ELECTRIC CIRCUITS ECSE-2010 Spring 2003 Class 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

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

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

PASSIVE CONVENTION

ACTIVE CONVENTION

MIDEAL SOURCES

OHM’S LAW

INITIAL CIRCUIT ELEMENTS

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

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

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

KIRCHHOFF EXAMPLE

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

KIRCHHOFF EXAMPLE

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

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

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

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!

KIRCHHOFF’S LAWS

ACTIVITY 2-1

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

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

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

RESISTORS IN SERIES

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 +….

RESISTORS IN SERIES

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

RESISTORS IN PARALLEL

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

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

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:

RESISTORS IN PARALLEL

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

ACTIVITY 2-2

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:

ACTIVITY 2-2

ACTIVITY 2-2

ACTIVITY 2-2

ACTIVITY 2-2

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

ACTIVITY 2-2

ACTIVITY 2-2

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