1. ELECTRIC CIRCUITS ECSE-2010 Spring 2003 Class 2

2. ASSIGNMENTS DUE Today (Tuesday/Wednesday): Thursday: Next Monday:
Activities 2-1, 2-2 (In Class)
Thursday:
Will do Experiment 1; Report Due Jan 27
Will also introduce PSpice
Activity 3-1 (In Class)
Next Monday:
No Classes! Martin Luther King Day
HW #1 Due Tuesday/Wednesday, 1/21, 22
See Syllabus for HW Assignments

3. REVIEW Current = i: Amps; ~Water Flow Voltage = v: Volts; ~Pressure
Power = p: Watts; p = v x i
Passive Convention:
Current Flows from + to -; p = vi = power absorbed
Active Convention:
Current Flows from - to +; p = vi = power supplied

4. REVIEW Passive Element = Load: Active Element = Source:
p = Power absorbed; p > 0
Active Element = Source:
p = Power Supplied; p > 0 OR p < 0
Initial Circuit Elements:
Ideal Voltage Source = Circle with + and -
Ideal Current Source = Circle with Arrow
Resistor = “Squiggle”= Passive Element
Ohm’s Law: v = i R
p = v i = v2/R = i2 R

5. PASSIVE CONVENTION

6. ACTIVE CONVENTION

7. MIDEAL SOURCES

8. OHM’S LAW

9. INITIAL CIRCUIT ELEMENTS

10. KIRCHHOFF’S LAWS Based on Conservation Laws
Conservation of Charge => KCL
Kirchhoff’s Current Law
Conservation of Energy => KVL
Kirchhoff’s Voltage Law
Will Use Again & Again & Again
Can ALWAYS Rely on KCL, KVL
Starting Point for most Circuit Analysis

11. KIRCHHOFF’S LAWS Kirchhoff’s Current Law:
The algebraic sum of the currents into (or out of) a node at any instant of time is zero
OR: Current Into a Node = Current Out of a Node

12. KIRCHHOFF’S LAWS Kirchhoff’s Voltage Law:
The algebraic sum of the voltages around any closed path in a circuit is zero for all time

13. KIRCHHOFF EXAMPLE

14. KIRCHHOFF EXAMPLE Choose Directions for i’s: Define NODES:
Polarity for v’s follow from Passive or Active Convention
Define NODES:
Region of Circuit that is All at Same Voltage
3 Nodes for this circuit
Can always choose 1 node as Reference
Choose Node b = 0 Volts

15. KIRCHHOFF EXAMPLE

16. KIRCHHOFF EXAMPLE Label Nodes: Node b = 0 V (Chosen as Reference)
Node a = 10 V (known)
Node c = v2; defines a Variable
Note: Node c also = v3
=> v2 = v3

17. KIRCHHOFF EXAMPLE Conserve Charge at Node c:
Current In = Current Out: KCL
i1 = i2 + i3
KCL Provides a Linear, Algebraic Equation Relating Currents to Each Other
1 Equation; 3 Unknowns; Cannot Solve Yet

18. KIRCHHOFF EXAMPLE Conserve Electrical Energy:
Sum of Voltages Around a Closed Path Must Be 0 => KVL
Start at b:
+10 - v1 - v2 = 0 => v1 + v2 = 10
v3 - v2 = => v2 = v3
6 ohm and 4 ohm Resistors have same voltage across them => Resistors are in PARALLEL

19. KIRCHHOFF EXAMPLE Now Have 3 Equations, But Need to Relate i’s to v’s:
=> Use Ohm’s Law!
i1 = v1/2; i2 = v2/4; i3 = v3/6
v2 = v3
Now Have 3 Equations; 3 Unknowns:
Solve for v1, v2, v3 => Calculate i1, i2, i3
v1 = 50/11 V; v2 = v3 = 60/11 V
i1 = 25/11 A; i2 = 15/11 A; i3 = 10/11 A
All there is to circuit analysis!

