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

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

(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?

(c) Graph this function $I = I_1$ (t), measuring I along the $y\text{-axis}$ and t along the $x\text{-axis}$.

(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 $I = I_2\left(t\right)$ on the same coordinate axes as $I_1(t)$.

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:

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

Calculation:

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

How much current $I_1$ is flowing after $0.3$ second $= \frac{120}{10}\left[1 - e^{-\left(\frac{10}{5}\right)0.3}\right]$

$= 12\left[1 - 0.5488\right]$

$= 12\left[0.4512\right]$

$= 5.4143$

How much current $I_1$ is flowing after $0.5$ second $= \frac{120}{10}\left[1 - e^{-\left(\frac{10}{5}\right)0.5}\right]$

$= 12\left[1 - 0.3679\right]$

$= 12\left[0.6321\right]$

$= 7.5855$

How much current $I_1$ is flowing after 1 second $= \frac{120}{10}\left[1 - e^{-\left(\frac{10}{5}\right)1}\right]$

$= 12\left[1 - 0.1353\right]$

$= 12\left[0.8647\right]$

$= 10.3760$

To determine

To find:

b. The maximum current.

Answer to Problem 113AYU

b. 12 maximum current when time approaches infinity.

Explanation of Solution

Given:

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

Calculation:

b. We obtain maximum current when time approaches infinity

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

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

$= 12\left[1 - 0\right]$

$= 12$

To determine

To find:

c. Graph the function $I = I_1\left(t\right),$ measuring $I$ along the $y\text{-axis}$ and $t$ along the $x\text{-axis}$.

Answer to Problem 113AYU

c.

Explanation of Solution

Given:

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

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 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:

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

Calculation:

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

How much current $I_2$ is flowing after $0.3$ second $= \frac{120}{5}\left[1 - e^{-\left(\frac{5}{10}\right)0.3}\right]$

$= 24\left[1 - 0.8607\right]$

$= 24\left[0.1393\right]$

$= 3.3430$

How much current $I_2$ is flowing after $0.5$ second $= \frac{120}{5}\left[1 - e^{-\left(\frac{5}{10}\right)0.5}\right]$

$= 24\left[1 - 0.7788\right]$

$= 2\left[0.2212\right]$

$= 5.3088$

How much current $I_2$ is flowing after 1 second $= \frac{120}{5}\left[1 - e^{-\left(\frac{5}{10}\right)1}\right]$

$= 24\left[1 - 0.6065\right]$

$= 24\left[0.3935\right]$

$= 9.4433$

To determine

To find:

e. The maximum current.

Answer to Problem 113AYU

e. 24 maximum current when time approaches infinity.

Explanation of Solution

Given:

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

Calculation:

e. We obtain maximum current when time approaches infinity.

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

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

$= 24\left[1 - 0\right]$

$= 24$

To determine

To find:

f. Graph the function $I= I_2\left(t\right),$ measuring $I$ along the $y\text{-axis}$ and $t$ along the $x\text{-axis}$.

Answer to Problem 113AYU

f.

Explanation of Solution

Given:

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

Calculation:

f.