Approximate the voltage with Kirchoff's law

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SUMMARY

The discussion centers on applying Kirchhoff's first law, represented by the equation E(t) = L * (di/dt) + R*i, to approximate voltage values at specific time intervals. Given the inductance L of 0.98 henries and resistance R of 0.142 ohms, the calculated voltage E(t) values for t = 1.00 to 1.04 seconds are 2.400, 2.403, 3.386, 5.352, and 7.320 volts respectively. The method used involves calculating di/dt using the slope formula and applying the three-point formula for better accuracy. The user seeks clarification on the calculation of di/dt at t = 1.04 and how to express di/dt in terms of t.

PREREQUISITES
  • Understanding of Kirchhoff's laws, specifically Kirchhoff's first law.
  • Familiarity with calculus concepts, particularly derivatives and slopes.
  • Knowledge of electrical components, including inductance and resistance.
  • Experience with numerical methods, such as the three-point formula for approximating derivatives.
NEXT STEPS
  • Study the application of Kirchhoff's laws in circuit analysis.
  • Learn how to derive expressions for di/dt in terms of time for dynamic systems.
  • Explore numerical methods for derivative approximation, including the three-point and n-point formulas.
  • Investigate the relationship between current, voltage, and resistance in electrical circuits.
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Students in electrical engineering, physics enthusiasts, and anyone looking to deepen their understanding of circuit analysis and numerical methods for approximating derivatives.

Ryuuken
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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.

<br /> \begin{tabular}{|c|c|c|c|c|c|}<br /> \hline<br /> $\emph{t}$ &amp; 1.00 &amp; 1.01 &amp; 1.02 &amp; 1.03 &amp; 1.04\\<br /> \hline<br /> $\emph{i}$ &amp; 3.10 &amp; 3.12 &amp; 3.14 &amp; 3.18 &amp; 3.24\\<br /> \hline<br /> \end{tabular}<br />

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: \frac{1}{2h}[f(x_{0}+h) - f(x_{1}-h)]

The Attempt at a Solution


The official solution is:

<br /> \begin{tabular}{|c|c|c|c|c|c|}<br /> \hline<br /> $\emph{t}$ &amp; 1.00 &amp; 1.01 &amp; 1.02 &amp; 1.03 &amp; 1.04\\<br /> \hline<br /> $\emph{E(t)}$ &amp; 2.400 &amp; 2.403 &amp; 3.386 &amp; 5.352 &amp; 7.320\\<br /> \hline<br /> \end{tabular}<br />

Using 0.98*di/dt+0.142*i and di/dt = (y_{1} - y_{0})/(x_{1} - x_{0})
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 \frac{1}{2h}[f(x_{0}+h) - f(x_{1}-h)]. Substitute f(x) for E(t) to get \frac{1}{2h}[E(t_{0}+h) - E(t_{1}-h)].

Thanks.
 
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Why do you give the "three point formula" when the problem says "using di/dt = (y_1-y_0)/(x_1- x_0)"?
 


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 = (y_1-y_0)/(x_1- x_0) works.

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

Thanks.
 

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