To determine To find: a. If E = 120 volts, R = 10 ohms and L = 5 henrys, how much current I 1 is flowing after 0.3 second? After 0.5 second? After 1 second? Answer a. 5.4143 current I 1 is flowing after 0.3 second, 7.5855 current I 1 is flowing after 0.5 second, 10.3760 current I 1 is flowing after 1 second. Explanation Given: I = E R [ 1 − e − ( R L ) t ] Calculation: a. E = 120 , R = 10 and L = 5 How much current I 1 is flowing after 0.3 second = 120 10 [ 1 − e − ( 10 5 ) 0.3 ] = 12 [ 1 − 0.5488 ] = 12 [ 0.4512 ] = 5.4143 How much current I 1 is flowing after 0.5 second = 120 10 [ 1 − e − ( 10 5 ) 0.5 ] = 12 [ 1 − 0.3679 ] = 12 [ 0.6321 ] = 7.5855 How much current I 1 is flowing after 1 second = 120 10 [ 1 − e − ( 10 5 ) 1 ] = 12 [ 1 − 0.1353 ] = 12 [ 0.8647 ] = 10.3760 To determine To find: b. The maximum current. Answer b. 12 maximum current when time approaches infinity. Explanation Given: I = E R [ 1 − e − ( R L ) t ] Calculation: b. We obtain maximum current when time approaches infinity = 120 10 [ 1 − e − ( 10 5 ) ∞ ] = 12 [ 1 − e − ∞ ] = 12 [ 1 − 0 ] = 12 To determine To find: c. Graph the function I = I 1 ( t ) , measuring I along the y -axis and t along the x -axis . Answer c. Explanation Given: I = E R [ 1 − e − ( R L ) t ] Calculation: c. To determine To find: d. If E = 120 volts, R = 5 ohms, and L = 10 henrys, how much current I 2 is flowing after 0.3 second? After 0.5 second? After 1 second? Answer d. 3.3430 current I 2 is flowing after 0.3 second, 5.3088 current I 2 is flowing after 0.5 second, 9.4433 current I 2 is flowing after 1 second, Explanation Given: I = E R [ 1 − e − ( R L ) t ] Calculation: d. E = 120 , R = 5 and L = 10 How much current I 2 is flowing after 0.3 second = 120 5 [ 1 − e − ( 5 10 ) 0.3 ] = 24 [ 1 − 0.8607 ] = 24 [ 0.1393 ] = 3.3430 How much current I 2 is flowing after 0.5 second = 120 5 [ 1 − e − ( 5 10 ) 0.5 ] = 24 [ 1 − 0.7788 ] = 2 [ 0.2212 ] = 5.3088 How much current I 2 is flowing after 1 second = 120 5 [ 1 − e − ( 5 10 ) 1 ] = 24 [ 1 − 0.6065 ] = 24 [ 0.3935 ] = 9.4433 To determine To find: e. The maximum current. Answer e. 24 maximum current when time approaches infinity. Explanation Given: I = E R [ 1 − e − ( R L ) t ] Calculation: e. We obtain maximum current when time approaches infinity. = 120 5 [ 1 − e − ( 5 10 ) ∞ ] = 24 [ 1 − e − ∞ ] = 24 [ 1 − 0 ] = 24 To determine To find: f. Graph the function I = I 2 ( t ) , measuring I along the y -axis and t along the x -axis . Answer f. Explanation Given: I = E R [ 1 − e − ( R L ) t ] Calculation: f.

9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Textbook Question

Chapter 5.3, Problem 113AYU

119. Current in an RL Circuit The equation governing the amount of current I (in amperes) after time t (in seconds) in a single RL circuit consisting of a resistance R (in ohms), an inductance L (in henrys), and an electromotive force E (in voles) is

$I = \frac{E}{R} [1 - e^{- (R / L) t}]$

Chapter 5.3, Problem 113AYU, 119. Current in an RL Circuit The equation governing the amount of current I (in amperes) after time

(a) If $E = 120$ volts, $R = 10$ ohms, and $L = 5$ henrys, how much current I₁ is flowing after 0.3 second? After 0.5 second? After 1 second?

(b) What is the maximum current?

(d) If $E = 120$ volts, $R = 5$ ohms, and $L = 10$ henrys, how much current I₂ is flowing after 0.3 second? After 0.5 second? After 1 second?

(e) What is the maximum current?

(f) Graph the function $I = I_{2} (t)$ on the same coordinate axes as $I_{1} (t)$ .

Expert Solution

To determine

To find:

a. If $E = 120$ volts, $R = 10$ ohms and $L = 5$ henrys, how much current $I_{1}$ is flowing after $0.3$ second? After $0.5$ second? After 1 second?

Answer to Problem 113AYU

a. $5.4143$ current $I_{1}$ is flowing after $0.3$ second, $7.5855$ current $I_{1}$ is flowing after $0.5$ second, $10.3760$ current $I_{1}$ is flowing after 1 second.

