Schedule… Date Day 8 Sept Mon 9 Sept Tue 10 Sept Wed 11 Sept Thu 12 Sept Fri 13 Sept Sat 14 Sept Sun 15 Sept Mon Class No. 2 Title Kirchoff’s Laws Chapters HW Due date 2.2 – 2.3 Lab Due date Exam NO LAB NO LAB 3 Power 2.4 – 2.5 NO LAB Recitation 4 Ohm’s Law HW 1 2.5 – 2.6 LAB 1 16 Sept ECEN 301 Tue Discussion #2 – Kirchhoff’s Laws 1 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. ECEN 301 Discussion #2 – Kirchhoff’s Laws 2 Lecture 2 – Kirchhoff’s Current and Voltage Laws ECEN 301 Discussion #2 – Kirchhoff’s Laws 3 Charge Elektron: Greek word for amber ~600 B.C. it was discovered that static charge on a piece of amber could attract light objects (feathers) Charge (q): fundamental electric quantity Smallest amount of charge is that carried by an electron/proton (elementary charges): 19 qe / q p / 1.602 10 C Coulomb (C): basic unit of charge. ECEN 301 Discussion #2 – Kirchhoff’s Laws 4 Electric Current Electric current (i): the rate of change (in time) of charge passing through a predetermined area (IE the cross-sectional area of a wire). Analogous to volume flow rate in hydraulics Current (i) refers to ∆q (dq) units of charge that flow through a crosssectional area (Area) in ∆t (dt) units of time i q dq i A t dt Area Ampere (A): electric current unit. 1 ampere = 1 coulomb/second (C/s) ECEN 301 Positive current flow is in the direction of positive charges (the opposite direction of the actual electron movement) Discussion #2 – Kirchhoff’s Laws 5 Charge and Current Example For a metal wire, find: The total charge (q) The current flowing in the wire (i) ECEN 301 Given Data: • wire length = 1m • wire diameter = 2 x 10-3m • charge density = n = 1029 carriers/m3 • charge of an electron = qe = -1.602 x 10-19 • charge carrier velocity = u = 19.9 x 10-6 m/s Discussion #2 – Kirchhoff’s Laws 6 Charge and Current Example For a metal wire, find: The total charge (q) The current flowing in the wire (i) • wire diameter = 2 x 10-3m • charge density = n = 1029 carriers/m3 • charge of an electron = qe = -1.602 x 10-19 • charge carrier velocity = u = 19.9 x 10-6 m/s Number of carriers volume carrier density Volume length area N V n L r 2 2 10 3 2 m (1m) 2 10 6 m3 2 ECEN 301 Given Data: • wire length = 1m carriers 10 6 m3 10 29 3 m 10 23 carriers Discussion #2 – Kirchhoff’s Laws 7 Charge and Current Example For a metal wire, find: The total charge (q) The current flowing in the wire (i) Given Data: • wire length = 1m • wire diameter = 2 x 10-3m • charge density = n = 1029 carriers/m3 • charge of an electron = qe = -1.602 x 10-19 • charge carrier velocity = u = 19.9 x 10-6 m/s Charge number of carriers charge/car rer q N qe 10 23 carriers 1.602 10 19 C / carrier 50.33 103 C ECEN 301 Discussion #2 – Kirchhoff’s Laws 8 Charge and Current Example For a metal wire, find: The total charge (q) The current flowing in the wire (i) Given Data: • wire length = 1m • wire diameter = 2 x 10-3m • charge density = n = 1029 carriers/m3 • charge of an electron = qe = -1.602 x 10-19 • charge carrier velocity = u = 19.9 x 10-6 m/s Current carrier charge density per unit length carrier ve locity q i (C / m) u (m / s) L 50.33 103 C / m 19.9 10 6 m / s 1A ECEN 301 Discussion #2 – Kirchhoff’s Laws 