1. Chapter 2 Resistive Circuits 1

2. Overview of Chapter 2 2.1 Series Resistors and Parallel Resistors 2.2Voltage Divider Circuit 2.3 Current Divider Circuit 2.4Voltage and Current Measurement 2.5Wheatstone Bridge 2.6 Wye-Delta Transformations. 2

3. Resistors 3

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7. 2.1 Series and Parallel Resistors (1)  Series Resistors  Two or more elements are in series if they are cascaded or connected sequentially and consequently carry the same current.  The equivalent resistance of any number of resistors connected in a series is the sum of the individual resistances. 7

8. Series Resistors in Circuit potential difference (voltage) The potential difference (voltage) between the terminals of the battery (V) equals the sum of the potential differences across the resistors.  KVL 8

9. 2.1 Series and Parallel Resistors (2) Parallel Resistor Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them. The equivalent resistance of a circuit with N resistors in parallel is: 9

10. Parallel Resistors in Circuit The supply current (I) equals the sum of the currents in the branches.  KCL 10

11. 2.1 Series and Parallel Resistors (3) Example Find Req in circuit below. 11

12. 2.1 Series and Parallel Resistors (3) 12

13. 2.1 Series and Parallel Resistors (4) Example Find Req in circuit below. 13 Answer: 11 ohm

14. 2.2Voltage Divider Circuit (1) The voltage divider for N series resistors can be expressed as 14

15. 2.2Voltage Divider Circuit (2) To determine the voltage across each resistor. 15

16. 2.2Voltage Divider Circuit (3) Apply KVL to the circuit; @ From Ohm’s law, From KVL, therefore, Solving equation for current, Apply Ohm’s law to determine voltage across each resistor; 16

17. 2.3Current Divider Circuit (1) The total current, I is shared by the resistors in inverse proportion to their resistances. The current divider can be expressed as: 17

18. 2.3Current Divider Circuit (2) To determine the current flow on each resistor. 18

19. 2.3Current Divider Circuit (3) Apply KCL, Using ohm’s law on each branch, therefore: Current flow through each branch, 19

20. Example (voltage divider) Determine v o and i o in the circuit below. 20 Answer: Vo=4V, io=4/3A

21. 2.4Voltage and Current Measurement an instrument designed to measure current and is placed in series with the circuit An ammeter is an instrument designed to measure current and is placed in series with the circuit element. an instrument designed to measure voltage and is placed in parallel with the element. A voltmeter is an instrument designed to measure voltage and is placed in parallel with the element. 21

22. 2.5Wheatstone Bridge A Wheatstone bridge circuit is an accurate device for measuring resistance 22

23. Under balance condition where no current flow between BD, Current in each resistance arm, Therefore, 23

24. 2.6 Wye-Delta Transformations 24 Wye-delta transformation occurs when the resistors are neither in parallel or in series. This circuit can be simplified to a three-terminal equivalent

25. Wye-Delta Transformations II Two topologies can be interchanged: Wye (Y) or tee (T) networks Delta (Δ) or pi (Π) networks Transforming between these two topologies often makes the solution of a circuit easier 25

26. Wye-Delta Transformations III The superimposed wye and delta circuits shown here will used for reference The delta consists of the outer resistors, labeled a,b, and c The wye network are the inside resistors, labeled 1,2, and 3 26

27. Y- Δ Transformations 27 RaRb Rc

28. Δ - Y Transformations 28

29. Example 1 Transform Y circuit to Δ circuit 29 Ra = 15 Rb = 10 Rc = 25

30. Example #2 Find the current and power supplied by the 40 V sources in the circuit shown below. 30

31. Example #2 Solution: We can find this equivalent resistance easily after replacing either the upper Δ (100Ω, 125Ω, 25Ω) or the lower Δ (40Ω, 25Ω, 37.5Ω) with its equivalent Y. We choose to replace the upper Δ. Thus, 31

32. Example #2 32

33. Substituting the Y-resistor into the circuit, Equivalent resistor, R eq: 33

34. Simplified circuit: Hence, current and power values are: 34

35. Another option: Another option is transform from Why to Delta.. You can try this as well. 35 100 40 Ra Rc Rb Rc

36. Example #3 36 (a) Find no load value of v o. (b) Find vo when RL = 150 kΩ (c) How much power is dissipated in the 25 kΩ resistor if the load terminals are short-circuited ?

37. Example #3 a) Find no load value of v o  Apply ohm law: b) Find vo when RL = 150 kΩ 37

38. Example #3 c) How much power is dissipated in the 25 kΩ resistor if the load terminals are short-circuited ? 38

39. Example #4 39 Find the power dissipated in the 6 Ω resistor.

40. Example #4 Solution: Equivalent resistance current i o, 40 R eq

41. Example #4 Note that i o is the current in the 1.6Ω resistor. Use current divider to get current in the 6Ω resistor, Then the power dissipated by the resistor is 41 i0i0

42. Example #5 42 Find the voltage of v o and v g.

43. Example #5 - Example #5 - Solution step 1  simplified the circuit 43 Current in resistor 30Ω (at Vo) is:

44. Voltage v 0 where i 0 is 15A and R 0 is 20 Total voltage at the resistor, V r : 44

45. Voltage v g 45 V 12 Vr

46. Example #6 Find the current of i g and i o in the circuit 46

47. Example #6 47 Find the current of i g and i o in the circuit. Solution: Equivalent resistance:

48. Example #6 The current values, Thus, 48

49. Exercise 1 49 Determine the value of i o

50. Exercise 2 Find i and V o 50

51. Exercise 3 Calculate the value of current; I. 51