Chapter 2, Problem 81CP

For a specific application, the circuit shown in Fig. 2.140 was designed so that I_L = 83.33 mA and that R_in = 5 kΩ. What are the values of R₁ and R₂?

Figure 2.140

To determine

Calculate the values of $R_1$ and $R_2$.

Answer to Problem 81CP

The values of $R_1$ and $R_2$ in Figure 2.140 are $\underset{\_}{6.667\text{k}\Omega \text{and}5\text{k}\Omega }$.

Explanation of Solution

Given Data:

Refer to Figure 2.140 in the textbook for the given circuit.

$I_L$ is $83.33\text{mA}$.

$R_{in}$ is $5\text{k}\Omega$.

Formula used:

Consider the expression for $N$ resistors connected in parallel.

$\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdots +\frac{1}{R_N}$

Here,

$R_1,R_2,R_3\cdots R_N$ are resistors.

Consider the expression for $N$ resistors connected in series.

$R_{\text{eq}}=R_1+R_2+R_3+\cdots +R_N$

Calculation:

Modify Figure 2.140 as shown in Figure 1.

In Figure 1, $R_{in}$ is connected in parallel with $5\text{k}\Omega$ resistor. Since the value of $R_{in}$ is $5\text{k}\Omega$ that is the flow of current in $R_{in}$ and $5\text{k}\Omega$ should be same which is half of the source current.

Therefore,

$\begin{array}{c}{I}_{R_{in}}=\frac{1\text{A}}{2}\\ =500\text{mA}\end{array}$

From Figure 1, write the expression for equivalent resistance $R_{in}$.

$R_{in}=\frac{10\text{k}\left[R_1+\left(\frac{R_2\left(10\text{k}\right)}{R_2+10\text{k}}\right)\right]}{10\text{k+}{R}_1+\left(\frac{R_2\left(10\text{k}\right)}{R_2+10\text{k}}\right)}$

Substitute $5\text{k}\Omega$ for $R_{in}$ as follows.

$5\text{k}\Omega =\frac{10\text{k}\left[R_1+\left(\frac{R_2\left(10\text{k}\right)}{R_2+10\text{k}}\right)\right]}{10\text{k+}{R}_1+\left(\frac{R_2\left(10\text{k}\right)}{R_2+10\text{k}}\right)}$ (1)

To maintain $R_{in}$ as $5\text{k}\Omega$, the parallel and series combination of $10\text{k}\Omega ,R_2,\text{and}{R}_1$ should be $10\text{k}\Omega$.

Modify the Figure as shown in Figure 2.

Therefore from current division rule, the current through $R_1$ resistor $I_{R_1}$ will be half of the current of $I_{R_{in}}$.

$\begin{array}{c}{I}_{R_1}=\frac{I_{R_{in}}}{2}\\ =\frac{500\text{mA}}{2}\left\{\because I_{R_{in}}=500\text{mA}\right\}\\ =250\text{mA}\end{array}$

From current division rule, write the expression for current $I_L$ in Figure 2.

$I_L=I_{R_1}\left(\frac{R_2}{R_2+10\text{k}\Omega }\right)$

Substitute $83.33\text{mA}$ for $I_L$ and $250\text{mA}$ for $I_{R_1}$ as follows.

$83.33\text{mA}=250\text{mA}\left(\frac{R_2}{R_2+10\text{k}\Omega }\right)$

Rearrange the equation as follows.

$\begin{array}{c}{R}_2=\left(\frac{83.33\text{mA}}{250\text{mA}}\right)\left(R_2+10\text{k}\Omega \right)\\ =0.33332\left(R_2+10\text{k}\Omega \right)\\ =0.33332R_2+3.3332\text{k}\Omega \end{array}$

Simplify the equation as follows.

$\begin{array}{l}{R}_2-0.33332R_2=3.3332\text{k}\Omega \\ R_2\left(1-0.33332\right)=3.3332\text{k}\Omega \\ R_2\left(0.66668\right)=3.3332\text{k}\Omega \\ R_2=\frac{3.3332\text{k}\Omega }{0.66668}\end{array}$

Simplify the equation as follows.

$\begin{array}{c}{R}_2=4.999\text{k}\Omega \\ \cong 5\text{k}\Omega \end{array}$

Substitute $5\text{k}\Omega$ for $R_2$ as follows.

$\begin{array}{c}5\text{k}=\frac{10\text{k}\left[R_1+\left(\frac{\left(5\text{k}\right)\left(10\text{k}\right)}{5\text{k}+10\text{k}}\right)\right]}{10\text{k+}{R}_1+\left(\frac{\left(5\text{k}\right)\left(10\text{k}\right)}{5\text{k}+10\text{k}}\right)}\\ =\frac{10\text{k}\left[R_1+\left(\frac{50\text{M}}{15\text{k}}\right)\right]}{10\text{k+}{R}_1+\left(\frac{50\text{M}}{15\text{k}}\right)}\\ =\frac{10\text{k}\left[R_1+3.333\text{k}\right]}{10\text{k+}{R}_1+3.333\text{k}}\\ =\frac{10\text{k}{R}_1+33.33\text{M}}{R_1+13.333\text{k}}\end{array}$

Simplify the equation as follows.

$\begin{array}{l}5\text{k}\left(R_1+13.333\text{k}\right)=10\text{k}{R}_1+33.33\text{M}\\ 5\text{k}{R}_1+66.66\text{M}=10\text{k}{R}_1+33.33\text{M}\\ 10\text{k}{R}_1-5\text{k}{R}_1=66.66\text{M}-33.33\text{M}\\ 5\text{k}{R}_1=33.33\text{M}\end{array}$

Simplify the equation to obtain the value of $R_1$.

$\begin{array}{l}{R}_1=\frac{33.33\text{M}}{5\text{k}}\\ R_1=6.666\text{k}\Omega \end{array}$

Conclusion:

Thus, the values of $R_1$ and $R_2$ in Figure 2.140 are $\underset{\_}{6.667\text{k}\Omega \text{and}5\text{k}\Omega }$.