Chapter 5.6, Problem 7E

a.

To determine

To find: the value of $\frac{d\text{V}}{dt}$ .

Answer to Problem 7E

The value of $\frac{d\text{V}}{dt}$ is $\text{1 volt/sec}$ .

Explanation of Solution

Given information:

  $\text{V}=\text{IR}$ ; V = volt, R = resistance, I = current.

V is increasing at the rate of $1\text{ volt/sec}$ while I is decreasing at the rate of $\frac{1}{3}\text{ ampere/sec}$ .

Calculation :

We have to find the value of $\frac{d\text{V}}{dt}$ .

Since, V is increasing at the rate of $1\text{ volt/sec}$

So,

  $\frac{d\text{V}}{dt}=1\text{ volt/sec}$ .

Hence, the value $\frac{d\text{V}}{dt}$ is $1\text{ volt/sec}$ .

b.

To determine

To find: the value of $\frac{d\text{I}}{dt}$ .

Answer to Problem 7E

The value of $\frac{d\text{I}}{dt}$ is $\frac{1}{3}\text{ ampere/sec}$ .

Explanation of Solution

Given information:

  $\text{V}=\text{IR}$ ; V = volt, R = resistance, I = current.

V is increasing at the rate of $1\text{ volt/sec}$ while I is decreasing at the rate of $\frac{1}{3}\text{ ampere/sec}$ .

Calculation :

Let $t$ denote time in seconds.

We have to find the value of $\frac{d\text{I}}{dt}$ .

Since I is decreasing at the rate of $\frac{1}{3}\text{ ampere/sec}$ ,

So,

  $\frac{d\text{I}}{dt}=-\frac{1}{3}\text{ ampere/sec}$ .

Hence, the value $\frac{d\text{I}}{dt}$ is $\frac{1}{3}\text{ ampere/sec}$ .

c.

To determine

To relate: $\frac{d\text{R}}{dt}$ to $\frac{d\text{V}}{dt}$ and $\frac{d\text{I}}{dt}$ .

Answer to Problem 7E

The value of $\frac{d\text{R}}{dt}$ is $\frac{1}{\text{I}}\left[\frac{d\text{V}}{dt}-\text{R}\frac{d\text{I}}{dt}\right]$ .

Explanation of Solution

Given information:

  $\text{V}=\text{IR}$

All variables are differentiable function of $t$ .

Calculation :

We have to relate $\frac{d\text{R}}{dt}$ to $\frac{d\text{V}}{dt}$ and $\frac{d\text{I}}{dt}$ .

Since all variables are differentiable function of $t$

Therefore,

  $\text{V}=\text{IR}$

Differentiate with respect to t .

  $\begin{array}{l}\frac{d\text{V}}{dt}=\text{I}\frac{d\text{R}}{dt}\text{+R}\frac{d\text{I}}{dt}\left[\text{Use product rule}\text{.}\right]\\ \text{I}\frac{d\text{R}}{dt}=\frac{d\text{V}}{dt}-\text{R}\frac{d\text{I}}{dt}\\ \frac{d\text{R}}{dt}=\frac{1}{\text{I}}\left[\frac{d\text{V}}{dt}-\text{R}\frac{d\text{I}}{dt}\right]\end{array}$

Hence, the value $\frac{d\text{R}}{dt}$ is $\frac{1}{\text{I}}\left[\frac{d\text{V}}{dt}-\text{R}\frac{d\text{I}}{dt}\right]$ .

d.

To determine

To find: R is increasing or decreasing.

Answer to Problem 7E

The value of $\frac{d\text{R}}{dt}$ is $\frac{3}{2}$ that is increasing.

Explanation of Solution

Given information:

  $\text{V}=\text{IR}$ ; V = 12 volts, I = 2 ampere.

  $\frac{dv}{dt}$ is 1 volt, $\frac{d\text{I}}{dt}$ is $\frac{1}{3}$ .

All variables are differentiable function of $t$ .

Calculation :

We have to find that R is increasing or decreasing.

Since all variables are differentiable function of $t$

Therefore,

  $\begin{array}{l}\text{V}=\text{IR}\\ \text{R}=\frac{\text{V}}{\text{I}}=\frac{12}{2}=6\left[\begin{array}{l}\text{V}=12\text{ volt}\text{.}\\ \text{I}=2\text{ ampere}\text{.}\end{array}\right]\end{array}$

So,

  $\text{V}=\text{IR}$

Differentiate with respect to t .

  $\begin{array}{l}\frac{d\text{V}}{dt}=\text{I}\frac{d\text{R}}{dt}\text{+R}\frac{d\text{I}}{dt}\left[\text{Use product rule}\text{.}\right]\\ \text{I}\frac{d\text{R}}{dt}=\frac{d\text{V}}{dt}-\text{R}\frac{d\text{I}}{dt}\\ \frac{d\text{R}}{dt}=\frac{1}{\text{I}}\left[\frac{d\text{V}}{dt}-\text{R}\frac{d\text{I}}{dt}\right]\end{array}$

Putting the value of $\text{I,}\frac{d\text{V}}{dt},\text{R and}\frac{d\text{I}}{dt}$ in above equation,

  $\begin{array}{l}\frac{d\text{R}}{dt}=\frac{1}{2}\left[1-6\left(-\frac{1}{3}\right)\right]\\ \frac{d\text{R}}{dt}=\frac{1}{2}\left[1+\frac{6}{3}\right]\\ \frac{d\text{R}}{dt}=\frac{1}{2}\left[1+2\right]\\ \frac{d\text{R}}{dt}=\frac{1}{2}\left[3\right]\\ \frac{d\text{R}}{dt}=\frac{3}{2}\text{ohm/sec}\text{.}\end{array}$

Hence, the value $\frac{d\text{R}}{dt}$ is $\frac{3}{2}$ that is increasing.