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Q76PE

Expert-verifiedFound in: Page 863

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**The 5.00 A current through a 1.50 H inductor is dissipated by a \({\rm{2}}{\rm{.00 \Omega }}\) resistor in a circuit like that in Figure 23.44 with the switch in position 2 . (a) What is the initial energy in the inductor? (b) How long will it take the current to decline to 5.00% of its initial value? (c) Calculate the average power dissipated, and compare it with the initial power dissipated by the resistor.**

- The initial energy stored in a coil is obtained as: 18.75 J.
- Time is taken by the current to decline is obtained as: 2.25 s.
- The average power dissipated is obtained as: 7.92 W and the initial power dissipated by the resistor is obtained as: 50 W.

The initial current in the circuit is I = 5.00 A

The self-inductance in the circuit is L = 1.50 H

The resistance of the circuit is $R=2.00\Omega $

**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 **1831** by Michael Faraday, and it was mathematically characterized as Faraday's law of induction by James Clerk Maxwell.**

The initial energy stored in the coil can be evaluated using the formula,

$E=\frac{L{I}^{2}}{2}$……………..(1)

Where L and I are the self-inductance and current of the circuit, respectively.

Substituting the given data in the equation (1), we get,

$E=\frac{\left(1.50H\right)\times {\left(5.00A\right)}^{2}}{2}\phantom{\rule{0ex}{0ex}}=18.75J$

Therefore, the initial energy stored in a coil is 18.75 J .

The current when a circuit is turned off can be evaluated using the formula,

$I={I}_{0}{e}^{-\left(\frac{T}{\tau}\right)}$ …………….(2)

It simply means that when we have a ratio to reach, in our case 0.05 , then we can simply write the time using equation (2) such that,

${e}^{-\left(\frac{t}{\tau}\right)}=\frac{I}{{I}_{0}}\phantom{\rule{0ex}{0ex}}{e}^{-\left(\frac{t}{\tau}\right)}=0.05\phantom{\rule{0ex}{0ex}}\frac{t}{\tau}=-\left(In\left(0.05\right)\right)\phantom{\rule{0ex}{0ex}}t=\tau \times 2.996$

We can calculate the value of the time constant such that,

$\tau =\frac{L}{R}\phantom{\rule{0ex}{0ex}}=\frac{1.50H}{2.00\Omega}\phantom{\rule{0ex}{0ex}}=0.75s$

Therefore we can calculate the value of time such that using the above expressions,

$t=\left(0.75s\right)\times 2.996\phantom{\rule{0ex}{0ex}}=2.25s$

Therefore, the time taken by the current to decline is 2.25 s .

The initial power can be evaluated as,

$P={I}^{2}R\phantom{\rule{0ex}{0ex}}={\left(5.00A\right)}^{2}\times \left(2.00\Omega \right)\phantom{\rule{0ex}{0ex}}=50W$

The average power for this case is that considering, when the power was reduced to 5% in the time of the value t , we will obtain,

${P}_{Average}=\frac{0.95E}{t}\phantom{\rule{0ex}{0ex}}=\frac{0.95\times 18.75J}{2.25s}\phantom{\rule{0ex}{0ex}}=7.92W$

Therefore, the average power dissipated is 7.92 W and the initial power dissipated by the resistor is 50 W .

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