20. KIRCHHOFF’S LAWS

21. ACTIVITY 2-1

22. ACTIVITY 2-1 See Circuit: Part a): KCL at Node a:
v = 18 V at T = 0o; v = 24 V at T = 100o
KCL at Node a:
6 – v/12 = i;
i in mA, v in V, R in kohms;
KVL around Resistors:
v – 2 i – R i = 0; => v/i = 2 + R

23. ACTIVITY 2-1 For v = 18 V and T = 0o: For v = 24 V and T = 100o:
i = 6 – 18/12 = 4.5 mA
2 + R = 18/4.5 = 4 kohms => R = 2 kohms
R0 (1 + 0) = 2 kohms => R0 = 2 kohms
For v = 24 V and T = 100o:
i = 6 – 24/12 = 4 mA
2 + R = 24/4 = 6 kohms => R = 4 kohms

24. ACTIVITY 2-1 Part b); Find v when T = 1000 0C:
R = 2 [1 + (.01)(1000)] = 22 kohms
v/i = 2 + R = 24 kohms; i = v/24
i = v/24 = 6 – v/12; => v = 48 V
Part c); Find T when v = 36 V
i = /12 = 3 mA
2 + R = v/i = 12 kohms => R = 10 kohms
10 = 2 [1 + (.01) T]
=> T = 400 0C

25. RESISTORS IN SERIES

26. RESISTORS IN SERIES KCL at a: i1 = i2 = i
Elements with Same Current => SERIES
KVL: v2 + v1 = v
Ohm’s Law: i2R2 + i1R1 = v
i(R1 + R2) = v
i Req = v; Req = Equivalent Resistance
Elements in Series: Req = R1 + R2 +….

27. RESISTORS IN SERIES

28. VOLTAGE DIVIDER RULE v1 = i R1 = v/Req x R1 v2 = i R2
v1 = [R1/(R1 + R2)] v
v2 = i R2
v2 = [R2/(R1 + R2)] v
Resistors in Series
v1 ~ R1
v2 ~ R2

29. RESISTORS IN PARALLEL

30. RESISTORS IN PARALLEL KCL at a: i = i1 + i2
KVL: v1 - v2 = 0 => v1 = v2
KVL: v1 - v = 0 => v1 = v2 = v
Elements with Same Voltage => PARALLEL
Ohm’s Law: i1R1 = i2R2 = v = i Req
i1R1 = (i - i1) R2 => i1 (R1 + R2) = i R2

31. CURRENT DIVIDER RULE i1 = [R2/(R1 + R2)] i i2 = [R1/(R1 + R2)] i
Resistors in Parallel
i1 ~ R2
i2 ~ R1

32. RESISTORS IN PARALLEL i1R1 = [R2/(R1 + R2)] i R1 = V = i Req
Req = R1R2/(R1 + R2); 2 Resistors in Parallel
For More than 2 Resistors:

33. RESISTORS IN PARALLEL

34. DUALS R’s in Series: v1 = [R1/(R1 + R2)] V
R’s in Parallel: i1 = [R2/(R1 + R2)] i
= [G1/(G1 + G2)] i
G = 1/R = Conductance (Siemons)
Replace v with i; R with G:
=> Same Equations
=> Circuits are DUALS of each other

35. ACTIVITY 2-2

36. ACTIVITY 2-2 This is called a LADDER Circuit:
See Circuit Diagram
Find Req “seen” by 30 V Source:
“Collapse” Ladder Using Series/Parallel Reduction:
Often must “Unfold” the ladder to find specific i’s and v’s:

37. ACTIVITY 2-2

38. ACTIVITY 2-2

39. ACTIVITY 2-2

40. ACTIVITY 2-2

41. ACTIVITY 2-2 Req = 15 ohms i = 30/15 = 2 A
Must “Unfold” ladder to see vx, ix

42. ACTIVITY 2-2

43. ACTIVITY 2-2

44. ACTIVITY 2-2 OR: Use Current Divider Rule ix = [4/(12 + 4)]i = i/4
i = 30/15 = 2A
ix = 2/4 = 0.5 A