Chapter 3, Problem 3.62HP

Find the Thé venin equivalent resistance seen byresistor $R_3$ in the circuit of Figure P3.5. Compute the Thé venin (open-circuit) voltage and the Norton(short-circuit) current from node A to node B when $R_3$ is the load.

To determine

The Thevenin equivalent resistance seen by the resistor $R_3$. Also the Thevenin(open circuit) voltage and the Norton current from node $A$ to node $B$ when the load is $R_3$ .

Answer to Problem 3.62HP

The Thevenin equivalent resistance is $\frac{R_1{R}_4}{R_1+R_4}+\frac{R_2{R}_5}{R_2+R_5}$, Thevenin open circuit voltage is $\left(\frac{R_1}{R_1+R_4}\right)V_{S1}+\left(\frac{R_2}{R_2+R_5}\right)V_{S2}$ and Norton short circuit current is $\frac{\frac{\left(V_{S1}\right)}{R_4}{R}_{eq1}+\frac{\left(V_{S2}\right)}{R_5}{R}_{eq2}}{R_{eq1}+R_{eq2}}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

 

To calculate the value of Thevenin equivalent resistance, short circuit the voltage source and open circuit the current source, open circuit the load resistance $R_3$ then redraw the circuit.

The required diagram is shown in Figure 2

 

From the above figure the expression for the equivalent resistance of the circuit is evaluated as,

  $R_{eq}=\frac{R_1{R}_4}{R_1+R_4}+\frac{R_2{R}_5}{R_2+R_5}$

The expression for the Thevenin resistance of the circuit is given by,

  $R_{Th}=R_{eq}$

Substitute $\frac{R_1{R}_4}{R_1+R_4}+\frac{R_2{R}_5}{R_2+R_5}$ for $R_{eq}$ in the above equation.

  $R_{Th}=\frac{R_1{R}_4}{R_1+R_4}+\frac{R_2{R}_5}{R_2+R_5}$

To calculate the open circuit voltage of the circuit, open circuit the load resistance and redraw the circuit.

The required diagram is shown in Figure 3

 

The expression to evaluate the voltage across the resistance $V_{R1}$ is given by,

  $V_{R1}=\left(\frac{R_1}{R_1+R_4}\right)V_{S1}$

The expression to evaluate the voltage across the resistance $V_{R2}$ is given by,

  $V_{R2}=\left(\frac{R_2}{R_2+R_5}\right)V_{S2}$

The expression for the open circuit voltage is given by,

  $V_{OC}=V_{R1}+V_{R2}$

Substitute $\left(\frac{R_1}{R_1+R_4}\right)V_{S1}$ for $V_{R1}$ and $\left(\frac{R_2}{R_2+R_5}\right)V_{S2}$ for $V_{R2}$ in the above equation.

  $V_{OC}=\left(\frac{R_1}{R_1+R_4}\right)V_{S1}+\left(\frac{R_2}{R_2+R_5}\right)V_{S2}$

Mark the values and draw the Thevenin equivalent seen by the loadresistance $R_3$ .

The required diagram is shown in Figure 4

 

The expression for the Norton equivalent resistance is given by,

  $R_N=R_{Th}$

Substitute $\frac{R_1{R}_4}{R_1+R_4}+\frac{R_2{R}_5}{R_2+R_5}$ for $R_{Th}$ in the above equation.

  $R_N=\frac{R_1{R}_4}{R_1+R_4}+\frac{R_2{R}_5}{R_2+R_5}$

To find the Norton current, short circuit the load resistance and redraw the circuit.

The required diagram is shown in Figure 5

 

To obtain the Norton current convert the voltage sources of the above circuit into current source and redraw the circuit..

The required diagram is shown in Figure 6

 

The resistance $R_4$ and $R_1$ are in parallel combination and the equivalent resistance is given by,

  $R_{eq1}=\frac{R_1}{R_1+R_4}$

The resistance $R_2$ and $R_5$ are in parallel combination and the equivalent resistance for the combination is given by,

  $R_{eq2}=\frac{R_2}{R_2+R_5}$

The expression for the Norton current is given by,

  $I_N=\frac{\frac{\left(V_{S1}\right)}{R_4}{R}_{eq1}+\frac{\left(V_{S2}\right)}{R_5}{R}_{eq2}}{R_{eq1}+R_{eq2}}$

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

The required diagram is shown in Figure 7

 

Conclusion:

Therefore, the Thevenin equivalent resistance is $\frac{R_1{R}_4}{R_1+R_4}+\frac{R_2{R}_5}{R_2+R_5}$, Thevenin open circuit voltage is $\left(\frac{R_1}{R_1+R_4}\right)V_{S1}+\left(\frac{R_2}{R_2+R_5}\right)V_{S2}$ and Norton short circuit current is $\frac{\frac{\left(V_{S1}\right)}{R_4}{R}_{eq1}+\frac{\left(V_{S2}\right)}{R_5}{R}_{eq2}}{R_{eq1}+R_{eq2}}$ .