.
Participants note that the original problem may not be fully represented in the equations provided, and there are concerns about consistency in notation and the interpretation of the equations from the referenced paper. Additionally, there is mention of homework constraints that may limit the information available for solving the problem.
Hina Gohar said:Homework Statement
it is capacitor charging expression..how to find its integration
Homework Equations
VL(t) = ∫_(T/2)^T▒〖Vme^(-T/2RC) 〗 dt
The Attempt at a Solution
result is 0.5...but how
Simon Bridge said:The expression you write inside the integral is a constant ... perhaps you mean that to be a t not a T in the exponential?
It's hard to make out, but it does not look like you have the expression for a charging capacitor there.
Are you trying to do $$V_L(t)=V_{m}\int_{\tau/2}^\tau(1-e^{-t/\tau})\;dt \qquad \tau=2RC$$
The symbols suggest you are looking for the voltage across an inductor though.
How about relate the complete problem so we know what you are trying to achieve?
... and do show us your working ;)
yes it is small part of eq(5)...2nd part I solved which is below while in 3rd part not getting how to integrate especially exponential with T instead of t..Simon Bridge said:Well, as near as I can make out, you wrote: $$V_L(t)=\int_{T/2}^T V_me^{-T/2RC}\;dt$$... I am sure that is not what you mean. I looked at the paper - and the integrand is only a small part of one case in eq(5)... which is:
$$V_L(t)=\begin{cases}V_m &: t=0\\ \\
V_me^{t/R_LC_2} &: 0<t<T/2\\ \\
V_m\left[e^{T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{T/2R_LC_2}\right)\right] &: T/2\leq t \leq T
\end{cases}$$
The case that ##T/2 \leq t \leq T## can be put in form: ##A+Be^{\lambda t}## ... please show your working. Remember it is the lower case ##t## that is being integrated against.
Don't forget to follow the instructions to substitute it into $$V_{L,ave}=\frac{1}{T}\int_{t_1}^{t_1+T} V_L(t)\;dt$$
... which you'll see is somewhat different from what you wrote.
... why would you want to? If the text suggests doing that I missed it.... in 3rd part not getting how to integrate especially exponential with T instead of t.
Simon Bridge said:... why would you want to? If the text suggests doing that I missed it.
[edit]
I managed to open your Word attachment[*]
In the attachment you wrote:
$$V_L(t)=\int_0^{T/2}V_m e^{-t/RC}\;dt$$ ... (dropping subscripts)
This equation does not appear to follow from anything in the attachment to post #3.
In that attachment, eq(5) gives you ##V_L(t)##, you then substitute eq(5) into eq(1) to get ##V_{L,ave}##.
I an guessing that this is what you intended to do.
Please be consistent in your notation.
You seem to be using eq(1) with ##t_1=0## so that:
$$V_{L,ave} = \int_0^{T/2} V_me^{-t/2R_LC_2}\;dt + \int_{T/2}^T V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right]\;dt$$ ... is that correct?
----------------------------
[*] ... please do not share information in proprietary formats.
I understand that writing equations online is annoying - the best practice is to use LaTeX.
If you must share a file - then PDF is better, since there are FOSS readers for it.
plus eq(5) I have to derive mathematically which are basically shows instantaneous load voltage expression for three different modes of this respective circuit.Mode 1 is simple in which no switch will be on so on output simple dc voltage appears..but for mode 2 T1 will on and for mode 3 T2 will on then how to find this whole expression.I tried so many timesSimon Bridge said:... why would you want to? If the text suggests doing that I missed it.
[edit]
I managed to open your Word attachment[*]
In the attachment you wrote:
$$V_L(t)=\int_0^{T/2}V_m e^{-t/RC}\;dt$$ ... (dropping subscripts)
This equation does not appear to follow from anything in the attachment to post #3.
In that attachment, eq(5) gives you ##V_L(t)##, you then substitute eq(5) into eq(1) to get ##V_{L,ave}##.
I an guessing that this is what you intended to do.
Please be consistent in your notation.
You seem to be using eq(1) with ##t_1=0## so that:
$$V_{L,ave} = \int_0^{T/2} V_me^{-t/2R_LC_2}\;dt + \int_{T/2}^T V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right]\;dt$$ ... is that correct?
----------------------------
[*] ... please do not share information in proprietary formats.
I understand that writing equations online is annoying - the best practice is to use LaTeX.
If you must share a file - then PDF is better, since there are FOSS readers for it.
Hina Gohar said:yes it is correct but I actually showed half part if that equation which I solved from 0 to T/2...while in remaining part I can not succeed.
... let's deal with only one problem at a time: you want to evaluate:Hina Gohar said:plus eq(5) I have to derive mathematically...
