C H A P T E R

1. 3 C H A P T E R Resistive Network Analysis

2. By KCL : i – i – i = 0. In the node 1 2 3 voltage method, we express KCL by v – v v – v v – v a b b c b d – – = 0 R R R 2 1 3 R R v 1 3 b v v a d i i i 1 2 3 R 2 v c Figure 3.2 Use of KCL in nodal analysis

3. Node a Node b R 2 i R R 1 3 S Node c v v a b R 2 i i 2 i 1 i 3 S i R R 1 3 S v = 0 c Figure 3.3 Illustration of nodal analysis Va/R1+(Va-Vb)/R2 =Is Vb/R3+(Vb-Va)/R2=0 Or Va(1/R1+1/R2)+Vb(-1/R2)=Is Va(-1/R2) +Vb(1/R2+1/R3)=0 or, in matrix form

4. Node 1 R 3 R 2 R R 1 4 I I 2 1 Node 2 R 3 R 2 R R 4 1 I I 2 1 Figure 3.5 Example 3.1 R1=1K, R2=2K, R3=10K,R4=2K I1=10mA, I2=50mA, V1/R1+(V1-V2)/R2+(V1-V2)/R3=I1 V2/R4+(V2-V1)/R2+(V2-V1)/R3=-I2 Or (1/R1+1/R2+1/R3)V1+ (-1/R2-1/R3)V2=I1 (-1/R2-1/R3)V1 + (1/R2+1/R3+1/R4)V2= I2 Plugging the numbers 1.6 V1- 0.6 V2=10 -0.5V1 +1.1 V2=-50 By solving the above Eq. V1=-13.57 V2=-52.86

5. v v v R R a b c 1 3 + v R R i _ S 2 4 S Figure 3.8 Nodal analysis with voltage sources Va=Vs (Vs-Vb)/R1-vb/R2-(Vb-Vc)/R3=0 (Vb-Vc)/R3+Is-Vc/R4=0 Or (1/R1+1/R2+1/R3)Vb+(-1/R3)Vc=Vs/R1 (-1/R3)Vb+ (1/R3+1/R4)Vc=Is Or in Matrix form

6. Mesh 1: KVL requires that i R v v v v , – – = 0, where = 1 S 1 2 1 1 v ) R . i – i = ( 2 1 2 1 R R 3 1 + – v 1 + + v _ i v i R R 2 S 1 2 2 4 – Figure 3.13 Assignment of currents and voltages around mesh 1

7. Mesh 2: KVL requires that v + v + v = 0 2 3 4 where v = ( i – i ) R , 2 2 1 2 v = i R , 3 2 3 v = i R 4 2 4 R R 1 3 + – v 3 – + + v i R v i R v _ S 1 2 2 4 2 4 + – Figure 3.14 Assignment of currents and voltages around mesh 3

8. R R 1 3 + _ v i R i R 1 2 S 2 4 Figure 3.12 A two-mesh circuit I1R1+(I1-I2)R2=Vs (I2-I1)R2 + I2R3 + I2R4=0 Or The advantage of Mesh Current Method is that it uses resistances in the equations, rather than conductances. But Node Voltage Method is physically more sensible.

9. 5 2 + 10 V 2 A 4 + v _ i x i 1 2 – Figure 3.18 Mesh analysis with current sources 5I1 +Vx =10 -Vx+2I2+4I2=0 I1-I2=2 Adding Eqs. 1 and 2 will delete Vx 5I1 +6 I2 =10 I1-I2=2 I1=2 A I2=0 P3.1-3.20

10. + + _ _ = + v v B 2 B 2 R R R i i i B 1 B 2 + + _ _ v v B 1 B 1 The net current through R is the sum of the in- dividual source currents: . i + i = i B 1 2 B Figure 3.26 The principle of superposition

11. 1.In order to set a voltage source equal to zero, we replace it with a short circuit. R R 1 1 + v i R i R _ S S S 2 2 A circuit The same circuit with v = 0 S 2. In order to set a current source equal tozero, we replace it with an open circuit. R R 1 1 + + v i v R R _ _ S S S 2 2 A circuit The same circuit with i = 0 S Figure 3.27 Zeroing voltage and current sources

12. i + Linear v network – i Figure 3.28 One-port network

13. i + + v v R R R _ S 1 2 3 – Source Load Figure 3.29 Illustration of equivalent-circuit concept

14. Figure 3.31 Illustration of Thevenin theorum i i R T + + + Load Source Load v v v _ T – –

15. i i Load Load + + Source i R v v N N – – – Figure 3.32 Illustration of Norton theorem

16. R 3 a R R 1 2 b R 3 a || R R R 2 1 T b Figure 3.34 Equivalent resistance seen by the load

17. What is the total resistance the i current will encounter in flowing S around the circuit? R a 3 + R R v i x S 1 2 – b R 3 R R i i 1 2 S S R = R || R + R T 1 2 3 Figure 3.35 An alternative method of determining the Thevenin resistance

18. R R 3 1 i L + v R R S 2 _ L Figure 3.46

19. R R 3 1 + + v v O C S R 2 _ – Figure 3.47

20. R R 1 3 + – + 0 V + + v R v v O C 2 O C S _ – – i Figure 3.48

21. R R R + R || R 1 3 3 1 2 i i L L R + + v R v R 2 R _ _ S 2 S L L R + R 1 2 A circuit Its Th é venin equivalent Figure 3.49 A circuit and its Thevenin equivalent

22. One-port i SC network i R = R i SC N T N Figure 3.57 Illustration of Norton equivalent circuit

23. Short circuit replacing the load v R R 3 1 R i + v 2 S S C _ i i 1 2 Figure 3.58 Computation of Norton current

24. R T One-port v + i R T network N T _ Th é venin equivalent Nortonequivalent Figure 3.63 Equivalence of Thevenin and Norton representations

25. R R 1 3 + v R i _ S 2 SC R 3 v S i R R 1 2 SC R 1 Figure 3.64 Effect of source transformation

26. Node a a a a R + or or R i i R v _ S S S + v _ S b b b Node b The venin subcircuits é Norton subcircuits Figure 3.65 Subcircuits amenable to source transformation

27. a Unknown Load network b An unknown network connected to a load a A Unknown network i “ ” r SC m b Network connected for measurement of short- circuit current a + Unknown V v r “ ” network m O C – b Network connected for measurement of open- circuit voltage Figure 3.71 Measurement of open-circuit voltage and short-circuit current

28. Practical source R L Load R T + v R _ T L i L Source equivalent , what value of Given v and R R T T L will allow for maximum power transfer? Figure 3.73 Power transfer between source and load

29. v + – i n t R T + v R _ T L i Source Load i + i n t i v R R N L T – Source Load Figure 3.74 Source loading effects

30. Nonlinear element as a load. We wish to solve for v and i . x x R T i x + Nonlinear + v v _ x element T – Figure 3.77 Representation of nonlinear element in a linear circuit

31. i X v T v 1 T R Load-line equation: i = – v + T x R R x T T – 1 R T v v T x Figure 3.78 Load line

32. i x curve of i-v “ exponential resistor ” v T R v T i = I e , v > 0 o Solution v 1 T Load-line equation: i = v + x x R R T T v v T x Figure 3.79 Graphical solution equations 3.48 and 3.49

33. R T i i x x + + Linear Nonlinear + Nonlinear v v v network _ x load x load T – – Figure 3.80 Transformation of nonlinear circuit of Thevenin equivalent