Chapter 2, Problem 2.42HP

For the circuits of Figure P2.42, determine the resistor values (including the power rating) necessaryto achieve the indicated voltages. Resistors are available in $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8^{-}$}\right.,\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4^{-}$}\right.,\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2^{-}$}\right.,$ and 1-W ratings.

To determine

(a)

The values of the resistors which is necessary to achieve the indicated voltage.

Answer to Problem 2.42HP

The value of all the given ratings is greater than $\frac{1}{25}\text{W}$ therefore any of the ratings among $\frac{1}{8}\text{W}$, $\frac{1}{4}\text{W}$, $\frac{1}{2}\text{W}$ and $1\text{W}$ can be used.

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1.

  

To calculate the value of the resistance $R_a$ in the above circuit apply voltage division rule in the above circuit.

  $\begin{array}{l}20\text{V}=\left(50\text{V}\right)\left(\frac{R_a}{R_a+15\text{k}\Omega }\right)\\ \frac{R_a}{R_a+15\text{k}\Omega }=0.4\\ R_a=0.4R_a+6\text{k}\Omega \\ R_a=10\text{k}\Omega \end{array}$

The total current in the circuit is calculated as,

  $\begin{array}{c}I=\frac{50\text{V}}{\left(10\text{k}\Omega +15\text{k}\Omega \right)}\\ =2\text{mA}\end{array}$

The expression to calculate the power rating across the resistance $R_a$ is given by,

  $P={\left(I\right)}^2{R}_a$

Substitute $10\text{k}\Omega$ for $R_a$ and $2\text{mA}$ for $I$ in the above equitation.

  $\begin{array}{c}P={\left(2\text{mA}\right)}^2\left(10\text{k}\Omega \right)\\ =0.04\text{W}\\ \text{=}\frac{1}{25}\text{W}\end{array}$

The available ratings of the resistance are $\frac{1}{8}\text{W}$, $\frac{1}{4}\text{W}$, $\frac{1}{2}\text{W}$ and $\text{1}\text{W}$, where all these ratings are greater than the rating $\frac{1}{25}\text{W}$ therefore, any of these ratings can be employed.

Conclusion:

Therefore, the value of all the given ratings is greater than $\frac{1}{25}\text{W}$ therefore any of the ratings among $\frac{1}{8}\text{W}$, $\frac{1}{4}\text{W}$, $\frac{1}{2}\text{W}$ and $1\text{W}$ can be used.

To determine

(b)

The values of the resistors which is necessary to achieve the indicated voltage.

Answer to Problem 2.42HP

The value of all the given ratings is greater than $0.618\text{mW}$ and $0.50625\text{mW}$ therefore any of the ratings among $\frac{1}{8}\text{W}$, $\frac{1}{4}\text{W}$, $\frac{1}{2}\text{W}$ and $1\text{W}$ can be used.

Explanation of Solution

Calculation:

The given diagram is shown in Figure 2

  

To calculate the value of the resistance $R_b$ in the above circuit apply voltage division rule in the above circuit.

  $2.25\text{V}=\left(5\text{V}\right)\left(\frac{R_a}{R_b+R_a}\right)$

Substitute $10\text{k}\Omega$ for $R_a$ in the above equation.

  $\begin{array}{l}2.25\text{V}=\left(5\text{V}\right)\left(\frac{10\text{k}\Omega }{R_b+10\text{k}\Omega }\right)\\ \frac{10\text{k}\Omega }{R_b+10\text{k}\Omega }=0.45\\ 4500+0.45R_b=10000\\ R_b=12.22\text{k}\Omega \end{array}$

The expression for the total current in the circuit is given by,

  $I=\frac{5\text{V}}{R_b+R_a}$

Substitute $10\text{k}\Omega$ for $R_a$ and $12.22\text{k}\Omega$ for $R_b$ in the above equation.

  $\begin{array}{c}I=\frac{50\text{V}}{\left(10\text{k}\Omega +12.22\text{k}\Omega \right)}\\ =0.225\text{mA}\end{array}$

The expression to calculate the power rating across the resistance $R_a$ is given by,

  $P={\left(I\right)}^2{R}_a$

Substitute $10\text{k}\Omega$ for $R_a$ and $0.225\text{mA}$ for $I$ in the above equitation.

  $\begin{array}{c}P={\left(0.225\text{mA}\right)}^2\left(10\text{k}\Omega \right)\\ =0.50625\text{mW}\end{array}$

The expression to calculate the power rating across the resistance $R_b$ is given by,

  $P={\left(I\right)}^2{R}_b$

Substitute $12.22\text{k}\Omega$ for $R_b$ and $0.225\text{mA}$ for $I$ in the above equitation.

  $\begin{array}{c}P={\left(0.225\text{mA}\right)}^2\left(12.22\text{k}\Omega \right)\\ =0.618\text{mW}\end{array}$

The available ratings of the resistance are $\frac{1}{8}\text{W}$, $\frac{1}{4}\text{W}$, $\frac{1}{2}\text{W}$ and $\text{1}\text{W}$, where all these ratings are greater than the rating $0.618\text{mW}$ and $0.50625\text{mW}$ therefore, any of the given ratings can be employed.

Conclusion:

Therefore, the value of all the given ratings is greater than $0.618\text{mW}$ and $0.50625\text{mW}$ therefore any of the ratings among $\frac{1}{8}\text{W}$, $\frac{1}{4}\text{W}$, $\frac{1}{2}\text{W}$ and $1\text{W}$ can be used.

To determine

(c)

The values of the resistors which is necessary to achieve the indicated voltage.

Answer to Problem 2.42HP

The power rating across the load must be $\text{1}\text{W}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 3.

  

To calculate the value of the resistance $R_L$ in the above circuit apply voltage division rule in the above circuit.

  $\begin{array}{l}28.3\text{V}=\left(110\text{V}\right)\left(\frac{2.7\text{k}\Omega }{R_L+2.7\text{k}\Omega +1\text{k}\Omega }\right)\\ R_L+2.7\text{k}\Omega +1\text{k}\Omega =\left(110\text{V}\right)\left(\frac{2.7\text{k}\Omega }{28.3\text{V}}\right)\\ R_L+3.7\text{k}\Omega =10.495\text{k}\Omega \\ R_L=6.8\text{k}\Omega \end{array}$

The expression for the total current in the circuit is given by,

  $I=\frac{50\text{V}}{R_L}$

Substitute $6.8\text{k}\Omega$ for $R_L$ in the above equation.

  $\begin{array}{c}I=\frac{50\text{V}}{6.8\text{k}\Omega }\\ =10.48\text{mA}\end{array}$

The expression to calculate the power rating across the resistance $R_L$ is given by,

  $P={\left(I\right)}^2{R}_L$

Substitute $6.8\text{k}\Omega$ for $R_L$ and $10.48\text{mA}$ for $I$ in the above equitation.

  $\begin{array}{c}P={\left(10.48\text{mA}\right)}^2\left(6.8\text{k}\Omega \right)\\ =0.747\text{W}\end{array}$

The power rating among the ratings $\frac{1}{8}\text{W}$, $\frac{1}{4}\text{W}$, $\frac{1}{2}\text{W}$ and $\text{1}\text{W}$ which is more than the derived power rating of $0.747\text{W}$ is $\text{1}\text{W}$ therefore, the suitable power rating across the load must be $\text{1}\text{W}$ .

Conclusion:

Therefore, the power rating across the load must be $\text{1}\text{W}$ .