Chapter 5, Problem 66E

(a)

To determine

Find the power absorbed by a $1\Omega$ resistor connected between the open terminals of the circuit.

Answer to Problem 66E

The power absorbed by a $1\Omega$ resistor connected between the open terminals of the circuit is $\text{175}\text{.45 nW}$.

Explanation of Solution

Given data:

The resistance of the load is $1\Omega$.

Formula used:

The expression for the equivalent resistor when resistors are connected in series is as follows:

$R_{eq}=R_1+R_2+\mathrm{......}+R_N$ (1)

Here,

$R_1$, $R_2$,…, $R_N$ are the resistances.

The expression for the equivalent resistor when resistors are connected in parallel is as follows:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\mathrm{......}+\frac{1}{R_N}$ (2)

Here,

$R_1$, $R_2$,…, $R_N$ are the resistances.

The $Y$-network and $\Delta$-network circuit diagram is given in Figure 1.

Refer to the redrawn Figure 1:

The expression for the conversion of $\Delta$-network into $Y$-network is as follows:

$R_1=\frac{R_b{R}_c}{R_a+R_b+R_c}$ (3)

$R_2=\frac{R_a{R}_c}{R_a+R_b+R_c}$ (4)

$R_3=\frac{R_a{R}_b}{R_a+R_b+R_c}$ (5)

Here,

$R_1$, $R_2$ and $R_3$ are the resistors connected in $Y$-network and

$R_a$, $R_b$ and $R_c$ are the resistors connected in $\Delta$-network.

Calculation:

The redrawn circuit diagram is given in Figure 2.

Refer to the redrawn Figure 2:

$\text{1 k}\Omega$ and $\text{7 k}\Omega$ resistor are connected in parallel.

Substitute $\text{1 k}\Omega$ for $R_1$ and $\text{7 k}\Omega$ for $R_2$ equation (2):

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\text{1 k}\Omega }+\frac{1}{\text{7 k}\Omega }\\ =\frac{8}{\text{7 k}\Omega }\end{array}$

Rearrange the equation for $R_{eq}$:

$\begin{array}{c}{R}_{eq}=\frac{7}{8}\text{ k}\Omega \\ =0.875\text{ k}\Omega \end{array}$

$\text{10 k}\Omega$ and $\text{2}\text{.2 k}\Omega$ resistor are connected in parallel.

Substitute $\text{10 k}\Omega$ for $R_1$ and $\text{2}\text{.2 k}\Omega$ for $R_2$ equation (2):

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\text{10 k}\Omega }+\frac{1}{\text{2}\text{.2 k}\Omega }\\ =\frac{61}{\text{110 k}\Omega }\end{array}$

Rearrange the equation for $R_{eq}$:

$\begin{array}{c}{R}_{eq}=\frac{110}{61}\text{ k}\Omega \\ =1.803\text{ k}\Omega \end{array}$

The simplified circuit diagram is given in Figure 3:

Refer to the redrawn Figure 3:

The $\text{4 k}\Omega$, $\text{10 k}\Omega$ and $\text{5 k}\Omega$ resistors are connected in $\Delta$-network.

Substitute $\text{4 k}\Omega$ for $R_a$, $\text{10 k}\Omega$ for $R_b$ and $\text{5 k}\Omega$ for $R_c$ in equation (3):

$\begin{array}{c}{R}_1=\frac{\left(10\text{ k}\Omega \right)\left(\text{5 k}\Omega \right)}{\text{4 k}\Omega +10\text{ k}\Omega +5\text{ k}\Omega }\\ =\frac{50}{19}\text{ k}\Omega \\ =2.632\text{ k}\Omega \end{array}$

Substitute $\text{4 k}\Omega$ for $R_a$, $\text{10 k}\Omega$ for $R_b$ and $\text{5 k}\Omega$ for $R_c$ in equation (4):

$\begin{array}{c}{R}_2=\frac{\left(\text{4 k}\Omega \right)\left(\text{5 k}\Omega \right)}{\text{4 k}\Omega +10\text{ k}\Omega +5\text{ k}\Omega }\\ =\frac{20}{19}\text{ k}\Omega \\ =1.053\text{ k}\Omega \end{array}$

Substitute $\text{4 k}\Omega$ for $R_a$, $\text{10 k}\Omega$ for $R_b$ and $\text{5 k}\Omega$ for $R_c$ in equation (5):

$\begin{array}{c}{R}_3=\frac{\left(\text{10 k}\Omega \right)\left(\text{4 k}\Omega \right)}{\text{4 k}\Omega +10\text{ k}\Omega +5\text{ k}\Omega }\\ =\frac{40}{19}\text{ k}\Omega \\ =2.105\text{ k}\Omega \end{array}$

The simplified circuit diagram is given in Figure 4.

