Chapter 5.3, Problem 113AYU

a.

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

Solution:

(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:

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

Calculation:

  

To determine

To find:

(b) The maximum current.

Answer to Problem 113AYU

Solution:

(b) $12$ maximum current when time approaches infinity.

Explanation of Solution

Given:

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

Calculation:

(b) We obtain maximum current when time approaches infinity

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

  $= 12\left[1-e^{-\infty }\right]$

  $= 12\left[1-0\right]$

  $=12$

b.

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

Solution:

(c)

  

Explanation of Solution

Given:

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

Calculation:

(c)

  

c.

To determine

To find: $I_2$ 9.4433$$ 0.5$$ I_2$$ 0.3$$ 3.3430$$

(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

Solution:

(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:

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

Calculation:

  

d.

To determine

To find:

(e) The maximum current.

Answer to Problem 113AYU

Solution:

(e) $24$ maximum current when time approaches infinity.

Explanation of Solution

Given:

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

Calculation:

(e) We obtain maximum current when time approaches infinity

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

  $= 24\left[1-e^{-\infty }\right]$

  $= 24\left[1-0\right]$

  $=24$

e.

To determine

To find:

(f) graph the function

  $I= I_2\left(t\right),$ measuring $I$ along the y -axis and t along the x -axis.

Answer to Problem 113AYU

Solution:

(f)

  

Explanation of Solution

Given:

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

Calculation:

(f)