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Expert-verified Found in: Page 863 ### College Physics (Urone)

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.

See the step by step solution

## Step 1: Define Electromagnetic Induction

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.

## Step 2 : Evaluating the time required to bring the current

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: $${\rm{24}}{\rm{.6 ms}}$$.

## Step 3 : Evaluating the time constants

(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.

## Step 4 : Calculating the error

(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. ### Want to see more solutions like these? 