Refer to the redrawn Figure 4.

$2.105\text{ k}\Omega$ and $0.875\text{ k}\Omega$ resistor are connected in series.

Substitute $2.105\text{ k}\Omega$ for $R_1$ and $0.875\text{ k}\Omega$ for $R_2$ equation (1):

$\begin{array}{c}{R}_{eq}=2.105\text{ k}\Omega +0.875\text{ k}\Omega \\ =2.98\text{ k}\Omega \end{array}$

$1.053\text{ k}\Omega$ and $1.803\text{ k}\Omega$ resistor are connected in series.

Substitute $1.053\text{ k}\Omega$ for $R_1$ and $1.803\text{ k}\Omega$ for $R_2$ equation (1):

$\begin{array}{c}{R}_{eq}=1.053\text{ k}\Omega +1.803\text{ k}\Omega \\ =2.856\text{ k}\Omega \end{array}$

The simplified circuit diagram is given in Figure 5.

Refer to the redrawn Figure 5:

$\text{2}\text{.98 k}\Omega$ and $\text{2}\text{.856 k}\Omega$ resistor are connected in parallel.

Substitute $\text{2}\text{.98 k}\Omega$ for $R_1$ and $\text{2}\text{.856 k}\Omega$ for $R_2$ equation (2):

$\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{\text{2}\text{.98 k}\Omega }+\frac{1}{\text{2}\text{.856 k}\Omega }\\ =0.6857{\left(\text{k}\Omega \right)}^{-}\end{array}$

Rearrange the equation for $R_{eq}$:

$\begin{array}{c}{R}_{eq}=\frac{1}{0.6857}\text{ k}\Omega \\ =1.458\text{ k}\Omega \end{array}$

The simplified circuit diagram is given in Figure 6.

Refer to the redrawn Figure 6:

$1.458\text{ k}\Omega$ and $2.632\text{ k}\Omega$ resistor are connected in series.

Substitute $1.458\text{ k}\Omega$ for $R_1$ and $2.632\text{ k}\Omega$ for $R_2$ equation (1):

$\begin{array}{c}{R}_{eq}=1.458\text{ k}\Omega +2.632\text{ k}\Omega \\ =4.090\text{ k}\Omega \end{array}$

The simplified circuit diagram is given in Figure 7.

Refer to the redrawn Figure 7:

So, the Thevenin equivalent resistance is $4.090\text{ k}\Omega$.

The simplified circuit diagram is given in Figure 8.

Refer to the redrawn Figure 8:

The $2.632\text{ k}\Omega$ resistor is open circuited, so, no current will flow through it.

The expression for the current flowing through $1.803\text{ k}\Omega$ resistor is as follows:

$i_0=i\left(\frac{R_2}{R_1+R_2+R_4+R_5}\right)$ (6)

Here,

$i$ is the current through $4\text{mA}$ independent source,

$i_0$ is the current flowing through $1.803\text{ k}\Omega$ resistor,

$R_1$ is the resistance of $2.105\text{ k}\Omega$ resistor,

$R_2$ is the resistance of $0.875\text{ k}\Omega$ resistor,

$R_4$ is the resistance of $1.053\text{ k}\Omega$ resistor and

$R_5$ is the resistance of $1.803\text{ k}\Omega$ resistor.

