Chapter 5, Problem 41E

(a)

To determine

Find the power dissipated by $1\text{ k}\Omega$ resistor connected between $a$ and $b$.

Answer to Problem 41E

The power dissipated by $1\text{ k}\Omega$ resistor connected between $a$ and $b$ is $1.37\mu \text{W}$.

Explanation of Solution

Given Data:

The load resistance is $1\text{ k}\Omega$.

Formula used:

The expression for the power dissipated by load resistor is as follows.

$p={\left(i_L\right)}^2{R}_L$ (1)

Here,

$p$ is the power dissipated by load resistor,

$i_L$ is the current flowing through the load resistor and

$R_L$ is the load resistor.

Calculation:

The redrawn circuit diagram is given in Figure 1,

Refer to the redrawn Figure 1,

Apply Kirchhoff’s current law at node 1.

$\frac{v_1}{R_1}+\frac{v_2}{R_2}+i=0\text{A}$ (2)

Here,

$v_1$ is the voltage at node 1,

$v_2$ is the voltage at node 2,

$i$ is the current through $0.0002v_1$ dependent source ,

$R_1$ is the resistance of $10\text{ k}\Omega$ resistor and

$R_2$ is the resistance of $20\text{ k}\Omega$ resistor,

Substitute $10\text{ k}\Omega$ for $R_1$, $20\text{ k}\Omega$ for $R_2$ and $0.0002v_1$ for $i$ in equation (2),

$\begin{array}{l}\frac{v_1}{10\text{ k}\Omega }+\frac{v_2}{20\text{ k}\Omega }+0.0002v_1=0\text{A}\\ \frac{v_1}{10\times {10}^3\Omega }+\frac{v_2}{20\times {10}^3\Omega }+0.0002v_1=0\text{A}\left\{\because 1\text{k}\Omega ={10}^3\Omega \right\}\\ \frac{2v_1+v_2+\left(20\times {10}^3\right)\left(0.0002v_1\right)}{20\times {10}^3\Omega }=0\text{A}\\ \frac{2v_1+v_2+4v_1}{20\times {10}^3\Omega }=0\text{A}\end{array}$

Rearrange for $v_1$ and $v_2$,

$\begin{array}{l}2v_1+v_2+4v_1=0\text{V}\\ 6v_1+v_2=0\text{V}\end{array}$

Rearrange for $v_2$,

$v_2=-6v_1$ (3)

Apply Kirchhoff’s voltage law between node 1 and 2.

$v_1-v_2=1\text{V}$ (4)

Substitute $-6v_1$ for $v_2$ in equation (4).

$\begin{array}{l}{v}_1-\left(-6v_1\right)=1\text{V}\\ v_1+6v_1=1\text{V}\\ 7v_1=1\text{V}\end{array}$

Rearrange for $v_1$,

$\begin{array}{c}{v}_1=\frac{1}{7}\text{V}\\ =0.1429\text{V}\end{array}$

So, the Thevenin voltage is $0.1429\text{V}$.

$1\text{A}$ test source is connected across the branch $a\text{-}b$ and short circuit the $1\text{V}$ independent source.

The redrawn circuit diagram is given in Figure 2,

Refer to the redrawn Figure 2.

Apply Kirchhoff’s current law at node 1.

$\frac{v_1}{R_1}+\frac{v_1}{R_2}+i=i_0$ (5)

Here,

$i_0$ is the $1\text{A}$ test current source.

Substitute $10\text{ k}\Omega$ for $R_1$, $20\text{ k}\Omega$ for $R_2$, $0.0002v_1$ for $i$ and $1\text{A}$ for $i_0$ in equation (5),

$\begin{array}{l}\left(\frac{v_1}{10\text{ k}\Omega }\right)+\left(\frac{v_1}{20\text{ k}\Omega }\right)+\left(0.0002v_1\right)=1\text{A}\\ \left(\frac{v_1}{10\times {10}^3\Omega }\right)+\left(\frac{v_1}{20\times {10}^3\Omega }\right)+\left(0.0002v_1\right)=1\text{A}\left\{\because 1\text{k}\Omega ={10}^3\Omega \right\}\\ \frac{2v_1+v_1+\left(20\times {10}^3\right)\left(0.0002v_1\right)}{20\times {10}^3\Omega }=1\text{A}\\ \frac{2v_1+v_1+4v_1}{20\times {10}^3\Omega }=1\text{A}\end{array}$

Rearrange for $v_1$,

$\begin{array}{l}7v_1=\left(1\text{A}\right)\left(20\times {10}^3\Omega \right)\\ v_1=\frac{20\times {10}^3}{7}\text{V}\\ v_1=2857.143\text{V}\end{array}$

The expression for the Thevenin equivalent resistance is as follows,

$R_{th}=\frac{v_1}{i_0}$ (6)

Here,

$R_{th}$ is the Thevenin equivalent resistance.

Substitute $2857.143\text{V}$ for $v_1$ and $1\text{A}$ for $i_0$,

$\begin{array}{c}{R}_{th}=\frac{2857.143\text{V}}{1\text{A}}\\ =2857.143\Omega \end{array}$

The redrawn circuit diagram is given in Figure 3,

Refer to the redrawn Figure 3,

The expression for the current flowing in the circuit is as follows.

