1. 1 ENGG 1015 Tutorial Circuit Analysis 5 Nov Learning Objectives  Analysis circuits through circuit laws (Ohm’s Law, KCL and KVL) News  HW1 deadline (5 Nov 23:55) Ack.: HKU ELEC1008 and MIT OCW 6.01

2. 2 Quick Checking NOT always true Always True If, then

3. 3 What is a Circuit? Circuits are connects of components  Through which currents flow  Across which voltages develop

4. 4 Rules Governing Flow and Voltages Rule 1: Currents flow in loops  The same amount of current flows into the bulb (top path) and out of the bulb (bottom path) Rule 2: Like the flow of water, the flow of electrical current (charged particles) is incompressible  Kirchoff’s Current Law (KCL): the sum of the currents into a node is zero Rule 3: Voltages accumulate in loops  Kirchoff’s Voltage Law (KVL): the sum of the voltages around a closed loop is zero

5. 5 Rules Governing Components Each component is represented by a relationship between the voltage (V) across the component to the current (I) through the component Ohm’s Law (V = IR)  R: Resistance

6. 6 Question 1: Current and Voltage If R = 0 ohm, I 1 = If R = 1 ohm, V 1 =

7. 7 Solution 1 If R = 0 ohm, I 1 = 6V/3 ohm = 2A If R = 1 ohm,

8. 8 Parallel/Series Combinations To simplify the circuit for analysis Series Parallel

9. 9 Voltage/Current Divider  Voltage Divider Current  Divider

10. 10 Question 2a: Voltage Calculation Find V 2 using single loop analysis  Without simplifying the circuit  Simplifying the circuit

11. 11 Solution 2a Choose loop current Apply KVL  Replace V 2 by R 2 I Find V 2

12. 12 Solution 2b Simplify the circuit with one voltage source and one resistor R eq. = R 1 + R 2 + R 3 = 7 ohm V eq. = V s1 + V s2 + V s3 = -2 + 2 + 2 = 2 V I = V eq. / R eq. = 2/7 A V 2 = 4/7 v

13. 13 Question 3: Potential Difference Assume all resistors have the same resistance, R. Determine the voltage v AB.

14. 14 Solution 3 Determine V AB We assign V G =0

15. For the circuit in the figure, determine i 1 to i 5. 15 Question 4: Current Calculation using Parallel/Series Combinations

16. 16 Solution 4 (i) (iii) (ii) (iv) We apply:  V = IR  Series / Parallel Combinations  Current Divider

17. 17 Solution 4 (v) (vi) (vii)

18. 18 Question 5: Resistance Calculation using Parallel/Series Combinations Find R eq and i o in the circuit of the figure.

19. 19 Solution 5 (i) (ii)

20. 20 Solution 5 (iii)

21. 21 Analyzing Circuits Assign node voltage variables to every node except ground (whose voltage is arbitrarily taken as zero) Assign component current variables to every component in the circuit Write one constructive relation for each component in terms of the component current variable and the component voltage Express KCL at each node except ground in terms of the component currents Solve the resulting equations Power = IV = I 2 R = V 2 /R

22. 22 R 1 = 80Ω, R 2 = 10Ω, R 3 = 20Ω, R 4 = 90Ω, R 5 = 100Ω Battery: V 1 = 12V, V 2 = 24V, V 3 = 36V Resistor: I 1, I 2, …, I 5 = ? P 1, P 2, …, P 5 = ? Question 6: Circuit Analysis (I)

23. 23 Solution 6a V N = 0 I 1 : M  R 5  V1  R 1  B I 2 : M  V 3  R 3  R 2  B I 4 : M  V 2  R 4  B Step 1, Step 2

24. 24 Solution 6a V M – V B = R 5 I 1 + V 1 + R 1 I 1 I 1 = (V M – V B – V 1 )/(R 5 + R 1 ) = (24 – V B )/180 Step 3

25. 25 Solution 6a V N – V B = R 2 I 2 + R 3 I 2 I 2 = (V N – V B )/(R 2 + R 3 ) = – V B /30 Step 3

26. 26 Solution 6a V M – V B = V 2 + R 4 I 4 I 4 = (V M – V B – V 2 )/R 4 = (12 – V B )/90 We get three relationships now (I 1, I 2, I 4 ) Step 3

27. 27 Solution 6a KCL of Node B: I 1 + I 4 + I 2 = 0 (24 – V B )/180 + (12 – V B )/90 – V B /30 = 0  V B = 16/3 V Step 4, Step 5

28. 28 Solution 6a I 1 = (24 – V B )/180 = 14/135 A = 0.104A I 4 = (12 – V B )/90 = 2/27 A = 0.074A I 2 = – V B /30 = – 8/45 A = – 0.178A Step 5

29. 29 Solution 6a P = I 2 R = P 1 = (0.104) 2 80 = 0.86528W P 4 = (0.074) 2 90 = 0.49284W = V R4 2 / R (6.66V, 90Ω)

30. 30 Solution 6b V M = 0 I 1 : B  R 1  V 1  R 5  M I 2 : B  R 2  R 3  V 3  M I 4 : B  R 4  V 2  M Let’s try another reference ground

31. 31 Quick Checking I 1 : B  R 1  V 1  R 5  M I 2 : B  R 2  R 3  V 3  M I 4 : B  R 4  V 2  M Different direction, different result?

32. 32 Solution 6b KCL of Node B: I 1 + I 2 + I 4 = 0 V B – V M = R 1 I 1 – V 1 + R 5 I 1 I 1 = (V B – V M + V 1 )/(R 1 + R 5 ) = (V B + 12)/180

33. 33 Solution 6b V B – V M = R 2 I 2 + R 3 I 2 – V 3 I 2 = (V B – V M + V 3 )/(R 2 + R 3 ) = (V B + 36)/30

34. 34 Solution 6b V B – V M = R 4 I 4 – V 2 I 4 = (V B – V M + V 2 )/R 4 = (V B + 24)/90

35. 35 Solution 6b KCL of Node B: I 1 + I 2 + I 4 = 0 (V B + 12)/180 + (V B + 36)/30 + (V B + 24)/90 = 0  V B = – 92/3 V

36. 36 Solution 6b I 1 = (V B + 12)/180 = –14/135 A = – 0.104A I 2 = (V B + 36)/30 = 8/45 A = 0.178A I 4 = (V B + 24)/90 = –2/27 A = – 0.074A

37. 37 Question 7: Circuit Analysis (II) Find v o in the circuit of the figure.

38. 38 Solution 7 Step 1: Define the node voltage (v 1,v 2,v 3 ) Step 2: Define the current direction

39. 39 Solution 7 Apply: 1) V = IR 2) KCL Step 3: Consider node 1

40. 40 Solution 7 Step 3: Consider node 2 Step 4, 5: From (1) and (2), v 1 = 30V, v 2 = 20V, v 0 = v 2 = 20V

41. 41 Quick Checking NOT always true Always True If, then

42. 42 Quick Checking NOT always true Always True If, then √ √ √ √ √