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### College Physics (Urone)

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.

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

## Step 1: Given Data

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 $R=2.00\Omega$

## Step 2: 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 1831 by Michael Faraday, and it was mathematically characterized as Faraday's law of induction by James Clerk Maxwell.

## Step 3: Evaluating the initial energy stored in a coil(a)

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,

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

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

## Step 3: Evaluating  the time taken by the current to decline(b)

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,

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

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,

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

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

## Step.4: Evaluating the average power dissipated and the initial power dissipated(c)

The initial power can be evaluated as,

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

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,

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

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