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Approximate the voltage with Kirchoff's law

  • Thread starter Ryuuken
  • Start date
  • #1
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Homework Statement


Kirchoff's first law gives the relationship E(t) = L * (di/dt) + R*i where L is the inductance, R is the resistance and i is the current.

[tex]
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$\emph{t}$ & 1.00 & 1.01 & 1.02 & 1.03 & 1.04\\
\hline
$\emph{i}$ & 3.10 & 3.12 & 3.14 & 3.18 & 3.24\\
\hline
\end{tabular}
[/tex]

Suppose t is measured in seconds, i is in amperes, the inductance L is a constant 0.98 henries and R is 0.142 ohms. Approximate the voltage E(t) when t = 1.00, 1.01, 1.02, 1.03, 1.04.

Homework Equations



Three-point formula: [tex]\frac{1}{2h}[f(x_{0}+h) - f(x_{1}-h)][/tex]

The Attempt at a Solution


The official solution is:

[tex]
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$\emph{t}$ & 1.00 & 1.01 & 1.02 & 1.03 & 1.04\\
\hline
$\emph{E(t)}$ & 2.400 & 2.403 & 3.386 & 5.352 & 7.320\\
\hline
\end{tabular}
[/tex]

Using 0.98*di/dt+0.142*i and di/dt = [tex](y_{1} - y_{0})/(x_{1} - x_{0})[/tex]
di/dt = 2, i = 3.10 I get 2.400.
di/dt = (2 + 2) / 2, i = 3.12 I get 2.403.
di/dt = (2 + 4) / 2, i = 3.14 I get 3.386.
di/dt = (4 + 6) / 2, i = 3.18 I get 5.352.

Now the last one is di/dt = 7 but how do I get that 7? For the other ones, I took the slope from both sides and divided by 2. The only way I can think of is to add all the previous slopes together and divide by 2, (2 + 2 + 4 + 6) / 2 = 7 but I'm not sure if that makes sense or why that would work.

Also, is there a way get some expression in terms of t for di/dt. So I can use a three-point formula like [tex]\frac{1}{2h}[f(x_{0}+h) - f(x_{1}-h)][/tex]. Substitute f(x) for E(t) to get [tex]\frac{1}{2h}[E(t_{0}+h) - E(t_{1}-h)][/tex].

Thanks.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Why do you give the "three point formula" when the problem says "using di/dt = [itex](y_1-y_0)/(x_1- x_0)[/itex]"?
 
  • #3
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Well my book says the results were given by the three-point formula (or some n-point formula) so I'm trying to figure out how to apply it to get the same result although using di/dt = [itex](y_1-y_0)/(x_1- x_0)[/itex] works.

Do you know if the way I got di/dt = 7 is correct? If so, why?

Thanks.
 

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