Explanation of Solution

Given:

$I = \frac{E}{R} [1 - e^{- (\frac{R}{L}) t}]$

Calculation:

a. $E = 120, R = 10$ and $L = 5$

How much current $I_{1}$ is flowing after $0.3$ second $= \frac{120}{10} [1 - e^{- (\frac{10}{5}) 0.3}]$

$= 12 [1 - 0.5488]$

$= 12 [0.4512]$

$= 5.4143$

How much current $I_{1}$ is flowing after $0.5$ second $= \frac{120}{10} [1 - e^{- (\frac{10}{5}) 0.5}]$

$= 12 [1 - 0.3679]$

$= 12 [0.6321]$

$= 7.5855$

How much current $I_{1}$ is flowing after 1 second $= \frac{120}{10} [1 - e^{- (\frac{10}{5}) 1}]$

$= 12 [1 - 0.1353]$

$= 12 [0.8647]$

$= 10.3760$

Expert Solution

To determine

To find:

b. The maximum current.

Answer to Problem 113AYU

b. 12 maximum current when time approaches infinity.

Explanation of Solution

Given:

$I = \frac{E}{R} [1 - e^{- (\frac{R}{L}) t}]$

Calculation:

b. We obtain maximum current when time approaches infinity

$= \frac{120}{10} [1 - e^{- (\frac{10}{5}) \infty}]$

$= 12 [1 - e^{- \infty}]$

$= 12 [1 - 0]$

$= 12$

Expert Solution

To determine

To find:

c. Graph the function $I = I_{1} (t),$ measuring $I$ along the $y -axis$ and $t$ along the $x -axis$ .

Answer to Problem 113AYU

Precalculus, Chapter 5.3, Problem 113AYU , additional homework tip 1

Explanation of Solution

Given:

$I = \frac{E}{R} [1 - e^{- (\frac{R}{L}) t}]$

Calculation:

Precalculus, Chapter 5.3, Problem 113AYU , additional homework tip 2

Expert Solution

To determine

To find:

d. If $E = 120$ volts, $R = 5$ ohms, and $L = 10$ henrys, how much current $I_{2}$ is flowing after $0.3$ second? After $0.5$ second? After 1 second?

Answer to Problem 113AYU

d. $3.3430$ current $I_{2}$ is flowing after $0.3$ second, $5.3088$ current $I_{2}$ is flowing after $0.5$ second, $9.4433$ current $I_{2}$ is flowing after 1 second,

Explanation of Solution

Given:

$I = \frac{E}{R} [1 - e^{- (\frac{R}{L}) t}]$

Calculation:

d. $E = 120, R = 5$ and $L = 10$

How much current $I_{2}$ is flowing after $0.3$ second $= \frac{120}{5} [1 - e^{- (\frac{5}{10}) 0.3}]$

$= 24 [1 - 0.8607]$

$= 24 [0.1393]$

$= 3.3430$

How much current $I_{2}$ is flowing after $0.5$ second $= \frac{120}{5} [1 - e^{- (\frac{5}{10}) 0.5}]$

$= 24 [1 - 0.7788]$

$= 2 [0.2212]$

$= 5.3088$

How much current $I_{2}$ is flowing after 1 second $= \frac{120}{5} [1 - e^{- (\frac{5}{10}) 1}]$

$= 24 [1 - 0.6065]$

$= 24 [0.3935]$

$= 9.4433$

Expert Solution

To determine

To find:

e. The maximum current.

Answer to Problem 113AYU

e. 24 maximum current when time approaches infinity.

Explanation of Solution

Given:

$I = \frac{E}{R} [1 - e^{- (\frac{R}{L}) t}]$

Calculation:

e. We obtain maximum current when time approaches infinity.

$= \frac{120}{5} [1 - e^{- (\frac{5}{10}) \infty}]$

$= 24 [1 - e^{- \infty}]$

$= 24 [1 - 0]$

$= 24$

Expert Solution

To determine

To find:

f. Graph the function $I = I_{2} (t),$ measuring $I$ along the $y -axis$ and $t$ along the $x -axis$ .

Answer to Problem 113AYU

Precalculus, Chapter 5.3, Problem 113AYU , additional homework tip 3

Explanation of Solution

Given:

$I = \frac{E}{R} [1 - e^{- (\frac{R}{L}) t}]$

Calculation:

Precalculus, Chapter 5.3, Problem 113AYU , additional homework tip 4

Chapter 5.3, Problem 112AYU

Chapter 5.3, Problem 114AYU

Chapter 5 Solutions

Precalculus

Ch. 5.1 - Find f( 3 ) if f( x )=4 x 2 +5x . (pp. 60-62)Ch. 5.1 - Find f(3x) if f(x)=42 x 2 . (pp. 60-62)Ch. 5.1 - Find the domain of the function f(x)= x 2 1 x 2 25...Ch. 5.1 - Prob. 4AYU Ch. 5.1 - Prob. 5AYU Ch. 5.1 - Prob. 6AYU Ch. 5.1 - In Problems 9 and 10, evaluate each expression...Ch. 5.1 - In Problems 9 and 10, evaluate each expression...Ch. 5.1 - In Problems 11 and 12, evaluate each expression...Ch. 5.1 - In Problems 11 and 12, evaluate each expression...

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