Chapter 5, Problem 33E

Determine the Norton equivalent of the circuit depicted in Fig. 5.74 as seen looking into the two open terminals. (b) Compute power dissipated in a 5 Ω resistor connected in parallel with the existing 5 Ω resistor. (c) Compute the current flowing through a short circuit connecting the two terminals.

To determine

Find the Norton equivalent of the circuit as seen looking into the two open terminals.

Answer to Problem 33E

The Norton current is $363.58\text{mA}$ and Norton resistance is $1.774\text{ Ω}$.

Explanation of Solution

Formula used:

The expression for voltage is as follows.

$v=iR$ (1)

Here,

$v$ is the voltage,

$i$ is the current and

$R$ is value of resistance.

The expression for series combination of resistance is as follows.

$R=\left(R_1+R_2\right)$ (2)

Here,

$R$ is the total resistances and

$R_1$ and $R_2$ are the resistances in the circuit.

The expression for parallel combination of resistance is as follows.

$\frac{1}{R}=\left(\frac{1}{R_1}+{\frac{1}{R}}_2\right)$ (3)

Here,

$R$ is the total resistances and

$R_1$ and $R_2$ are the resistances in the circuit.

Calculation:

The circuit diagram is redrawn as shown in Figure 1.

Refer to redrawn Figure 1,

The expression for KVL in super mesh $BACEFDB$ is as follows.

$v_1+i_1{R}_1+i_2{R}_2+(i_2-i_3)R_3=0$ (4)

Here,

$v_1$ is voltage in loop $BACDB$,

$R_1$, $R_2$ and $R_3$ are the resistances in the circuit,

$i_1$ is the current flowing in loop $BACDB$,

$i_2$ is the current flowing in loop $DCEFD$ and

$i_3$ is the current flowing in loop $FEGHF$.

Substitute $2\text{ V}$ for $v_1$, $2\text{ Ω}$ for $R_1$, $5\text{ Ω}$ for $R_2$ and $1\text{ Ω}$ for $R_3$ in equation (4).

$\begin{array}{l}2\text{ V}+i_1\left(2\text{ Ω}\right)+i_2\left(5\text{ Ω}\right)+\left(i_2-i_3\right)\left(1\text{ Ω}\right)=0\\ 2\text{ V}+i_1\left(2\text{ Ω}\right)+i_2\left(5\text{ Ω}\right)+\left(1\text{ Ω}\right)i_2-\left(1\text{ Ω}\right)i_3=0\end{array}$

$2\text{ V}+i_1\left(2\text{ Ω}\right)+i_2\left(\text{6 Ω}\right)-\left(1\text{ Ω}\right)i_3=0$ (5)

The expression for KVL in mesh $FEGHF$ is as follows.

$(i_3-i_2)R_3+i_3{R}_4+v_2=0$ (6)

Here,

$R_3$ and $R_4$ are the resistances in the circuit,

$v_2$ is voltage in loop $FEGHF$,

$i_2$ is the current flowing in loop $DCEFD$ and

$i_3$ is the current flowing in loop $FEGHF$.

Substitute $1\text{ Ω}$ for $R_3$, $3\text{ Ω}$ for $R_4$ and $4\text{ V}$ for $v_2$ in the equation (6).

$\begin{array}{l}\left(i_3-i_2\right)\left(1\text{ Ω}\right)+i_3\left(3\text{ Ω}\right)+4\text{ V}=0\\ \left(1\text{ Ω}\right)i_3-\left(1\text{ Ω}\right)i_2+i_3\left(3\text{ Ω}\right)+4\text{ V}=0\end{array}$

$\left(\text{4 Ω}\right)i_3-\left(1\text{ Ω}\right)i_2+4\text{ V}=0$ (7)

The expression for current $i$ is as follows.

$i_4=i_2-i_1$ (8)

Here,

$i_4$ is current through independent current source,

$i_2$ is the current flowing in loop $DCEFD$ and

$i_1$ is the current flowing in loop $BACDB$.

Substitute $2\text{ A}$ for $i_4$ in equation (8).

$2\text{ A}=i_2-i_1$ (9)

Rearrange equations(5),(7) and (9).