I also did the same thing which I showed in word file..main problem I thing is in equations writing.Simon Bridge said:... let's deal with only one problem at a time: you want to evaluate:
$$V_{L,ave} = \frac{1}{T}\int_0^{T/2} V_me^{-t/R_LC_2}\;dt + \frac{1}{T}\int_{T/2}^T V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right]\;dt \qquad\text{...(1)}$$
There seems to be a communication problem.
In your attachment you told me that: $$\int_0^{T/2} V_me^{-t/R_LC_2}\;dt=\frac{V_m(1-e^{-T/2RC})RC}{T}$$ ... but this is not correct.
The correct working is:$$\begin{align}
\frac{1}{T}\int_0^{T/2} V_me^{-t/R_LC_2}\;dt &=\left.-\frac{V_m R_LC_2}{T}e^{-t/2R_LC_2}\right|_{0}^{T/2}\\
&= \frac{V_mR_LC_2}{T}\left(1-e^{-T/2R_LC_2}\right)
\end{align}$$I suspect that this is what you meant but the trouble is that it is not what you said. In your version the "divide by T" part appears out of nowhere.
It is important to say what you mean in maths if you are not to get confused and if you don't want to confuse other people.
So much for the first integral in the expression (1).
Like I already told you, the second integral can be written in form: $$\int_{T/2}^T A+Be^{\lambda t}\; dt$$ ... I do not understand why you are having trouble with this integral.
Please find the values of A, B, and λ; and show me your working for the integration.
If you do not follow suggestions, I cannot help you.
If you do not provide needed information, I cannot help you.
Please bear in mind that it is against the rules for me to do your work for you.
... do not get confused between T and t in the expression.Problem I have in 2nd part is integration of exponential power of 'T'.This is T not t so it should be treated as constant?
Simon Bridge said:... do not get confused between T and t in the expression.
The "dt" at the end of the expression means that everything that is not a lower case "t" is treated as a constant.
So "T" is a constant, "t" is a variable.
This is how it should have worked out when you followed the suggestion:
Please find the values of A, B, and λ; and show me your working for the integration.... in post #9 which was repeated from earlier.
If you will not follow suggestions I cannot help you.
... that last line is not an error message.how to write eq(5) last expression in Latex.. I tried this but error comes
V_{L}(t)=int_0^\T/2V_{m} \mathrm{e}^{-\frac{t}{R_{L}C_{2}}}\,\mathrm{d}t
Simon Bridge said:... that last line is not an error message.
If you click the "quote" button on one of my posts, you will see how I did it.
That last line renders as:
$$V_{L}(t)=int_0^\T/2V_{m} \mathrm{e}^{-\frac{t}{R_{L}C_{2}}}\,\mathrm{d}t$$
You need a backslash before the "int" and curly-brackets around the upper limit.
But that won't produce an error.$$V_{L}(t)=\int_0^{T/2}V_{m} \mathrm{e}^{-\frac{t}{R_{L}C_{2}}}\,\mathrm{d}t$$... better. But it usually renders better if you don't use "frac" in an exponent. If you must, then use \exp{} instead of e^{}
v_L(t)=\begin{cases}
V_m &: t=0\\ \\
V_me^{-t/R_LC_2} &: 0<t<T/2\\ \\
V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right] &: T/2\leq t \leq T
\end{cases}V_{L,ave}=\frac{1}{T}\int_{t_1}^{t_1+T} v_L(t)\;\mathrm{d}tHina Gohar said:yeh now it is working..if kindly again post eq(5) so I would come to know how to write in latex
When I wrote these equations in separate lines in latex after compilation in pdf file they appear on same line then I put $$ before and after these equations to be in separate lines but it shows error.Simon Bridge said:It's in post #4 ... Oh OK then:
This is equation (5)
$$v_L(t)=\begin{cases}
V_m &: t=0\\ \\
V_me^{-t/R_LC_2} &: 0<t<T/2\\ \\
V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right] &: T/2\leq t \leq T
\end{cases}$$
Here's the markup I used:Code:v_L(t)=\begin{cases} V_m &: t=0\\ \\ V_me^{-t/R_LC_2} &: 0<t<T/2\\ \\ V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right] &: T/2\leq t \leq T \end{cases}
And this is equation (1)
$$V_{L,ave}=\frac{1}{T}\int_{t_1}^{t_1+T} v_L(t)\;\mathrm{d}t$$
... and the markup for that:
Code:V_{L,ave}=\frac{1}{T}\int_{t_1}^{t_1+T} v_L(t)\;\mathrm{d}t
Simon Bridge said:I am using pdfTeX, Version 3.1415926-1.40.10 (TeX Live 2009/Debian) and it works fine.
You can see that the above code also renders fine in these forums.
I cannot comment further without the text of the error message.