9 Kirchhoff’s Current Law (KCL) KCL: charge must be conserved – the sum of the currents at a node must equal zero. N i Node 1 i i1 + 1.5_V i2 i3 n 1 n 0 At Node 1: -i + i1 + i2 + i3 = 0 OR: i - i1 - i2 - i3 = 0 i NB: a circuit must be CLOSED in order for current to flow ECEN 301 Discussion #2 – Kirchhoff’s Laws 10 Kirchhoff’s Current Law (KCL) Potential problem of too many branches on a single node: not enough current getting to a branch Node 1 i i1 + 1.5_V i ECEN 301 i2 i3 i4 i5 i6 Suppose: • all lights have the same resistance • i4 needs 1A What must the value of i be? Discussion #2 – Kirchhoff’s Laws 11 Kirchhoff’s Current Law (KCL) Potential problem of too many branches on a single node: not enough current getting to a branch -i + i1 + i2 + i3 + i4 + i5 + i6 = 0 Node 1 i i1 + 1.5_V i ECEN 301 i2 i3 i4 i5 i6 BUT: since all resistances are the same: i1 = i2 = i3 = i4 = i5 = i6 = i n -i + 6in 6in 6(1A) i =0 =i =i = 6A Discussion #2 – Kirchhoff’s Laws 12 Kirchhoff’s Current Law (KCL) Example1: find i0 and i4 is = 5A, i1 = 2A, i2 = -3A, i3 = 1.5A i0 i1 i2 i3 i4 is + Vs _ ECEN 301 Discussion #2 – Kirchhoff’s Laws 13 Kirchhoff’s Current Law (KCL) Example1: find i0 and i4 is = 5A, i1 = 2A, i2 = -3A, i3 = 1.5A Node a i0 i1 i2 Node b is + Vs i3 i4 _ 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 Node c ECEN 301 Discussion #2 – Kirchhoff’s Laws 14 Kirchhoff’s Current Law (KCL) Example1: find i0 and i4 is = 5A, i1 = 2A, i2 = -3A, i3 = 1.5A Node a Find i0 at Node a : i0 i1 i2 Node b is + Vs i3 i4 i0 i1 i2 0 i0 i1 i2 23 1A Find i4 at Node c : i4 is i3 0 i4 i3 is 1.5 5 3.5 A _ Node c ECEN 301 Discussion #2 – Kirchhoff’s Laws 15 Kirchhoff’s Current Law (KCL) Example2: using KCL find is1 and is2 i3 = 2A, i5 = 0A, i2 = 3A, i4 = 1A is2 is1 Vs1 R2 i2 + _ Vs2 R4 + _ R3 ECEN 301 i4 i3 i5 R5 Discussion #2 – Kirchhoff’s Laws 16 Kirchhoff’s Current Law (KCL) Example2: using KCL find is1 and is2 i3 = 2A, i5 = 0A, i2 = 3A, i4 = 1A is2 is1 Vs1 R2 Supernode i2 + _ Vs2 KCL at supe rnode : is1 i3 i5 0 R4 + _ R3 ECEN 301 20 i4 i3 is1 i3 i5 i5 R5 Discussion #2 – Kirchhoff’s Laws 2A 17 Kirchhoff’s Current Law (KCL) Example2: using KCL find is1 and is2 i3 = 2A, i5 = 0A, i2 = 3A, i4 = 1A Node a is1 Vs1 R2 i2 + _ Vs2 R4 + _ R3 ECEN 301 is2 KCL at Node a : is 2 is1 i2 0 is 2 i2 is1 3 2 i4 i3 i5 R5 Discussion #2 – Kirchhoff’s Laws 1A 18 Voltage Moving charges in order to produce a current requires work Voltage: the work (energy) required to move a unit charge between two points Volt (V): the basic unit of voltage (named after Alessandro Volta) Volt (V): voltage unit. 1 Volt = 1 joule/coulomb (J/C) ECEN 301 Discussion #2 – Kirchhoff’s Laws 19 Voltage Voltage is also called potential difference Very similar to gravitational potential energy Voltages are relative _ vba + • voltage at one node is measured relative to the voltage at another node • Convenient to set the reference voltage to be zero a + _ vab vab = va - vb ECEN 301 b vab => the work required to move a positive charge from terminal a to terminal b vba => the work required to move a positive charge from terminal b to terminal a vba = - vab Discussion #2 – Kirchhoff’s Laws 20 Voltage Polarity of voltage direction (for a given