Chapter 4.6, Problem 41E

(a)

To determine

To find: The rational equation to model the situation.

Answer to Problem 41E

The rational equation to model the situationis $\frac{1}{2R_2}+\frac{1}{R_2}=\frac{1}{10}-\frac{1}{20}$ .

Explanation of Solution

Given:

The general equation is $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ and In the circuit $R_1$ is equal to twice of $R_2$ and $R_3$ is $20\text{ohms}$ and the equivalent resistance equals to $10\text{ohms}$ .

Calculation:

Substitute $10\text{ohms}$ for $R$ and $20\text{ohms}$ for $R_3$ , $2R_2$ for $R_1$ in the equation $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ to find the model.

  $\begin{array}{c}\frac{1}{10}=\frac{1}{2R_2}+\frac{1}{R_2}+\frac{1}{20}\\ \frac{1}{2R_2}+\frac{1}{R_2}=\frac{1}{10}-\frac{1}{20}\end{array}$

Therefore, the rational equation to model the situationis $\frac{1}{2R_2}+\frac{1}{R_2}=\frac{1}{10}-\frac{1}{20}$ .

(b)

To determine

To find: The value of $R_1$ and $R_2$ .

Answer to Problem 41E

The value of $R_1$ and $R_2$ are $30\text{ohms}$ and $15\text{ohms}$ respectively.

Explanation of Solution

Given:

The general equation is $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ and In the circuit $R_1$ is equal to twice of $R_2$ and $R_3$ is $20\text{ohms}$ and the equivalent resistance equals to $10\text{ohms}$ .

Calculation:

Let the velocity of wind is $v$ mph and both planes flying for $t\text{hr}$ .

The airspeed of both planes is 200 mph in still air.

The airspeed of 1st plane moving along the direction of wind will be $\left(200+v\right)\text{mph}$ .

The distance traversed by 1st plane moving along the direction of wind in t hr will be $\left(200+v\right)t\text{ mile}$ .

The airspeed of 2nd plane moving against the direction of wind will be $\left(200-v\right)\text{mph}$ .

The distance traversed by 2ndplane moving against the direction of wind in t hr will be $\left(200-v\right)t\text{ mile}$ .

So, the problems are,

  $\begin{array}{c}(200+v)t=\mathrm{1062......}(1)\\ (200-v)t=\mathrm{738........}(2)\end{array}$

Divide the equation (1) by equation (2).

  $\begin{array}{c}\frac{200+v}{200-v}=\frac{1062}{738}\\ 200\times 738+738v=1062\times 200-1062v\\ 1062v+738v=1062\times 200-200\times 738\\ 1800v=1062\times 200-200\times 738\end{array}$

Further, solve the above problem.

  $\begin{array}{c}v=\frac{64800}{1800}\\ =36\text{mph}\end{array}$

Therefore, the speed of the wind is $36\text{mph}$ .