$\begin{array}{c}2i_1+6i_2-i_3=-2\\ 0i_1-i_2+4i_3=-4\\ -i_1+i_2+0i_3=2\end{array}$

The equations so formed can be written in matrix form as,

$\left(\begin{array}{ccc}2& 6& -1\\ 0& -1& 4\\ -1& 1& 0\end{array}\right)\left(\begin{array}{c}{i}_1\\ i_2\\ i_3\end{array}\right)=\left(\begin{array}{c}-2\\ -4\\ 2\end{array}\right)$

Therefore, by Cramer’s rule,

The determinant of coefficient matrix is as follows.

$\begin{array}{c}\Delta =\left|\begin{array}{ccc}2& 6& -1\\ 0& -1& 4\\ -1& 1& 0\end{array}\right|\\ =-31\end{array}$

The 1^st determinant is as follows.

$\begin{array}{c}{\Delta }_1=\left|\begin{array}{ccc}-2& 6& -1\\ -4& -1& 4\\ 2& 1& 0\end{array}\right|\\ =58\end{array}$

The 2^nd determinant is as follows,

$\begin{array}{c}{\Delta }_2=\left|\begin{array}{ccc}2& -2& -1\\ 0& -4& 4\\ -1& 2& 0\end{array}\right|\\ =-4\end{array}$

The 3rd determinant is as follows.

$\begin{array}{c}{\Delta }_3=\left|\begin{array}{ccc}2& 6& -2\\ 0& -1& -4\\ -1& 1& 2\end{array}\right|\\ =30\end{array}$

Simplify for $i_1$.

$\begin{array}{c}{i}_1=\frac{{\Delta }_1}{\Delta }\\ =\frac{58}{-31}\\ =-1.87\text{ A}\end{array}$

Simplify for $i_2$.

$\begin{array}{c}{i}_2=\frac{{\Delta }_2}{\Delta }\\ =\frac{-4}{-31}\\ =0.129\text{ A}\\ =1\text{29 mA }\left\{\because \text{ 1 A}={\text{10}}^3\text{ mA}\right\}\end{array}$

Simplify for $i_3$.

$\begin{array}{c}{i}_3=\frac{{\Delta }_3}{\Delta }\\ =\frac{30}{-31}\\ =-0.9677\text{ A}\end{array}$

Substitute $129\text{mA}$ for $i$ and $5\text{ Ω}$ for $R$ in equation (1).

$\begin{array}{c}v=\left(129\text{mA}\right)\times \left(5\text{ Ω}\right)\\ =\left(129\times {10}^{-3}\text{A}\right)\times \left(5\text{ Ω}\right)\left\{\because 1\text{mA}={10}^{-3}\text{A}\right\}\\ =645\times {10}^{-3}\text{V }\\ =645\text{mV}\left\{\because {10}^{-3}\text{V}=1\text{mV}\right\}\end{array}$

So, the Thevenin voltage is $645\text{mV}$.

The circuit diagram is redrawn as shown in Figure 2.

As $R_3$ and $R_4$ are in parallel therefore from equation (3).

$\begin{array}{c}\frac{\text{1}}{R_5}=\left(\frac{\text{1}}{R_3}\text{+}\frac{\text{1}}{R_4}\right)\\ \frac{1}{R_5}=\left(\frac{\text{1}}{1\text{Ω}}\text{+}\frac{\text{1}}{3\text{Ω}}\right)\\ \frac{1}{R_5}=\frac{\text{1}}{0.75\text{ Ω}}\end{array}$

Rearrange for $R_5$,

$R_5=0.75\text{ Ω}$

The circuit diagram is redrawn as shown in Figure 3.

As $R_1$ and $R_5$ are in series therefore from equation (2).

$\begin{array}{c}{R}_6=R_1+R_5\\ =2\Omega +0.75\Omega \\ =2.75\Omega \end{array}$

The circuit diagram is redrawn as shown in Figure 4.

Refer to redrawn Figure 4,

As $R_2$ and $R_6$ are in parallel therefore from equation (3),

$\begin{array}{c}\frac{\text{1}}{R_{th}}\text{=}\left(\frac{\text{1}}{R_2}\text{+}\frac{\text{1}}{R_6}\right)\\ \frac{\text{1}}{R_{th}}\text{=}\left(\frac{\text{1}}{5\text{Ω}}\text{+}\frac{\text{1}}{2.75\text{Ω}}\right)\\ \frac{\text{1}}{R_{th}}=\frac{\text{1}}{1.774\text{ Ω}}\end{array}$

Rearrange for $R_{th}$,

$R_{th}=1.774\text{ Ω}$

So, the Thevenin equivalent resistance across the branch $XY$ is $1.774\text{ Ω}$.

