1. ECEN 301Discussion #2 – Kirchhoff’s Laws1 DateDayClass No. TitleChaptersHW Due date Lab Due date Exam 8 SeptMon2Kirchoff’s Laws2.2 – 2.3 NO LAB 9 SeptTueNO LAB 10 SeptWed3Power2.4 – 2.5 11 SeptThu NO LAB 12 SeptFri RecitationHW 1 13 SeptSat 14 SeptSun 15 SeptMon4Ohm’s Law2.5 – 2.6 LAB 1 16 SeptTue Schedule…

2. ECEN 301Discussion #2 – Kirchhoff’s Laws2 Divine Source 2 Nephi 25:26 26 And we talk of Christ, we rejoice in Christ, we preach of Christ, we prophesy of Christ, and we write according to our prophecies, that our children may know to what source they may look for a remission of their sins.

3. ECEN 301Discussion #2 – Kirchhoff’s Laws3 Lecture 2 – Kirchhoff’s Current and Voltage Laws

4. ECEN 301Discussion #2 – Kirchhoff’s Laws4 Charge uElektron: Greek word for amber Ù~600 B.C. it was discovered that static charge on a piece of amber could attract light objects (feathers) uCharge (q): fundamental electric quantity ÙSmallest amount of charge is that carried by an electron/proton (elementary charges): Coulomb (C): basic unit of charge.

5. ECEN 301Discussion #2 – Kirchhoff’s Laws5 Electric Current uElectric current (i): the rate of change (in time) of charge passing through a predetermined area (IE the cross-sectional area of a wire). ÙAnalogous to volume flow rate in hydraulics ÙCurrent (i) refers to ∆q (dq) units of charge that flow through a cross- sectional area (Area) in ∆t (dt) units of time i Area Ampere (A): electric current unit. 1 ampere = 1 coulomb/second (C/s) Positive current flow is in the direction of positive charges (the opposite direction of the actual electron movement)

6. ECEN 301Discussion #2 – Kirchhoff’s Laws6 Charge and Current Example uFor a metal wire, find: ÙThe total charge (q) ÙThe current flowing in the wire (i) Given Data: wire length = 1m wire diameter = 2 x 10 -3 m charge density = n = 10 29 carriers/m 3 charge of an electron = q e = -1.602 x 10 -19 charge carrier velocity = u = 19.9 x 10 -6 m/s

7. ECEN 301Discussion #2 – Kirchhoff’s Laws7 Charge and Current Example uFor a metal wire, find: ÙThe total charge (q) ÙThe current flowing in the wire (i) Given Data: wire length = 1m wire diameter = 2 x 10 -3 m charge density = n = 10 29 carriers/m 3 charge of an electron = q e = -1.602 x 10 -19 charge carrier velocity = u = 19.9 x 10 -6 m/s

8. ECEN 301Discussion #2 – Kirchhoff’s Laws8 Charge and Current Example uFor a metal wire, find: ÙThe total charge (q) ÙThe current flowing in the wire (i) Given Data: wire length = 1m wire diameter = 2 x 10 -3 m charge density = n = 10 29 carriers/m 3 charge of an electron = q e = -1.602 x 10 -19 charge carrier velocity = u = 19.9 x 10 -6 m/s

9. ECEN 301Discussion #2 – Kirchhoff’s Laws9 Charge and Current Example uFor a metal wire, find: ÙThe total charge (q) ÙThe current flowing in the wire (i) Given Data: wire length = 1m wire diameter = 2 x 10 -3 m charge density = n = 10 29 carriers/m 3 charge of an electron = q e = -1.602 x 10 -19 charge carrier velocity = u = 19.9 x 10 -6 m/s

10. ECEN 301Discussion #2 – Kirchhoff’s Laws10 Kirchhoff’s Current Law (KCL) uKCL: charge must be conserved – the sum of the currents at a node must equal zero. + _ 1.5 V i i i1i1 i2i2 i3i3 Node 1 At Node 1: -i + i 1 + i 2 + i 3 = 0 OR: i - i 1 - i 2 - i 3 = 0 NB: a circuit must be CLOSED in order for current to flow

11. ECEN 301Discussion #2 – Kirchhoff’s Laws11 Kirchhoff’s Current Law (KCL) uPotential problem of too many branches on a single node: Ùnot enough current getting to a branch Suppose: all lights have the same resistance i 4 needs 1A What must the value of i be? + _ 1.5 V i i i1i1 i2i2 i6i6 i3i3 i4i4 i5i5 Node 1

