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Q77PE

Expert-verifiedFound in: Page 863

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) Use the exact exponential treatment to find how much time is required to bring the current through an **\({\rm{80}}{\rm{.0 mH}}\)** inductor in series with a **\({\rm{15}}{\rm{.0 \Omega }}\)** resistor to **\({\rm{99}}{\rm{.0\% }}\)**of its final value, starting from zero. (b) Compare your answer to the approximate treatment using integral numbers of **\({\rm{\tau }}\)**. (c) Discuss how significant the difference is.**

(a.) The time required to bring the current is obtained as: \({\rm{24}}{\rm{.6 ms}}\).

(b.) The time is said to be between \({\rm{4}}\) and \({\rm{5}}\) time constants.

(c.) The difference is estimated to be 9%, which is greater than the projected time necessary.

**The creation of an electromotive force across an electrical conductor in a changing magnetic field is known as electromagnetic or magnetic induction. Induction was discovered in **\({\rm{1831}}\)** by Michael Faraday, and it was mathematically characterised as Faraday's law of induction by James Clerk Maxwell.**

The current in a \({\rm{RL}}\) circuit when it is being turned on is evaluated using the equation:

\(I(t) = {I_0}\left( {1 - {e^{ - \frac{t}{r}}}} \right)\)

In this case, the factor is \({\rm{0}}{\rm{.99}}\), which means that:

\(\begin{aligned} 1 - {e^{ - \frac{t}{r}}} &= 0.99\\ &= {e^{ - \frac{t}{r}}}\\ &= 0.01\end{aligned}\)

We will then have, solving, for the time will be:

\(\begin{aligned} - \frac{t}{r} &= ln(0.01)\\ &= t\\ &= - \tau In(0.01)\end{aligned}\)

Nothing that the time constant is:

\(\begin{aligned} T{\rm{ }} &= {\rm{ }}L/R\\ &= {\rm{ }}0.08/15\end{aligned}\)

Then, we will have:

\(\begin{aligned} t &= - \frac{{0.08}}{{15}} \times In(0.01)\\ &= 24.6ms\end{aligned}\)

Therefore, time required to bring the current is: \({\rm{24}}{\rm{.6 ms}}\).

(b)

The easiest way to do the comparison is to calculate the value of \({\rm{T}}\) numerically and then do the comparison. We will then have:

\(\begin{aligned} \tau &= \frac{L}{R}\\ &= \frac{{0.08}}{{15}}\\ &= 5.33\,s\end{aligned}\)

It means that the time required is between four and five times of the time constant.

Therefore, time required is between \({\rm{4}}\) and \({\rm{5}}\) time constants.

(c)

The difference is evaluated using the formula:

\(\begin{aligned} {\rm{Error }}&=\frac{{{\rm{Compared value - Original value}}}}{{{\rm{Original value}}}}\\ &= \frac{{26.65 - 24.54}}{{24.54}}\\ &= \frac{{2.11}}{{24.54}}\end{aligned}\)

The difference or the error is evaluated as:

\(\begin{aligned} {\rm{Error }} &= {\rm{ }}0.859\\ &= 0.09\\ &= 9\% \end{aligned}\)

Therefore, the difference is evaluated to be \({\rm{9\% }}\), which exceeds the calculated value of the time required.

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