current direction) indicates whether energy is being absorbed or supplied i i a + _ b a _ vab vba • Since i is going from + to – energy is being absorbed by the element (passive element) ECEN 301 b + • Since i is going from – to + energy is being supplied by the element (active element) Discussion #2 – Kirchhoff’s Laws 21 Voltage Polarity of voltage direction (for a given current direction) indicates whether energy is being absorbed or supplied i i a + _ a b vab ECEN 301 + a + + 1.5_V v1 _ _ vba i + Absorbing energy (load) (passive element) POSITIVE voltage b vab _ i b Discussion #2 – Kirchhoff’s Laws Supplying energy (source) (active element) NEGATIVE voltage 22 Voltage Ground: represents a specific reference voltage Most often ground is physically connected to the earth (the ground) Convenient to assign a voltage of 0V to ground The ground symbol we’ll use (earth ground) ECEN 301 Another ground symbol (chasis ground) Discussion #2 – Kirchhoff’s Laws 24 Kirchhoff’s Voltage Law (KVL) KVL: energy must be conserved – the sum of the voltages in a closed circuit must equal zero. N v i n 1 a + v1 + + vab 1.5_V _ _ n 0 vab v1 0 vab va vb vab v1 va vab 0 1.5V i b Use Node b as the reference voltage (ground): vb = 0 ECEN 301 Discussion #2 – Kirchhoff’s Laws 1.5 0 1.5V 25 Kirchhoff’s Voltage Law (KVL) Example3: using KVL, find v2 vs1 = 12V, v1 = 6V, v3 = 1V + v2 – i Vs1 + v1 – + _ + v3 – Source: loop travels from – to + terminals • Sources have negative voltage Load: loop travels from + to – terminals • Loads have positive voltage ECEN 301 Discussion #2 – Kirchhoff’s Laws 26 Kirchhoff’s Voltage Law (KVL) Example3: using KVL, find v2 vs1 = 12V, v1 = 6V, v3 = 1V + v2 – i Vs1 + v1 – + _ + v3 – Source: loop travels from – to + terminals • Sources have negative voltage Load: loop travels from + to – terminals • Loads have positive voltage ECEN 301 Discussion #2 – Kirchhoff’s Laws 27 Kirchhoff’s Voltage Law (KVL) Example3: using KVL, find v2 vs1 = 12V, v1 = 6V, v3 = 1V + v2 – vs1 v1 v2 v3 0 i Vs1 + v1 – + _ + v2 vs1 v1 v3 v3 12 6 1 – 5V NB: v2 is the voltage across two elements in parallel branches. The voltage across both elements is the same: v2 ECEN 301 Discussion #2 – Kirchhoff’s Laws 28 Kirchhoff’s Voltage Law (KVL) Example4: using KVL find v1 and v4 vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V + v1 – + + _ v2 Vs1 + _ – + v3 – ECEN 301 Vs2 + v4 – + v5 – Discussion #2 – Kirchhoff’s Laws 29 Kirchhoff’s Voltage Law (KVL) Example4: using KVL find v1 and v4 vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V + v1 – + Loop1 Vs1 + _ v2 Loop2 – + + v4 – v3 Loop3 – ECEN 301 + _ Vs2 + v5 – Discussion #2 – Kirchhoff’s Laws 30 Kirchhoff’s Voltage Law (KVL) Example4: using KVL find v1 and v4 vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V + v1 – Loop 1 : + Loop1 Vs1 + _ v2 Loop2 – + + v4 – v3 Loop3 – ECEN 301 + _ Vs2 vs1 v1 v2 v3 0 v1 vs1 v2 v3 12 2 6 + 4V v5 – Discussion #2 – Kirchhoff’s Laws 31 Kirchhoff’s Voltage Law (KVL) Example4: using KVL find v1 and v4 vs1 = 12V, vs2 = -4V, v2 = 2V, v3 = 6V, v5 = 12V + v1 – Loop 2 : + Loop1 Vs1 + _ v2 Loop2 – + + v4 – v3 Loop3 – ECEN 301 + _ Vs2 v s 2 v4 v2 0 v4 v s 2 v2 4 2 + 6V v5 – Discussion #2 – Kirchhoff’s Laws 32