12. ECEN 301Discussion #2 – Kirchhoff’s Laws12 Kirchhoff’s Current Law (KCL) uPotential problem of too many branches on a single node: Ùnot enough current getting to a branch -i + i 1 + i 2 + i 3 + i 4 + i 5 + i 6 = 0 BUT: since all resistances are the same: i 1 = i 2 = i 3 = i 4 = i 5 = i 6 = i n -i + 6i n = 0 6i n = i 6(1A) = i i = 6A + _ 1.5 V i i i1i1 i2i2 i6i6 i3i3 i4i4 i5i5 Node 1

13. ECEN 301Discussion #2 – Kirchhoff’s Laws13 Kirchhoff’s Current Law (KCL) uExample1: find i 0 and i 4 Ùi s = 5A, i 1 = 2A, i 2 = -3A, i 3 = 1.5A + _ VsVs isis i0i0 i1i1 i2i2 i3i3 i4i4

14. ECEN 301Discussion #2 – Kirchhoff’s Laws14 Kirchhoff’s Current Law (KCL) uExample1: find i 0 and i 4 Ùi s = 5A, i 1 = 2A, i 2 = -3A, i 3 = 1.5A + _ VsVs isis i0i0 i1i1 i2i2 i3i3 i4i4 Node a Node c Node b NB: First thing to do – decide on unknown current directions. If you select the wrong direction it won’t matter a negative current indicates current is flowing in the opposite direction. Must be consistent Once a current direction is chosen must keep it

15. ECEN 301Discussion #2 – Kirchhoff’s Laws15 Kirchhoff’s Current Law (KCL) uExample1: find i 0 and i 4 Ùi s = 5A, i 1 = 2A, i 2 = -3A, i 3 = 1.5A + _ VsVs isis i0i0 i1i1 i2i2 i3i3 i4i4 Node a Node c Node b

16. ECEN 301Discussion #2 – Kirchhoff’s Laws16 Kirchhoff’s Current Law (KCL) uExample2: using KCL find i s1 and i s2 Ùi 3 = 2A, i 5 = 0A, i 2 = 3A, i 4 = 1A V s1 + _ V s2 + _ R2R2 R3R3 R4R4 R5R5 i s1 i2i2 i s2 i4i4 i3i3 i5i5

17. ECEN 301Discussion #2 – Kirchhoff’s Laws17 Kirchhoff’s Current Law (KCL) uExample2: using KCL find i s1 and i s2 Ùi 3 = 2A, i 5 = 0A, i 2 = 3A, i 4 = 1A V s1 + _ V s2 + _ R2R2 R3R3 R4R4 R5R5 Supernode i s1 i2i2 i s2 i4i4 i3i3 i5i5

18. ECEN 301Discussion #2 – Kirchhoff’s Laws18 Kirchhoff’s Current Law (KCL) V s1 + _ V s2 + _ R2R2 R3R3 R4R4 R5R5 i s1 i2i2 i s2 i4i4 i3i3 i5i5 Node a uExample2: using KCL find i s1 and i s2 Ùi 3 = 2A, i 5 = 0A, i 2 = 3A, i 4 = 1A

19. ECEN 301Discussion #2 – Kirchhoff’s Laws19 Voltage uMoving charges in order to produce a current requires work uVoltage: the work (energy) required to move a unit charge between two points uVolt (V): the basic unit of voltage (named after Alessandro Volta) Volt (V): voltage unit. 1 Volt = 1 joule/coulomb (J/C)

20. ECEN 301Discussion #2 – Kirchhoff’s Laws20 Voltage uVoltage is also called potential difference ÙVery similar to gravitational potential energy ÙVoltages are relative voltage at one node is measured relative to the voltage at another node Convenient to set the reference voltage to be zero + v ab _ a b _ + v ba v ab => the work required to move a positive charge from terminal a to terminal b v ba => the work required to move a positive charge from terminal b to terminal a v ba = - v ab v ab = va - vb

21. ECEN 301Discussion #2 – Kirchhoff’s Laws21 Voltage uPolarity of voltage direction (for a given current direction) indicates whether energy is being absorbed or supplied + v ab _ a b i v ba _ a b i + Since i is going from + to – energy is being absorbed by the element (passive element) Since i is going from – to + energy is being supplied by the element (active element)