Thevenin equivalent resistance is same as the Norton equivalent resistance,

Hence,

$R_N=1.774\Omega$

The circuit diagram is redrawn as shown in Figure 5,

Refer to redrawn Figure 5,

Norton current is the current in the load resistor when load resistance is replaced by a short circuit.

The expression for the Norton current flowing in the circuit is as follows,

$I_N=\frac{V_{th}}{R_{th}}$ (10)

Here,

$I_N$ is the Norton current,

$V_{th}$ is the Thevenin voltage and

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

Substitute $1.774\text{ Ω}$ for $R_{th}$ and $645\text{mV}$ for $V_{th}$ in equation (10),

$\begin{array}{c}{I}_N=\frac{\left(645\text{mV}\right)}{\left(1.774\text{ Ω}\right)}\\ =363.58\text{mA}\end{array}$

Conclusion:

Thus the Norton current is $363.58\text{mA}$ and Norton resistance is $1.774\text{ Ω}$.

To determine

Find the power dissipated in a $5\Omega$ resistor connected in parallel with the existing $5\Omega$ resistor.

Answer to Problem 33E

The power dissipated in a $5\Omega$ resistor connected in parallel with the existing $5\Omega$ resistor is $\text{45}\text{.33}\text{m}\text{}\text{W}$.

Explanation of Solution

Given Data:

The load resistance is $5\Omega$.

Formula used:

The expression for the power dissipated in a resistor is as follows,

$p={\left(i_L\right)}^2{R}_L$ (`11)

Here,

$p$ is the power dissipated in a resistor,

$R_L$ is the load resistance and

$i_L$ is the current flowing through the load.

Calculation:

The circuit is drawn as shown in Figure 6,

Refer to redrawn Figure 6,

The simplified Norton equivalent of the circuit is drawn as shown in Figure 7,

Refer to redrawn Figure 7,

The expression for the current flowing through the load is as follows,

$i_L=I_N\left(\frac{R_N}{R_N+R_L}\right)$ (12)

Substitute $363.58\text{mA}$ for $I_N$, $1.774\text{ Ω}$ for $R_N$ and $5\Omega$ for $R_L$ in equation (12),

$\begin{array}{c}{i}_L=\left(363.58\text{mA}\right)\left(\frac{1.774\text{ Ω}}{\left(1.774\text{ Ω}\right)+\left(\text{5 Ω}\right)}\right)\\ =\left(363.58\text{mA}\right)\left(0.26188\right)\\ =95.22\text{mA}\end{array}$

Substitute $95.22\text{mA}$ for $i_L$ and $5\Omega$ for $R_L$ in equation (11),

$\begin{array}{c}p={\left(95.22\text{mA}\right)}^2\left(5\Omega \right)\\ ={\left(95.22\times {10}^{-3}\text{A}\right)}^2\left(5\Omega \right)\left\{\because 1\text{mA}={10}^{-3}\text{A}\right\}\\ =\left(9066.023\times {10}^{-6}\right)\left(5\right)\text{W}\\ =\text{45330}\text{.115}\times {10}^{-6}\text{}\text{W}\end{array}$

$p=\text{45}\text{.33}\text{m}\text{}\text{W}\left\{\because {10}^{-6}\text{A}={10}^{-3}\text{mA}\right\}$

Conclusion:

Thus, the power dissipated in a $5\Omega$ resistor connected in parallel with the existing $5\Omega$ resistor is $\text{45}\text{.33}\text{m}\text{}\text{W}$.

To determine

Find the current flowing through a short circuit connecting the two terminals.

Answer to Problem 33E

The current flowing through a short circuit connecting the two terminals is $363.58\text{ mA}$.

Explanation of Solution

Calculation:

The current flowing through a short circuit connecting the two terminals is Norton current which is $363.58\text{mA}$ as calculated in sub-part (a).

So, the current flowing through a short circuit connecting the two terminals is $363.58\text{mA}$

Conclusion:

Thus, the current flowing through a short circuit connecting the two terminals is $363.58\text{mA}$.