$i_L=\frac{V_{th}}{R_{th}+R_L}$ (7)

Here,

$i_L$ is the current flowing in the circuit,

$V_{th}$ is the Thevenin voltage and

$R_L$ is the load resistance.

Substitute $2857.143\Omega$ for $R_{th}$, $1\text{ k}\Omega$ for $R_L$ and $0.1429\text{V}$ for $v$ in equation (7),

$\begin{array}{c}{i}_L=\frac{0.1429\text{V}}{2857.143\Omega +1\text{ k}\Omega }\\ =\frac{0.1429\text{V}}{2857.143\Omega +1\times {10}^3\Omega }\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =3.7048\times {10}^{-5}\text{A}\end{array}$

Substitute $3.7048\times {10}^{-5}\text{A}$ for $i_L$ and $1\text{ k}\Omega$ for $R_L$ in equation (1),

$\begin{array}{c}p={\left(3.7048\times {10}^{-5}\text{A}\right)}^2\left(1\text{ k}\Omega \right)\\ ={\left(3.7048\times {10}^{-5}\text{A}\right)}^2\left(1\times {10}^3\Omega \right)\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =1.37\times {10}^{-6}\text{W}\\ =1.37\mu \text{W }\left\{\because {10}^{-6}\text{W}=1\mu \text{W}\right\}\end{array}$

Conclusion:

Thus, the power dissipated by $1\text{ k}\Omega$ resistor connected between $a$ and $b$ is $1.37\mu \text{W}$.

(b)

To determine

The power dissipated by $\text{4}\text{.7 k}\Omega$ resistor connected between $a$ and $b$.

Answer to Problem 41E

The power dissipated by $\text{4}\text{.7 k}\Omega$ resistor connected between $a$ and $b$ is $1.68\mu \text{W}$.

Explanation of Solution

Given Data:

The load resistance is $\text{4}\text{.7 k}\Omega$.

Calculation:

Refer to the redrawn Figure 3,

Substitute $2857.143\Omega$ for $R_{th}$, $\text{4}\text{.7 k}\Omega$ for $R_L$ and $0.1429\text{V}$ for $v$ in equation (7),

$\begin{array}{c}{i}_L=\frac{0.1429\text{V}}{2857.143\Omega +\text{4}\text{.7 k}\Omega }\\ =\frac{0.1429\text{V}}{2857.143\Omega +4.7\times {10}^3\Omega }\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =1.8909\times {10}^{-5}\text{A}\end{array}$

Substitute $1.8909\times {10}^{-5}\text{A}$ for $i_L$ and $\text{4}\text{.7 k}\Omega$ for $R_L$ in equation (1),

$\begin{array}{c}p={\left(1.8909\times {10}^{-5}\text{A}\right)}^2\left(\text{4}\text{.7 k}\Omega \right)\\ ={\left(1.8909\times {10}^{-5}\text{A}\right)}^2\left(4.7\times {10}^3\Omega \right)\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =1.68\times {10}^{-6}\text{W}\\ =1.68\mu \text{W }\left\{\because {10}^{-6}\text{W}=1\mu \text{W}\right\}\end{array}$

Conclusion:

Thus, the power dissipated by $\text{4}\text{.7 k}\Omega$ resistor connected between $a$ and $b$ is $1.68\mu \text{W}$.

(c)

To determine

The power dissipated by $\text{10}\text{.54 k}\Omega$ resistor connected between $a$ and $b$.

Answer to Problem 41E

The power dissipated by $\text{10}\text{.54 k}\Omega$ resistor connected between $a$ and $b$ is $1.199\mu \text{W}$.

Explanation of Solution

Given Data:

The load resistance is $\text{10}\text{.54 k}\Omega$.

Calculation:

Refer to the redrawn Figure 3,

Substitute $2857.143\Omega$ for $R_{th}$, $\text{10}\text{.54 k}\Omega$ for $R_L$ and $0.1429\text{V}$ for $v$ in equation (7).

$\begin{array}{c}{i}_L=\frac{0.1429\text{V}}{2857.143\Omega +\text{10}\text{.54 k}\Omega }\\ =\frac{0.1429\text{V}}{2857.143\Omega +10.54\times {10}^3\Omega }\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =1.0667\times {10}^{-5}\text{A}\end{array}$

Substitute $1.0667\times {10}^{-5}\text{A}$ for $i_L$ and $\text{10}\text{.54 k}\Omega$ for $R_L$ in equation (1).

$\begin{array}{c}p={\left(1.0667\times {10}^{-5}\text{A}\right)}^2\left(\text{10}\text{.54 k}\Omega \right)\\ ={\left(1.0667\times {10}^{-5}\text{A}\right)}^2\left(10.54\times {10}^3\Omega \right)\left\{\because 1\text{ k}\Omega ={10}^3\Omega \right\}\\ =1.199\times {10}^{-6}\text{W}\\ =1.199\mu \text{W }\left\{\because {10}^{-6}\text{W}=1\mu \text{W}\right\}\end{array}$

Conclusion:

Thus, the power dissipated by $\text{10}\text{.54 k}\Omega$ resistor connected between $a$ and $b$ is $1.199\mu \text{W}$.