Chapter 4.6, Problem 7E

$\text{(a)}$

To determine

To find: The value of $dV/dt$ .

Answer to Problem 7E

The answer: $\frac{dV}{dt}=1\text{ volt }/\text{sec}$

Explanation of Solution

Given information:

The equation is $V=IR$ .

Calculation:

  $V$ is increasing at $1\text{ volt/sec}$ .

Therefore, the required value of $\frac{dV}{dt}=1\text{ volt }/\text{sec}$ .

$\text{(b)}$

To determine

To find: The value of $dI/dr$ .

Answer to Problem 7E

The answer: $\frac{dI}{dt}=-1/3\text{amp}/\text{sec}$

Explanation of Solution

Given information:

The equation is $V=IR$ .

Calculation:

  $I$ is decreasing at $1/3\text{amp}/\text{sec}$ .

Therefore, the required value of $\frac{dI}{dt}=-1/3\text{amp}/\text{sec}$ .

$(c)$

To determine

To find: The equation that relates $dR/dt$ to $dV/dt$ and $dI/dt$ .

Answer to Problem 7E

The answer: $\frac{dR}{dt}=\frac{1}{I}\left(\frac{dV}{dt}-R\frac{dI}{dt}\right)$

Explanation of Solution

Given information:

The equation is $V=IR$ .

And the figure is:

  

Calculation:

  $\begin{array}{c}V=IR\\ \to \frac{dV}{dt}=I\frac{dR}{dt}+R\frac{dI}{dt}\\ \to I\frac{dR}{dt}=\frac{dV}{dt}-R\frac{dI}{dt}\\ \to \frac{dR}{dt}=\frac{1}{I}\left(\frac{dV}{dt}-R\frac{dI}{dt}\right)\end{array}$

Therefore, the required equation that relates $dR/dt$ to $dV/dt$ and $dI/dt$ is:

  $\frac{dR}{dt}=\frac{1}{I}\left(\frac{dV}{dt}-R\frac{dI}{dt}\right)$

$(d)$

To determine

To find: The value of $R$ and it is increasing or decreasing.

Answer to Problem 7E

The answer: $\frac{dR}{dt}=\frac{3}{2}\text{ ohms }/\text{sec}$

And $R$ is increasing.

Explanation of Solution

Given information:

The equation is $V=IR$ .

And the figure is:

  

  $R$ is changing when

  $V=12\text{ volts}$

And

  $I=2\text{ amps}$

Calculation:

From part $\text{(c)}$ , the answer is positive, so $R$ is increasing.

  $\begin{array}{c}\frac{dV}{dt}=1\\ \frac{dI}{dt}=-1/3\\ V=12\\ I=2\\ R=\frac{V}{I}=12/2\\ =6\\ \frac{dR}{dt}=\frac{1}{I}\left(\frac{dV}{dt}-R\frac{dI}{dt}\right)\\ =\frac{1}{2}\left(1-6\left(-\frac{1}{3}\right)\right.\\ =\frac{3}{2}\end{array}$

Therefore, the required $\frac{dR}{dt}=\frac{3}{2}\text{ ohms }/\text{sec}$ and $R$ is increasing.