22. ECEN 301Discussion #2 – Kirchhoff’s Laws22 Voltage uPolarity of voltage direction (for a given current direction) indicates whether energy is being absorbed or supplied + v ab _ a b i v ba _ a b i + + _ 1.5 V i i v1v1 a b v ab ++ _ _ Supplying energy (source) (active element) NEGATIVE voltage Absorbing energy (load) (passive element) POSITIVE voltage

23. ECEN 301Discussion #2 – Kirchhoff’s Laws23 Voltage uWork (W) done in moving a charge q across terminals a and b: + v ab _ a b i + _ 1.5 V v1v1 a b + _

24. ECEN 301Discussion #2 – Kirchhoff’s Laws24 Voltage uGround: represents a specific reference voltage ÙMost often ground is physically connected to the earth (the ground) ÙConvenient to assign a voltage of 0V to ground The ground symbol we’ll use (earth ground) Another ground symbol (chasis ground)

25. ECEN 301Discussion #2 – Kirchhoff’s Laws25 Kirchhoff’s Voltage Law (KVL) uKVL: energy must be conserved – the sum of the voltages in a closed circuit must equal zero. + _ 1.5 V i i v1v1 a b v ab ++ _ _ Use Node b as the reference voltage (ground): v b = 0

26. ECEN 301Discussion #2 – Kirchhoff’s Laws26 Kirchhoff’s Voltage Law (KVL) uExample3: using KVL, find v 2 Ùv s1 = 12V, v 1 = 6V, v 3 = 1V V s1 + _ i + v 1 – +v3–+v3– + v 2 – Source: loop travels from – to + terminals Sources have negative voltage Load: loop travels from + to – terminals Loads have positive voltage

27. ECEN 301Discussion #2 – Kirchhoff’s Laws27 Kirchhoff’s Voltage Law (KVL) uExample3: using KVL, find v 2 Ùv s1 = 12V, v 1 = 6V, v 3 = 1V V s1 + _ i + v 1 – +v3–+v3– + v 2 – Source: loop travels from – to + terminals Sources have negative voltage Load: loop travels from + to – terminals Loads have positive voltage

28. ECEN 301Discussion #2 – Kirchhoff’s Laws28 Kirchhoff’s Voltage Law (KVL) uExample3: using KVL, find v 2 Ùv s1 = 12V, v 1 = 6V, v 3 = 1V V s1 + _ i + v 1 – +v3–+v3– + v 2 – NB: v 2 is the voltage across two elements in parallel branches. The voltage across both elements is the same: v 2

29. ECEN 301Discussion #2 – Kirchhoff’s Laws29 Kirchhoff’s Voltage Law (KVL) uExample4: using KVL find v 1 and v 4 Ùv s1 = 12V, v s2 = -4V, v 2 = 2V, v 3 = 6V, v 5 = 12V V s1 + _ V s2 + _ +v3–+v3– + v 4 – + v 1 – +v5–+v5– +v2–+v2–

30. ECEN 301Discussion #2 – Kirchhoff’s Laws30 Kirchhoff’s Voltage Law (KVL) uExample4: using KVL find v 1 and v 4 Ùv s1 = 12V, v s2 = -4V, v 2 = 2V, v 3 = 6V, v 5 = 12V V s1 + _ V s2 + _ +v3–+v3– + v 4 – + v 1 – +v5–+v5– +v2–+v2– Loop1 Loop2 Loop3

31. ECEN 301Discussion #2 – Kirchhoff’s Laws31 Kirchhoff’s Voltage Law (KVL) uExample4: using KVL find v 1 and v 4 Ùv s1 = 12V, v s2 = -4V, v 2 = 2V, v 3 = 6V, v 5 = 12V V s1 + _ V s2 + _ +v3–+v3– + v 4 – + v 1 – +v5–+v5– +v2–+v2– Loop1 Loop2 Loop3

32. ECEN 301Discussion #2 – Kirchhoff’s Laws32 Kirchhoff’s Voltage Law (KVL) uExample4: using KVL find v 1 and v 4 Ùv s1 = 12V, v s2 = -4V, v 2 = 2V, v 3 = 6V, v 5 = 12V V s1 + _ V s2 + _ +v3–+v3– + v 4 – + v 1 – +v5–+v5– +v2–+v2– Loop1 Loop2 Loop3