09-Circuit Theorems Text: , 4.8. ECEGR 210 Electric Circuits I

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1 09Circuit Theorems Text: , 4.8 ECEGR 210 Electric Circuits I

2 Overview Introduction Linearity Superposition Maximum Power Transfer Dr. Louie 2

3 Introduction Nodal and mesh analysis can be tedious to apply to large circuits, especially with multiple sources Linearity of the circuits considered can be exploited to simplify analysis Dr. Louie 3

4 Linearity Circuits considered in the this class are linear An equation f(x) or system is linear iff Homogeneity: f(ax) = af(x) Additivity: f(x y) = f(x) f(y) Dr. Louie 4

5 f(x) Linearity Is f(x) = 2x 1 linear? Check homogeneity: f(ax) = af(x) 2ax 1 a(2x 1) Check additivity : f(x y) = f(x) f(y) 2(x y) 1 2x 1 2y 1 No. This is an affine function x Dr. Louie 5

6 Linearity What about Ohm s Law (V=IR)? Let I be the argument so that f(x) = V(I) Check homogeneity: f(ax) = af(x) air a(ir) Check additivity: f(x y) = f(x) f(y) (Ia I b)r IaR IbR Yes. A circuit is linear if it consists only of linear elements, linear dependent sources and independent sources Dr. Louie 6

7 What about P=IV? Linearity Careful. V = IR, so P = I 2 R Check homogeneity: f(ax) = af(x) 2 2 (ai) R ai R Check additivity: f(x y) = f(x) f(y) (I I ) R I R I R No. The power dissipated by a resistor is not linear. Dr. Louie 7

8 Significance of Linearity The mesh currents for the shown circuit can be written as (by inspection): R1 R2 R 2 i1 V1 V2 R2 R2 R3 R 4 i 2 V 2 R 1 R 3 A (matrix) R 2 i 1 i 2 R 4 i1 V1 V2 A i V 2 2 i1 1 V1 V2 i A V 2 2 Solving for i 1, i 2 V 1 V 2 Dr. Louie 8

9 Observing that: Significance of Linearity i1 1 V1 V2 1 V V2 i A V A 0 A V R 1 R 3 R 1 R 3 V 1 R 2 i 1 i 2 R 4 We can solve two easier circuits, and add the results! R 2 i 1 i 2 V 2 R 4 Dr. Louie 9

10 Significance of Linearity Find the current out of the dependent voltage source if V s = 12V, and if V s = 24V Should we use nodal or mesh analysis? 2W 8W V x 4W 4W 6W V s 3V x Dr. Louie 10

11 since V s = 12V Significance of Linearity 0 12i 4i V (mesh 1) 1 2 s 0 4i 16i 3V (mesh 2) v 2i x x 2W 8W V x 4W 4W i 1 i2 6W V s 3V x Dr. Louie 11

12 Significance of Linearity V s = 12V 0 12i 4i V 1 2 s 0 10i 16i adding i 12i 0 i 1 2 6i 1 2 via substitution 76i V 0 2 s V s i A 6W 2W V x i 1 4W V s i2 8W 4W 3V x Dr. Louie 12

13 Significance of Linearity By linearity: if V s = 24V (voltage doubled) i 2 = 2x0.158A = 0.316A (current doubled) No need to resolve the circuit Behold, the power of linearity! 2W 8W V x 4W 4W i 1 i2 6W V s 3V x Dr. Louie 13

14 Superposition One of the most important circuit analysis benefits of linearity is that superposition can be applied Very useful for circuits with multiple independent sources Basic idea: analyze the circuit considering only one source at a time, and sum the results once all sources have been considered Valid for all circuit analysis techniques: Nodal, mesh, Ohm s Law, etc. Dr. Louie 14

15 Superposition Simple example with V 1 = 10V, V 2 = 15V Find I I V R V 1 2 1A I V 1 5W V 2 Dr. Louie 15

16 Superposition Now solve using superposition Consider V 1 only Remove contribution from V 2 (set source to 0, a short circuit) I V1 = 2A I V 1 5W V 2 = 0 Dr. Louie 16

17 Consider V 2 only Superposition Remove contribution from V 1 (set source to 0, a short circuit) I V2 = 3A I V 1 = 0 5W V 2 = 15 Dr. Louie 17

18 Combining Superposition I = I V2 I V2 = 1A Same result as solving the circuit simultaneously Dr. Louie 18

19 Superposition Analyze circuit with only one independent source at a time Independent voltage sources are ignored by replacing them with a short (voltage source where V s = 0) Independent current sources are ignored by replacing them with an open circuit (current source where I s = 0) Dependent sources are not ignored Sum individual results to analyze circuit Dr. Louie 19

