Q. 113 - Precalculus Enhanced with Graphing Utilities

Question

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 volts) is 

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

Part (a): If $E = 120 volts, R = 10 ohms, and L = 5 henrys$, how much current $l_1$ is flowing after $0.3$ second? After $0.5$ second? After $1$ second?

Part (b): What is the maximum current?

Part (c): Graph this function $I = I_1\left(t\right)$, measuring I along the y-axis and t along the x-axis.

Part (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?

Part (e): What is the maximum current?

Part (f): Graph the function $I=I_2\left(t\right)$ on the same coordinate axes as $I_1\left(t\right)$.

Step-by-Step Solution

Answer

Part (a): The current flowing after $0.3$secs will be $5.41amp$.

The current flowing after $0.5$ secs will be $7.59amp$.

The current flowing after $1$ secs will be $10.38amp$.

Part (b): The maximum current is $12amp$.

Part (c): The graph of the function $I=I_1\left(t\right)$ is given below,

Part (d): The current flowing after $0.3$ secs will be $3.34amp$.

The current flowing after $0.5$ secs will be $5.31amp$.

The current flowing after $1$ secs will be $9.44amp$.

Part (e): The maximum current is $24amp$.

Part (f): The graph of the function $I=I_2\left(t\right)$ is given below,

Part (a) Step 1. Determine the amount of current after 0 . 3 s e c s .

Consider the given question,

$$\begin{gathered} I=\frac{E}{R}[1-e^{-\left(R/L\right)t}], \\ E = 120 volts, \\ R = 10 ohms, \\ L = 5 henrys, \\ t=0.3secs \end{gathered}$$

Substitute the values in the function,

$$\begin{gathered} I_1=\frac{120}{10}[1-e^{-\left(10/5\right)\times 0.3}] \\ =12[1-e^{-\frac{3}{5}}] \\ =12\left(1-0.549\right) \\ \approx 5.41 \end{gathered}$$

Part (a) Step 2. Determine the amount of current after 0 . 5 s e c s .

Consider the given question,

$$\begin{gathered} I=\frac{E}{R}[1-e^{-\left(R/L\right)t}], \\ E = 120 volts, \\ R = 10 ohms, \\ L = 5 henrys, \\ t=0.5secs \end{gathered}$$

Substitute the values in the function,

$$\begin{gathered} I_1=\frac{120}{10}[1-e^{-\left(10/5\right)\times 0.5}] \\ =12[1-e^{-\frac{5}{5}}] \\ =12\left(1-0.3679\right) \\ \approx 7.59 \end{gathered}$$

Part (a) Step 3. Determine the amount of current after 1 s e c s .

Consider the given question,

$$\begin{gathered} I=\frac{E}{R}[1-e^{-\left(R/L\right)t}], \\ E = 120 volts, \\ R = 10 ohms, \\ L = 5 henrys, \\ t=1secs \end{gathered}$$

Substitute the values in the function,

$$\begin{gathered} I_1=\frac{120}{10}[1-e^{-\left(10/5\right)\times 1}] \\ =12[1-e^{-\frac{10}{5}}] \\ =12\left(1-0.1353\right) \\ \approx 10.38 \end{gathered}$$

Part (b) Step 1. Determine the maximum current.

Consider the given question,

$$\begin{gathered} E=120, \\ R=10 \end{gathered}$$

Maximum current is given below,

$I=\frac{E}{R}$

Substitute the values in the equation,

$$\begin{gathered} I=\frac{120}{10} \\ =12 \end{gathered}$$

Part (c) Step 1. Plot the function.

Consider the given question,

$\left(0.3,5.41\right),\left(0.5,7.59\right),\left(1,10.38\right)$

Plot the points on the graph,

Part (d) Step 1. Determine the amount of current after 0 . 3 s e c s .

Consider the given question,

$$\begin{gathered} I=\frac{E}{R}[1-e^{-\left(R/L\right)t}], \\ E = 120 volts, \\ R = 5 ohms, \\ L = 10 henrys, \\ t=0.3secs \end{gathered}$$

Substitute the values in the function,

$$\begin{gathered} I_2=\frac{120}{5}[1-e^{-\left(5/10\right)\times 0.3}] \\ =24[1-e^{-\frac{1.5}{5}}] \\ =24\left(1-0.8607\right) \\ \approx 3.34 \end{gathered}$$

Part (d) Step 2. Determine the amount of current after 0 . 5 s e c s .

Consider the given question,

$$\begin{gathered} I=\frac{E}{R}[1-e^{-\left(R/L\right)t}], \\ E = 120 volts, \\ R = 5 ohms, \\ L = 10 henrys, \\ t=0.5secs \end{gathered}$$

Substitute the values in the function,

$$\begin{gathered} I_2=\frac{120}{5}[1-e^{-\left(5/10\right)\times 0.5}] \\ =24[1-e^{-\frac{2.5}{10}}] \\ =24\left(1-0.7788\right) \\ \approx 5.31 \end{gathered}$$

Part (d) Step 3. Determine the amount of current after 1 s e c s .

Consider the given question,

$$\begin{gathered} I=\frac{E}{R}[1-e^{-\left(R/L\right)t}], \\ E = 120 volts, \\ R = 5 ohms, \\ L = 10 henrys, \\ t=1secs \end{gathered}$$

Substitute the values in the function,

$$\begin{gathered} I_2=\frac{120}{5}[1-e^{-\left(5/10\right)\times 1}] \\ =24[1-e^{-0.5}] \\ =24\left(1-0.6065\right) \\ \approx 9.44 \end{gathered}$$

Part (e) Step 1. Determine the maximum current.

Consider the given question,

$$\begin{gathered} E=120, \\ R=5 \end{gathered}$$

Maximum current is given below,

$I=\frac{E}{R}$

Substitute the values in the equation,

$$\begin{gathered} I=\frac{120}{5} \\ =24 \end{gathered}$$

Part (f) Step 1. Plot the function.

Consider the given question,

$\left(0.3,3.34\right),\left(0.5,5.31\right),\left(1,9.44\right)$

Plot the points on the graph,