Thevenin theorem

Thevenin theorem
statement
numerical problems

1. Thevenin’s Theorem By- Rajni Maurya MITS, Gwalior [email protected]

2. Thevenin’s Theorem Statement Thévenin’s theorem states the following: Any two-terminal dc network can be replaced by an equivalent circuit consisting solely of a voltage source and a series resistor as shown below. Fig. 1 Thévenin equivalent circuit

3. Thévenin’s Theorem Procedure PRELIMINARY: STEP 1. Remove that portion of the network where the Thévenin equivalent circuit is found. In Fig. 2 (a), this requires that the load resistor be temporarily removed from the network. STEP 2. Mark the terminals of the remaining two-terminal network.   : STEP 3. Calculate by first setting all sources to zero (voltage sources are replaced by short circuits and current sources by open circuits) and then finding the resultant resistance between the two marked terminals.  

4. Thévenin’s Theorem Procedure : STEP 4. Calculate by first returning all sources to their original position and finding the open-circuit voltage between the marked terminals.   Conclusion: STEP 5 . Draw the Thévenin equivalent circuit with the portion of the circuit previously removed replaced between the terminals of the equivalent circuit. This step is indicated by the placement of the resistor between the terminals of the Thévenin equivalent circuit as shown in Fig. 2 (b)  

5. Fig. 2 (a) & (b) Substituting the Thévenin equivalent circuit for a complex network Thévenin’s Theorem Procedure

6. Example 1. Find the Thévenin equivalent circuit for the network in the shaded area of the network in Fig. 3. Then find the current through RL for values of 2 Ω & 10 Ω . Fig. 3

7. Solution: Steps 1 and 2: These produce the network in Fig. 4. Note that the load resistor has been removed and the two terminals are a and b.   Fig. 4 Example 1 .

8.   Fig.5 Determining for the network   Step 3: Replacing the voltage source E 1 with a short-circuit equivalent yields the network in Fig. 5, where   Example 1 .

9. Step 4: Replace the voltage source (Fig.6) . For this case, the open circuit voltage is the same as the voltage drop across the 6 Ω resistor. Applying the voltage divider rule gives   Fig.6: Determining for the network in fig 4   Example 1 .

10. ; ;   Fig.7: Substituting the Thévenin equivalent circuit for the network external to RL in fig 3 Example 1 .

11. EXAMPLE 2. Find the Thévenin equivalent circuit for the network in the shaded area of the network in Fig. 8 Fig. 8

12. Steps 1 and 2: The load resistor has been removed and the two terminals are a and b.   Fig. 9 Example 2 . Solution:

13. Step 3: See Fig.10. The current source has been replaced with an open-circuit equivalent and the resistance determined between terminals a and b. R 1 and R 2 are in series and the Thévenin resistance is the sum of the two, = R 1 + R 2 = 4 + 2 = 6 Ω   Fig.10 Determining RTh for the network Example 2 .

14. = (0) R 2 = 0 V and = = (12 A)(4 Ω ) = 48 V   Fig. 11 Determining for the network   Example 2 . Step 4: See Fig. 11. In this case, since an open circuit exists between the terminals a & b, the current is zero between these terminals and through the 2 Ω resistor. The voltage drop across R 2 is, therefore,

15. Step 5: See Fig. 12 Fig. 12 Substituting the Thévenin equivalent circuit in the network external to the resistor R 3 in Fig. 8 Example 2 .

16. Example 3 : F ind the Thevenin equivalent of the circuit. Solution : In order to find the Thevenin equivalent circuit for the circuit shown in Figure 1 3 , calculate the open circuit voltage, Vab. Note that when the a, b terminals are open, there is no current flow to 4Ω resistor. Therefore, the voltage vab is the same as the voltage across the 3A current source, labeled v1. To find the voltage v1, solve the equations for the singular node voltage. By choosing the bottom right node as the reference node, 25 V 20 + - v 1 3 A 5 4 + v ab - b a Fig. 13

17. By solving the equation, V 1 = 32 V. Therefore, the Thevenin voltage V Th for the circuit is 32 V. The next step is to short circuit the terminals and find the short circuit current for the circuit shown in Figure 14 . ( Note that the current is in the same direction as the falling voltage at the terminal. ) 3 0 5 20 v 1 25 v 1 25 V 20 + - v 2 3 A 5 - + v ab 4 a b i sc Figure 14 Example 3

18. 4 v 2 3 20 v 2 25 v 2 5 Current i sc can be found if v 2 is known. By using the bottom right node as the reference node, the equationfor v 2 becomes By solving the above equation, v 2 = 16 V. Therefore, the short C ircuit curr ent i sc The Thevenin resistance R Th is Example 3

19. Figure 15 : Thevenin equivalent circuit for the Figure 1 Example 3

20. THANKS