Substitute $4\text{mA}$ for $i$, $2.105\text{ k}\Omega$ for $R_1$, $0.875\text{ k}\Omega$ for $R_2$, $1.053\text{ k}\Omega$ for $R_4$ and $1.803\text{ k}\Omega$ for $R_5$ in equation (6):

$\begin{array}{c}{i}_0=\left(4\text{mA}\right)\left(\frac{0.875\text{ k}\Omega }{2.105\text{ k}\Omega +0.875\text{ k}\Omega +1.053\text{ k}\Omega +1.803\text{ k}\Omega }\right)\\ =\left(4\text{mA}\right)\left(\frac{0.875\text{ k}\Omega }{5.836\text{ k}\Omega }\right)\\ =\left(4\text{mA}\right)\left(0.1499\right)\\ =0.6\text{mA}\end{array}$

The expression for the Thevenin voltage is as follows:

$v_{TH}=i_0\left(R_4+R_5\right)$ (7)

Here,

$v_{TH}$ is the Thevenin voltage.

Substitute $0.6\text{mA}$ for $i_o$, $1.053\text{ k}\Omega$ for $R_4$ and $1.803\text{ k}\Omega$ for $R_5$ in equation (7):

$\begin{array}{c}{v}_{TH}=\left(0.6\text{mA}\right)\left(1.053\text{ k}\Omega +1.803\text{ k}\Omega \right)\\ =\left(0.6\times {10}^{-3}\text{A}\right)\left(2.856\text{ k}\Omega \right)\left\{\because 1\text{ mA}={10}^{-3}\text{ A}\right\}\\ =\left(0.6\times {10}^{-3}\text{A}\right)\left(2.856\times {10}^3\Omega \right)\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =1.7136\text{ V}\end{array}$

So, the Thevenin voltage of the circuit is $1.7136\text{ V}$.

The redrawn circuit diagram is given in Figure 9.

Refer to the redrawn Figure 9:

The expression for the power absorbed by the load resistor is as follows:

$p={\left(\frac{v_{TH}}{R_{TH}+R_L}\right)}^2{R}_L$ (8)

Here,

$p$ is the power absorbed by the load resistor.

Substitute $1\Omega$ for $R_L$, $1.7136\text{ V}$ for $v_{TH}$ and $4.090\text{ k}\Omega$ for $R_{TH}$ in equation (8):

$\begin{array}{c}p={\left(\frac{1.7136\text{ V}}{\left(4.090\text{ k}\Omega \right)+\left(1\Omega \right)}\right)}^2\left(1\Omega \right)\\ ={\left(\frac{1.7136\text{ V}}{\left(4.090\times {10}^3\Omega \right)+\left(1\Omega \right)}\right)}^2\left(1\Omega \right)\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =\left(1.7545\times {10}^{-7}\right)\left(1\right)\text{ W}\\ \text{=175}\text{.45 nW }\left\{{10}^{-9}\text{ W}=1\text{ nW}\right\}\end{array}$

Conclusion:

Thus, the power absorbed by a $1\Omega$ resistor connected between the open terminals of the circuit is $\text{175}\text{.45 nW}$.

(b)

To determine

Find the power absorbed by a $100\Omega$ resistor connected between the open terminals of the circuit.

Answer to Problem 66E

The power absorbed by a $100\Omega$ resistor connected between the open terminals of the circuit is $\text{16725}\text{.9 nW}$.

Explanation of Solution

Given data:

The resistance of the load is $100\Omega$.

Calculation:

Refer to the redrawn Figure 9:

Substitute $100\Omega$ for $R_L$, $1.7136\text{ V}$ for $v_{TH}$ and $4.090\text{ k}\Omega$ for $R_{TH}$ in equation (8):

$\begin{array}{c}p={\left(\frac{1.7136\text{ V}}{\left(4.090\text{ k}\Omega \right)+\left(100\Omega \right)}\right)}^2\left(100\Omega \right)\\ ={\left(\frac{1.7136\text{ V}}{\left(4.090\times {10}^3\Omega \right)+\left(100\Omega \right)}\right)}^2\left(100\Omega \right)\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =\left(1.6725\times {10}^{-7}\right)\left(100\right)\text{ W}\\ \text{=1}\text{.67259}\times {\text{10}}^{-5}\text{ W}\end{array}$

$p=\text{16}\text{.725 }\mu \text{W }\left\{{10}^{-6}\text{ W}=1\mu \text{W}\right\}$

Conclusion:

Thus, the power absorbed by a $100\Omega$ resistor connected between the open terminals of the circuit is $\text{16}\text{.725 }\mu \text{W}$.

