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ELECTRIC CIRCUITS ECSE-2010 Spring 2003 Class 2 PowerPoint Presentation
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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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slide2

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
slide3

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
slide4

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
slide10

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
slide11

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
slide12

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
slide14

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
slide16

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
slide17

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
slide18

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
slide19

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!
slide22

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
slide23

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
slide24

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
slide26

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 +….
slide28

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
slide30

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
slide31

CURRENT DIVIDER RULE

  • i1 = [R2/(R1 + R2)] i
  • i2 = [R1/(R1 + R2)] i
  • Resistors in Parallel
    • i1 ~ R2
    • i2 ~ R1
slide32

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:
slide34

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
slide36

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:
slide41

ACTIVITY 2-2

  • Req = 15 ohms
  • i = 30/15 = 2 A
  • Must “Unfold” ladder to see vx, ix
slide44

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