Lecture - 6 Circuit Theorems - PowerPoint Presentation

1. Lecture - 6 Circuit Theorems 1

2. Outline Source transformation. Thevenin and Norton Equivalents. Maximum Power Transfer. Superposition2

3. Source transformation Source transformations allow us to exchange a voltage source (vs) and a series resistor (R) for a current source (is) and a parallel resistor (R) and vice versa.The combinations must be equivalent in terms of their terminal voltage and current. Terminal equivalence holds provided that:3

4. Example 1a) For the circuit shown, find the power associated with the 6 V source.b) State whether the 6 V source is absorbing or delivering the power calculated in (a).-----------------------------------------a) We must reduce the circuit in a way that preserves the identity of the branch containing the 6 V source. We have no reason to preserve the identity of the branch containing the 40 V source. The circuit reduction is shown in the following figures: 4

5. Example 1The current in the direction of the voltage drop across the 6 V source is (19.2 - 6)/16, or 0.825 A. Therefore, the power associated with the 6 V source is:p6V = (0.825)(6) = 4.95 W.b) The voltage source is absorbing power.5

6. Example 2a) Use source transformations to find the voltage vo in the circuit shown in the figure b) Find the power developed by the 250 V voltage source.c) Find the power developed by the 8 A current source.6

7. Example 2We begin by removing the 125 ꭥ and 10 ꭥ resistors, because the 125 ꭥ resistor is connected across the 250 V voltage source and the 10 ꭥ resistor is connected in series with the 8 A current source. We also combine the series-connected resistors into a single resistance of 20 ꭥ . 7We now use a source transformation to replace the 250 V source and 25 ꭥ resistor with a 10 A source in parallel with the 25 ꭥ resistor.

8. Example 2We can now simplify the circuit by using Kirchhoffs current law to combine the parallel current sources into a single source. The parallel resistors combine into a single resistor.Hence, vo =20Vb) The current supplied by the 250 V source equals the current in the 125 ꭥ resistor plus the current in the 25 ꭥ resistor. Thus: = 11.2 ATherefore, the power developed by the voltage source isp250v(developed) = (250)(11.2) = 2800 W. 8

9. Example 2c) To find the power developed by the 8 A current source, we first find the voltage across the source. If we let vs represent the voltage across the source, positive at the upper terminal of the source, we obtainvs + 8(10) = vovo = 20, or vs = -60 V,and the power developed by the 8 A source is 480 W. Note that the 125 ꭥ and 10 ꭥ resistors do not affect the value of vo but do affect the power calculations.9

10. Thevenin and Norton Equivalents Thevenin equivalents and Norton equivalents allow us to simplify a circuit comprised of sources and resistors into an equivalent circuitThe equivalent circuit will be consisting of a voltage source and a series resistor (Thevenin) or a current source and a parallel resistor (Norton). 10

11. Thevenin and Norton Equivalents The simplified circuit and the original circuit must be equivalent in terms of their terminal voltage and current. Thus, keep in mind that : (1) the Thevenin voltage (VTh) is the open-circuit voltage across the terminals of the original circuit, (2) the Thevenin resistance (RTh) is the ratio of the Thevenin voltage to the short-circuit current across the terminals of the original circuit; (3) the Norton equivalent is obtained by performing a source transformation on a Thevenin equivalent. 11

12. Example 3Find the Thevenin equivalent for the circuit shown.--------------------------------------------- The current labeled ix must be zero. (Note the absence of a return path for ix to enter the left-hand portion of the circuit) The open-circuit, or Thevenin, voltage will be the voltage across the 25Ω resistor. With ix = 0,VTh = vab = (-20i)(25) = -500i- The current i is:12

13. Example 3To calculate the short-circuit current, we place a short circuit across a,b. When the terminals a,b are shorted together, the control voltage v is reduced to zero.With the short circuit shunting the 25 Ω resistor, all the current from the dependent current source appears in the short, soisc= -20i As the voltage controlling the dependent voltage source has been reduced to zero, the current controlling the dependent current source is:i=5/2000= 2.5mA13

14. Example 3- Combining these two equations yields a short-circuit current of : isc = -20(2.5) = -50 mA.- From isc and VTh we get:14

15. Maximum Power TransferMaximum power transfer is a technique for calculating the maximum value of p that can be delivered to a load, RL. Maximum power transfer occurs when RL = Rth, the Thevenin resistance as seen from the resistor RL.15

16. Example 4For the circuit shown, find the value of RL that results in maximum power being transferred to RL.Calculate the maximum power that can be delivered to RLc) When RL is adjusted for maximum power transfer, what percentage of the power delivered by the 360 V source reaches RL.16

17. Example 4The Thevenin voltage for the circuit to the left of the terminals a,b is The Thevenin resistance isReplacing the circuit to the left of the terminals a,b with its Thevenin equivalent gives us the circuit shown, which indicates that RL must equal 25 Ω for maximum power transfer. 17

18. Example 4b) The maximum power that can be delivered to RL is:c) When RL equals 25 Ω, the voltage vab is:- when vab equals 150 V, the current in the voltage source in the direction of the voltage rise across the source is:- Therefore, the source power is  ps=-is(360) = -2520W - The percentage of the source power delivered to the load is18

19. SuperpositionThe superposition principle states that whenever a linear system is excited, or driven, by more than one independent source of energy, the total response is the sum of the individual responses.In a circuit with multiple independent sources, superposition allows us to activate one source at a time and sum the resulting voltages and currents to determine the voltages and currents that exist when all independent sources are active. The current source is replaced with an open circuit and the voltage source by short circuit. Dependent sources are never deactivated when applying superposition.19

20. Example 5 Use the principle of superposition to find vo in the circuit.20

21. Example 5 We begin by finding the component of vo resulting from the 10 V source. With the 5 A source deactivated, v'∆ must equal to (-0.4 v'∆)(10).Hence, v'∆ must be zero, the branch containing the two dependent sources is open, andv'o=10*(20/(25))= 8V21

22. Example 5 Then we finding the component of vo resulting from the 5 A source by deactivating the 10 V source, Then solve the resulting circuit by node voltage method. Summing the currents away from node a yields:22

23. Example 5 Summing the currents away from node b gives:We now use vb = 2i’’ ∆ + v’’∆To find the value for v’’∆, , thus: 5 v’’∆, =50  v’’∆= 10V From the node a equation: 5v’’o = 80  v’’o = 16 VThe value of vo is the sum of v’o and v’’o or 24 V.23

24. SummarySource transformations allow us to exchange a voltage source (vs) and a series resistor (R) for a current source (is) and a parallel resistor (R) and vice versa.Thevenin equivalents and Norton equivalents allow us to simplify a circuit comprised of sources and resistors into an equivalent circuit.Maximum power transfer is a technique for calculating the maximum value of p that can be delivered to a load, RL. superposition allows us to activate one source at a time and sum the resulting voltages and currents to determine the voltages and currents that exist when all independent sources are active.24

25. Drill MCQ questionsSelect an option that makes the statement TRUE • In the superposition, the current source is replaced with ….. :an open circuit.a short circuit .a or b. D. None of the above. • In the superposition, the voltage source is replaced with ….. :an open circuit.a short circuit .a or b. D. None of the above25

26. Homeworka) Use a series of source transformations to find the voltage v in the circuit shown.b) How much power does the 120 V source deliver to the circuit?26

27. HomeworkFind the Norton equivalent circuit with respect to the terminals a,b for the circuit shown.27

28. Homeworka) Find the value of R that enables the circuit shown to deliver maximum power to the terminals a,b.b) Find the maximum power delivered to R.28