Chapter 3, Problem 3.65HP

Find the Thé’cnin equivalent resistance seen by $R_3$ in Figure P3.23. Compute the Thévenin (open-circuit)voltage $V_T$ and the Norton (short-circuit) current $I_N$ from node a o node b when $R_3$ is the load.

To determine

The Thevenin equivalent resistance seen by $R_3$. Also the Thevenin open circuit voltage $V_T$ and the Norton current $I_N$ from node $a$ to node $b$ when $R_3$ is the load.

Answer to Problem 3.65HP

The Thevenin equivalent resistance seen by the resistance $R_3$ is $-0.5437\text{V}$ and value of Thevenin voltage is $-4.7142\text{V}$ and the value of the Norton current between the nodes $a$ and $b$ is $-0.5437\text{mA}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

To calculate the Thevenin resistance, open circuit current source and short circuit the voltage source, the redraw the circuit.

The required diagram is shown in Figure 2

  

From above the equivalent resistance of the circuit is evaluated as,

  $\begin{array}{c}{R}_{Th}=\frac{\left(12\text{k}\Omega \right)\left(12\text{k}\Omega \right)}{\left(12\text{k}\Omega \right)+\left(12\text{k}\Omega \right)}+\frac{\left(3\text{k}\Omega \right)\left(24\text{k}\Omega \right)}{\left(3\text{k}\Omega \right)+\left(24\text{k}\Omega \right)}\\ =8.67\text{k}\Omega \end{array}$

Mark the values and redraw the given circuit.

The required diagram is shown in Figure 3

  

From the above figure the node voltage $V_1$ is given by,

  $V_1=24\text{V}$

The expression for the voltage $V_{R_3}$ is given by,

  $V_{R_3}=V_2-V_3$ ....... (1)

Apply KVL at node $V_3$

  $\frac{V_3-V_2}{10}+\frac{V_3-24}{3}+\frac{V_3}{24}=0$

Subsitue $V_2-V_3$ for $V_{R_3}$ in the above equation.

  $\frac{-V_{R_3}}{10}+\frac{V_3-24}{3}+\frac{V_3}{24}=0$

Substitute $-2.524\text{V}$ for $V_{R_3}$ in the above equation.

  $\begin{array}{l}\frac{-2.524\text{V}}{10}+\frac{V_3-24}{3}+\frac{V_3}{24}=0\\ -0.2524+\frac{V_3}{3}-8+0.375V_3=0\\ V_3=20.66\text{V}\end{array}$

Substitute $20.66\text{V}$ for $V_3$ and $-2.524\text{V}$ for $V_{R_3}$ in equation (1)

  $\begin{array}{l}-2.524\text{V}=V_2-20.66\text{V}\\ V_2=18.13\text{V}\end{array}$

Apply KVL to the node $V_2$ .

  $\frac{V_2-V_3}{10}+\frac{V_2-V_{S2}}{12}+\frac{V_2-24}{12}=0$

Substitute $18.13\text{V}$ for $V_2$ and $20.66\text{V}$ for $V_3$ in the above equation.

  $\begin{array}{l}\frac{18.13\text{V}-20.66\text{V}}{10}+\frac{18.13{\text{V}}_2-V_{S2}}{12}+\frac{18.13\text{V}-24}{12}=0\\ -0.2524+1.5108-\frac{V_{S2}}{12}-0.4891=0\\ \frac{V_{S2}}{12}=0.7693\\ V_{S2}=9.2316\text{V}\end{array}$

To obtain the equivalent Thevenin resistance of the circuit open circuit the load terminals, short circuit the voltage source and redraw the circuit.

The required diagram is shown in Figure 4

  

From the above circuit, the equivalent resistance is calculated as,

  $\begin{array}{c}{R}_{Th}=\left(\frac{\left(12\text{k}\Omega \right)\left(12\text{k}\Omega \right)}{\left(12\text{k}\Omega \right)+\left(12\text{k}\Omega \right)}\right)+\left(\frac{\left(3\text{k}\Omega \right)\left(24\text{k}\Omega \right)}{\left(3\text{k}\Omega \right)+\left(24\text{k}\Omega \right)}\right)\\ =8.67\text{k}\Omega \end{array}$

To obtain the Thevenin equivalent voltage between the terminals, open circuit the load terminals and redraw the circuit.

The required diagram is shown in Figure 5

  

From the above circuit, the node voltage $v_b$ is calculated as,

  $\begin{array}{c}{v}_b=\left(\frac{24\text{k}\Omega }{24\text{k}\Omega +3\text{k}\Omega }\right)\left(24\text{V}\right)\\ =21.33\text{V}\end{array}$

The current through the resistance of $12\text{k}\Omega$ is calculated as,

  $\begin{array}{c}{i}_{12\text{k}\Omega }=\frac{24\text{V}-9.2316\text{V}}{12\text{k}\Omega +12\text{k}\Omega }\\ =0.61535\text{mA}\end{array}$

From above circuit, the node voltage $v_a$ is calculated as,

  $v_a=i_{12\text{k}\Omega }\left(12\text{k}\Omega \right)+9.2316\text{V}$

Substitute $0.61535\text{mA}$ for $i_{12\text{k}\Omega }$ in the above equation.

  $\begin{array}{c}{v}_a=\left(0.61535\text{mA}\right)\left(12\text{k}\Omega \right)+9.2316\text{V}\\ =\text{7}\text{.3842}+\text{9}\text{.2316}\\ =16.6158\text{V}\end{array}$

The expression to calculate the Thevenin equivalent voltage is given by,

  $v_{Th}=v_a-v_b$

Substitute $16.6158\text{V}$ for $v_a$ and $21.33\text{V}$ for $v_b$ in the above equation.

  $\begin{array}{c}{v}_{Th}=16.6158\text{V}-21.33\text{V}\\ =-4.7142\text{V}\end{array}$

The expression to calculate the Norton equivalent current is given by,

  $i_N=\frac{v_{Th}}{R_{Th}}$

Substitute $-4.7142\text{V}$ for $v_{Th}$ and $8.67\text{k}\Omega$ for $R_{Th}$ in the above equation.

  $\begin{array}{r}{i}_N=\frac{-4.7142\text{V}}{8.67\text{k}\Omega }\\ =-0.5437\text{mA}\end{array}$

Conclusion:

Therefore, the Thevenin equivalent resistance seen by the resistance $R_3$ is $-0.5437\text{V}$ and the value of Thevenin voltage is $-4.7142\text{V}$ and the value of the Norton current between the nodes $a$ and $b$ is $-0.5437\text{mA}$ .