Chapter 3, Problem 3.71HP

To determine

The Norton and Thevenin equivalent networks from node $a$ to $b$ .

Answer to Problem 3.71HP

The Thevenin equivalent Network is shown in Figure 6 and the Norton equivalent network is shown in Figure 4

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

 

To determine the Norton equivalent, short circuit the output terminals, Mark the values, current direction, and redraw the circuit.

The required diagram is shown in Figure 2

 

Apply KVL to the first loop.

  $\begin{array}{l}12\left(I_1-3\right)+5I_1=0\\ 17I_1-36=0\\ I_1=2.12\text{A}\end{array}$   ....... (1)

Apply KVL to the second loop.

  $\begin{array}{l}10\left(I_2-3\right)+2I_2=0\\ 12I_2-30=0\\ I_2=2.5\text{A}\end{array}$

The equation for the Norton current is given by,

  $I_N=I_2-I_1$

Substitute $2.12\text{A}$ for $I_1$ and $2.5\text{A}$ for $I_1$ in the above equation.

  $\begin{array}{c}{I}_N=2.5\text{A}-2.12\text{A}\\ =\text{0}\text{.38}\text{A}\end{array}$

To obtain the Norton resistance of the circuit, short circuit the voltage source, open circuit the current source and redraw the circuit.

The required diagram is shown in Figure 3

 

Form the above circuit the Norton impedance of the circuit is calculated as,

  $Z_N=\left(R_1+R_3\right)\left|\right|\left(R_2+R_3\right)$

Substitute $12\Omega$ for $R_1$, $10\Omega$ for $R_2$, $5\Omega$ for $R_3$ and $2\Omega$ for $R_4$ in the above equation.

  $\begin{array}{c}{Z}_N=\left(12\Omega +5\Omega \right)\left|\right|\left(10\Omega +2\Omega \right)\\ =\left(17\Omega \right)\left|\right|\left(12\Omega \right)\\ =7.03\Omega \end{array}$

Mark the values and draw the Norton equivalent circuit.

The required diagram is shown in Figure 4

 

To calculate the Thevenin voltage, remove the load resistance and redraw the circuit.

The required diagram is shown in Figure 5

 

From the above circuit the value of the current $I_1$ is given by,

  $I_1=I_S\left(\frac{R_3+R_4}{R_1+R_2+R_3+R_4}\right)$

Substitute $3\text{A}$ for $I_S$ $12\Omega$ for $R_1$, $10\Omega$ for $R_2$, $5\Omega$ for $R_3$ and $2\Omega$ for $R_4$ in the above equation.

  $\begin{array}{c}{I}_1=\left(3\text{A}\right)\left(\frac{5\Omega +10\Omega }{12+10+5\Omega +2\Omega }\right)\\ =0.724\text{A}\end{array}$

The expression for the value of the current $I_2$ is given by,

  $I_2=I_S\left(\frac{R_1+R_2}{R_1+R_2+R_3+R_4}\right)$

Substitute $3\text{A}$ for $I_S$

  $12\Omega$ for $R_1$, $10\Omega$ for $R_2$, $5\Omega$ for $R_3$ and $2\Omega$ for $R_4$ in the above equation.

  $\begin{array}{c}{I}_2=\left(3\text{A}\right)\left(\frac{12\Omega +10\Omega }{12+10+5\Omega +2\Omega }\right)\\ =2.276\text{A}\end{array}$

The expression for the voltage across the resistor $R_2$ is given by,

  $V_{R2}=I_1{R}_2$

Substitute $10\Omega$ for $R_2$ and $0.724\text{A}$ for $I_1$ in the above equation.

  $\begin{array}{c}{V}_{R2}=\left(0.724\text{A}\right)\left(10\Omega \right)\\ =7.24\text{V}\end{array}$

The expression for the voltage across the resistor $R_4$ is given by,

  $V_{R4}=I_2{R}_4$

Substitute $2\Omega$ for $R_4$ and $2.276\text{A}$ for $I_2$ in the above equation.

  $\begin{array}{c}{V}_{R4}=\left(2.276\text{A}\right)\left(2\Omega \right)\\ =4.552\text{V}\end{array}$

From Figure (5), the expression for the Thevenin voltage is given by,

  $V_{Th}=V_{R2}-V_{R4}$

Substitute $4.552\text{V}$ for $V_{R4}$ and $7.24\text{V}$ for $V_{R2}$ in the above equation.

  $\begin{array}{c}{V}_{Th}=7.24\text{V}-4.552\text{V}\\ =\text{2}\text{.668}\text{V}\end{array}$

The Thevenin and the Norton resistance of the circuit are equal, thus the expression for the Thevenin resistance is given by,

  $Z_{Th}=Z_N$

Substitute $7.03\Omega$ for $Z_N$ in the above equation.

  $Z_{Th}=7.03\Omega$

Mark the values and draw the Thevenin equivalent of the circuit.

The required diagram is shown in Figure 6

 

Conclusion:

Therefore, the Thevenin equivalent Network is shown in Figure 6 and the Norton equivalent network is shown in Figure 4