20 Example Find V R using superposition. 8W 6V 4W V R 3A Dr. Louie 20

21 6V Example Consider the voltage source first. Set current source to 0A (open circuit) By KVL: 12i 1 6 = 0 i 1 = 0.5A V 1 = 2V (V R due to voltage source) 8W 4W V R 3A 6V 8W 4W i 1 V 1 Dr. Louie 21

22 Example Now consider the current source. Set voltage source to 0V (short circuit) By current division: i 2 = (3x8)/(12) = 2A V 2 = 8V (V R due to current source) 6V 8W 4W V R 3A 8W 4W i 2 V 2 3A Dr. Louie 22

23 Now sum results: V R = V 1 V 2 = 10V Note: i R = i 1 i 2 = 2.5A Example P R = i 2 R R = (i 1 i 2 ) 2 R =31.25W 2 2 P P P R 1 2 Any power calculations must be done after all sources considered Dr. Louie 23

24 Example Use superposition to find V x 20W 20V 4A V x 4W 0.1V x Dr. Louie 24

25 Dr. Louie 26

26 Example First consider voltage source. Use nodal analysis Trying to find node voltage Only one node with unknown voltage Many elements in parallel 20W i 1 i 2 20V V x1 4W 0.1V x1 Dr. Louie 28

27 KCL of node i i i x1 1 2 via substitution: Vx1 20 Vx1 0.1Vx V 0.4V 0.2V 4 V x1 x1 x1 x1 5V Example 20W i 1 i 2 20V i x1 V x1 4W 0.1V x1 Dr. Louie 29

28 Example Now consider current source. Rearrange parallel elements for clarity Use nodal analysis Trying to find node voltage Only one node with unknown voltage Many elements in parallel One node 4A 20W V x 4W 0.1V x Dr. Louie 30

29 Example Now consider current source: 4 0.1V i i x2 20 x2 via substitution: x2 x V x x2 V V 0.25V 0.1V V 20V V x2 x2 x2 One node 4A 20W V x 4W 0.1V x Dr. Louie 31

30 Summing results: V x = V x1 V x2 = 25V Example Dr. Louie 32

31 Significance of Linearity Consider past example R1 R2 R 2 i1 V1 V2 R2 R2 R3 R 4 i 2 V 2 R 1 R 3 A (matrix) R 2 i 1 i 2 R 4 V 1 V 2 Dr. Louie 33

32 Significance of Linearity Assume that we solved for i 1 and i 2 If the voltage sources are increased by a factor of x, what happens to i 1 and i 2? new i V 1 1 V2 A x new i V 2 2 new i 1 1 V1 V2 x new A i V 2 2 new i 1 i 1 x new i2 i2 i 1 i 2 Current also increases by x Dr. Louie 34

33 Significance of Linearity If all the voltages are proportionally increased/decreased, then all current will also proportionally increase/decrease What happens if all the resistances are proportionally changed? Useful if you have already solved the circuit and want to examine how changes will affect the solution Dr. Louie 35

34 Maximum Power Transfer Often interested in designing a circuit so that the maximum power is delivered (transferred) to a load Consider an 11V battery connected to a heater by way of a cable What resistance should R load be to maximize the power it consumes? R s = 5W V s R load =? Dr. Louie 36

35 Maximum Power Transfer Idea: try a small resistance in order to increase current (P = I 2 R) Let R load = 0.5W Compute power to the load resistor and power consumed by the cable R s = 5W V s R load = 0.5W Dr. Louie 37

36 Maximum Power Transfer V 11 P I R ( ) R ( ) 0.5 2W 2 s 2 2 load load load Rload Rs Ps I R load ( ) 5 10W More power is consumed by the cable than the load Only small portion of applied voltage is across the load V load = 1V R s = 5W V s R load = 0.5W Dr. Louie 38

37 Maximum Power Transfer New idea: try large resistance so a large voltage appears across the load (P = V 2 /R) Let R load = 50W Compute power to the load resistor and power consumed by the cable R s = 5W V s R load = 0.5W Dr. Louie 39

38 Maximum Power Transfer Pload I R load ( ) 50 2W Ps I R load ( ) 5 10W 5 50 More power is consumed by the cable than the load Small amount of current flows I = 0.2A R s = 5W V s R load = 50W Dr. Louie 40

39 Maximum Power Transfer Tradeoff between voltage across load and current through load resistor Try R load = R s Dr. Louie 41

40 Maximum Power Transfer Pload I R load ( ) W Ps I R load ( ) W 5 5 Power to load is increased V load = 5.5V I = 1.1A R s = 5W V s R load = 5W Dr. Louie 42

41 Maximum Power Transfer Maximum power transfer occurs when R load = R s Load and source (with series impedance) are said to be matched Observations Voltage across load is one half applied power Relationship between R load and P load must be nonlinear Dr. Louie 43

43 Power (W) Maximum Power Transfer R load 7 too small R load too large R Load (W) Dr. Louie 45

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