Chapter 5, Problem 1RP

(a) Show how you would rewire the protoboard in Figure 5-4(a) so that all the odd-numbered resistors come first followed by the even-numbered ones. (b) Determine the resistance value of each resistor.

FIGURE 5-4

To determine

Show the rewired diagram of the protoboard in given Figure so that all the odd-numbered resistors come first followed by the even-numbered ones.

Explanation of Solution

Discussion:

Refer to given Figure 5-4 in the textbook.

The circuit in given Figure is redrawn as Figure 1 as per the given data.

Given that, the odd-numbered resistors are connected first, and then followed by the even-numbered resistors. That is, the resistors $R_1$, $R_3$, $R_5$ are connected first followed by the resistors $R_2$, $R_4$.

The schematic circuit of Figure 1 is drawn as Figure 2.

Conclusion:

Thus, the rewired diagram of the given protoboard is drawn.

To determine

Find the resistance value for each of the resistors in the given Figure.

Answer to Problem 1RP

The resistance value for the resistor $R_1$ is $1\text{k}\Omega$, $R_2$ is $33\text{k}\Omega$, $R_3$ is $39\text{k}\Omega$, $R_4$ is $470\Omega$ and $R_5$ is $22\text{k}\Omega$.

Explanation of Solution

Calculation:

Using the standard color codes, the value of the resistors can be calculated. The resistor schematic with 4-band code is drawn as shown in Figure 3.

For Resistor $R_1$:

The color code for brown color is digit $1$, black color is digit $0$ and red color code is 2 which is represented as power of multiplier 10.

Therefore, the value of the resistor $R_1$ is,

$\begin{array}{c}{R}_1=10\times {10}^2\Omega \\ =1\times {10}^3\Omega \\ =1\text{k}\Omega \left\{\because 1\text{k}={10}^3\right\}\end{array}$

For Resistor $R_2$:

The color code for the first orange color line is digit $3$, the second orange color line is digit $3$ and the third orange color code is 3 which is represented as power of multiplier 10.

Therefore, the value of the resistor $R_2$ is,

$\begin{array}{c}{R}_2=33\times {10}^3\Omega \\ =33\text{k}\Omega \left\{\because 1\text{k}={10}^3\right\}\end{array}$

For Resistor $R_3$:

The color code for orange color is digit $3$, white color is digit $9$ and orange color code is 3 which is represented as power of multiplier 10.

Therefore, the value of the resistor $R_3$ is,

$\begin{array}{c}{R}_3=39\times {10}^3\Omega \\ =39\text{k}\Omega \left\{\because 1\text{k}={10}^3\right\}\end{array}$

For Resistor $R_4$:

The color code for yellow color is digit $4$, violet color is digit $7$ and brown color code is 1 which is represented as power of multiplier 10.

Therefore, the value of the resistor $R_4$ is,

$\begin{array}{c}{R}_4=47\times {10}^1\Omega \\ =470\Omega \end{array}$

For Resistor $R_5$:

The color code for the first red color line is digit $2$, the second red color line is digit $2$ and the orange color code is 3 which is represented as power of multiplier 10.

Therefore, the value of the resistor $R_5$ is,

$\begin{array}{c}{R}_5=22\times {10}^3\Omega \\ =22\text{k}\Omega \left\{\because 1\text{k}={10}^3\right\}\end{array}$

Conclusion:

Thus, the resistance value for the resistor $R_1$ is $1\text{k}\Omega$, $R_2$ is $33\text{k}\Omega$, $R_3$ is $39\text{k}\Omega$, $R_4$ is $470\Omega$ and $R_5$ is $22\text{k}\Omega$.