E E 2315 Circuits I Lecture 7 - UT Arlington - UTA

E E 2315 Circuits I Lecture 7 - UT Arlington - UTA

E E 2315 Lecture 07 - Network Theorems, Maximum Power Transfer Theorem Given a Thvenin equivalent circuit of a network, the maximum power that can be transferred to an external load resistor is obtained if the load resistor equals the Thvenin resistance of the network. Any load resistor smaller or larger than the Thvenin resistance will receive less power. Maximum Power Demonstrated (1/2) R Th VTh a I

VTh I RTh RL RL b dPL 0 PL is max if dRL and 2 Th PL V RL

RTh RL 2 Maximum Power Demonstrated (2/2) R Th VTh a I RL b dPL 0 dRL RL RTh

and Maximum Power Transfer Theorem applied when matching loads to output resistances of amplifiers Efficiency is 50% at maximum power transfer Superposition Theorem (1/2) In a linear system, the linear responses of linear independent sources can be combined in a linear manner. This allows us to solve circuits with one independent source at a time and then combine the solutions. If an independent voltage source is not present it is replaced by a short circuit. If an independent current source is not present it is replaced by an open circuit. Superposition Theorem (2/2)

If dependent sources exist, they must remain in the circuit for each solution. Nonlinear responses such as power cannot be found directly by superposition Only voltages and currents can be found by superposition Superposition Example 1 (1/4) 20 10 V a I1 5 I2 5A

b Find I1, I2 and Vab by superposition Superposition Example 1 (2/4) Step 1: Omit current source. 20 10 V a I 11 I 21 5 5A b

By Ohms law and the voltage divider rule: I11 I 21 0.4 A Superposition Example 1 (3/4) Step 2: Omit voltage source. 20 a I 12 I 22 5 b By the current divider rule and Ohms law :

5A I12 1.0 A Vab 2 20V Superposition Example 1 (4/4) 20 10 V a I1 5 I2 5A

b Combining steps 1 & 2, we get: I1 I11 I12 0.6 A Vab Vab1 Vab 2 22V Superposition Example 2 (1/5) 16 16 A 6 Ix 64 V 16 A Find Ix by superposition

10 Superposition Example 2 (2/5) 16 16 A 6 I xa 16 A 10 Activate only the 16 A Current source at the left. Then use Current Divider Rule: 10 16 I xa

16 A 13 A 6 10 16 Superposition Example 2 (3/5) 16 16 A 6 I xb 16 A 10 Activate only the 16 A Current source at the right. Then use Current Divider Rule: 10 I xb

16 A 5 A 6 10 16 Superposition Example 2 (4/5) 16 16 A 6 I xc 64 V 16 A 10 Activate only the 64 V voltage source at the bottom. Then use Ohms Law: 64V

I xc 2 A 6 10 16 Superposition Example 2 (5/5) 16 16 A 6 Ix 64 V 16 A 10 Sum the partial currents due to each of the sources: I x I xa I xb I xc 13 A 5 A 2 A

Superposition Example 3 (1/4) 4 12 V a ix 2 1.5ix 30 V b Solve for Vab by Superposition method Dependent source must remain in circuit for both steps

Superposition Example 3 (2/4) 4 a ix 2 1.5ix 30 V b Vab1 Vab1 30 0 1.5ix 4 2

1 1 1 9 15 Vab1 1.5 Vab1 4 2 8 4 40 Vab1 V 3 Superposition Example 3 (3/4) 4 12 V a ix 2 1.5ix

b Vab 2 Vab 2 12 0 1.5ix 2 4 1 1.5 1 9 7.5 Vab 2 Vab 2 2 4 4 8 Vab 2 20 3

Superposition Example 3 (4/4) 4 12 V a ix 2 1.5ix b 30 V Vab Vab1 Vab 2 40 20 Combining the solutions: Vab V V

3 3 Vab

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