(c)

To determine

Find the power absorbed by a $2.65\text{ k}\Omega$ resistor connected between the open terminals of the circuit.

Answer to Problem 66E

The power absorbed by a $2.65\text{ k}\Omega$ resistor connected between the open terminals of the circuit is $\text{171}\text{.29 }\mu \text{W}$.

Explanation of Solution

Given Data:

The resistance of the load is $2.65\text{ k}\Omega$.

Calculation:

Refer to the redrawn Figure 9:

Substitute $2.65\text{ k}\Omega$ for $R_L$, $1.7136\text{ V}$ for $v_{TH}$ and $4.090\text{ k}\Omega$ for $R_{TH}$ in equation (8):

$\begin{array}{c}p={\left(\frac{1.7136\text{ V}}{\left(4.090\text{ k}\Omega \right)+\left(2.65\text{ k}\Omega \right)}\right)}^2\left(2.65\text{ k}\Omega \right)\\ ={\left(\frac{1.7136\text{ V}}{\left(4.090\times {10}^3\Omega \right)+\left(2.65\times {10}^3\Omega \right)}\right)}^2\left(2.65\times {10}^3\Omega \right)\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =\left(6.464\times {10}^{-8}\right)\left(2.65\times {10}^3\right)\text{ W}\\ \text{=1}\text{.7129}\times {\text{10}}^{-4}\text{ W}\end{array}$

$p=\text{171}\text{.29 }\mu \text{W }\left\{{10}^{-6}\text{ W}=1\mu \text{W}\right\}$

Conclusion:

Thus, the power absorbed by a $2.65\text{ k}\Omega$ resistor connected between the open terminals of the circuit is $\text{171}\text{.29 }\mu \text{W}$.

(d)

To determine

Find the power absorbed by a $\text{1}\text{.13 M}\Omega$ resistor connected between the open terminals of the circuit.

Answer to Problem 66E

The power absorbed by a $\text{1}\text{.13 M}\Omega$ resistor connected between the open terminals of the circuit is $\text{2}\text{.58 }\mu \text{W}$.

Explanation of Solution

Given Data:

The resistance of the load is $\text{1}\text{.13 M}\Omega$.

Calculation:

Refer to the redrawn Figure 9:

Substitute $\text{1}\text{.13 M}\Omega$ for $R_L$, $1.7136\text{ V}$ for $v_{TH}$ and $4.090\text{ k}\Omega$ for $R_{TH}$ in equation (8):

$\begin{array}{c}p={\left(\frac{1.7136\text{ V}}{\left(4.090\text{ k}\Omega \right)+\left(\text{1}\text{.13 M}\Omega \right)}\right)}^2\left(\text{1}\text{.13 M}\Omega \right)\\ ={\left(\frac{1.7136\text{ V}}{\left(4.090\times {10}^3\Omega \right)+\left(\text{1}\text{.13 M}\Omega \right)}\right)}^2\left(\text{1}\text{.13 M}\Omega \right)\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ ={\left(\frac{1.7136\text{ V}}{\left(4.090\times {10}^3\Omega \right)+\left(\text{1}\text{.13}\times {\text{10}}^6\Omega \right)}\right)}^2\left(\text{1}\text{.13}\times {\text{10}}^6\Omega \right)\left\{\because 1\text{ M}\Omega ={10}^6\Omega \right\}\\ =\left(2.28\times {10}^{-12}\right)\left(\text{1}\text{.13}\times {\text{10}}^6\right)\text{ W}\end{array}$

$\begin{array}{c}p=\text{=2}\text{.58}\times {\text{10}}^{-6}\text{ W}\\ =2.\text{58 }\mu \text{W }\left\{{10}^{-6}\text{ W}=1\mu \text{W}\right\}\end{array}$

Conclusion:

Thus, the power absorbed by a $\text{1}\text{.13 M}\Omega$ resistor connected between the open terminals of the circuit is $\text{2}\text{.58 }\mu \text{W}$.