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ELECTRIC CIRCUITS ECSE-2010 Spring 2003 Class 2

ELECTRIC CIRCUITS ECSE-2010 Spring 2003 Class 2. ASSIGNMENTS DUE. Today (Tuesday/Wednesday): 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

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ELECTRIC CIRCUITS ECSE-2010 Spring 2003 Class 2

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  1. ELECTRIC CIRCUITSECSE-2010Spring 2003Class 2

  2. ASSIGNMENTS DUE • Today (Tuesday/Wednesday): • 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: • 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: • 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 EquationRelating 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 = 0 => 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): • 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: • 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 = 6 - 36/12 = 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 • 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

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