Schedule… Date Day Class Title Chapters HW Lab Exam No. Due date Due date Kirchoff’s Laws 2.2 – 2.3 8 Sept Mon 2 NO LAB 9 Sept Tue NO LAB 2.4 – 2.5 10 Sept Wed 3 Power 11 Sept Thu NO LAB 12 Sept Fri Recitation HW 1 13 Sept Sat 14 Sept Sun Ohm’s Law 2.5 – 2.6 15 Sept Mon 4 LAB 1 16 Sept Tue Discussion #2 – Kirchhoff’s Laws ECEN 301 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. Discussion #2 – Kirchhoff’s Laws ECEN 301 2
Lecture 2 – Kirchhoff’s Current and Voltage Laws Discussion #2 – Kirchhoff’s Laws ECEN 301 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 q / q / 1 . 602 10 C e p Coulomb (C) : basic unit of charge. Discussion #2 – Kirchhoff’s Laws ECEN 301 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 cross- sectional area ( Area ) in ∆t ( dt ) units of time i q dq i A t dt Area Ampere ( A ) : electric current unit. Positive current flow is in the direction 1 ampere = 1 coulomb/second (C/s) of positive charges (the opposite direction of the actual electron movement) Discussion #2 – Kirchhoff’s Laws ECEN 301 5
Charge and Current Example For a metal wire, find: Given Data : The total charge ( q ) • wire length = 1m The current flowing in the • wire diameter = 2 x 10 -3 m • charge density = n = 10 29 carriers/m 3 wire ( i ) • charge of an electron = q e = -1.602 x 10 -19 • charge carrier velocity = u = 19.9 x 10 -6 m/s Discussion #2 – Kirchhoff’s Laws ECEN 301 6
Charge and Current Example For a metal wire, find: Given Data : The total charge ( q ) • wire length = 1m The current flowing in the • wire diameter = 2 x 10 -3 m • charge density = n = 10 29 carriers/m 3 wire ( i ) • charge of an electron = q e = -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 2 L r carriers 6 3 29 2 10 m 10 3 2 10 3 m 2 ( 1 m ) m 2 23 10 carriers 6 3 10 m Discussion #2 – Kirchhoff’s Laws ECEN 301 7
Charge and Current Example For a metal wire, find: Given Data : The total charge ( q ) • wire length = 1m The current flowing in the • wire diameter = 2 x 10 -3 m • charge density = n = 10 29 carriers/m 3 wire ( i ) • charge of an electron = q e = -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 q e 23 19 10 carriers 1 . 602 10 C / carrier 3 50 . 33 10 C Discussion #2 – Kirchhoff’s Laws ECEN 301 8
Charge and Current Example For a metal wire, find: Given Data : The total charge ( q ) • wire length = 1m The current flowing in the • wire diameter = 2 x 10 -3 m • charge density = n = 10 29 carriers/m 3 wire ( i ) • charge of an electron = q e = -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 3 6 50 . 33 10 C / m 19 . 9 10 m / s 1 A Discussion #2 – Kirchhoff’s Laws ECEN 301 9
Kirchhoff’s Current Law (KCL) KCL : charge must be conserved – the sum of the currents at a node must equal zero. N 0 i n n 1 Node 1 i At Node 1 : i 1 i 3 i 2 - i + i 1 + i 2 + i 3 = 0 + OR: 1.5 V _ i - i 1 - i 2 - i 3 = 0 i NB : a circuit must be CLOSED in order for current to flow Discussion #2 – Kirchhoff’s Laws ECEN 301 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 i 1 i 6 i 2 i 3 i 4 i 5 Suppose : • all lights have the same resistance + • i 4 needs 1A 1.5 V _ What must the value of i be? i Discussion #2 – Kirchhoff’s Laws ECEN 301 11
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 + i 1 + i 2 + i 3 + i 4 + i 5 + i 6 = 0 i BUT : since all resistances are the same: i 1 i 6 i 2 i 3 i 4 i 5 i 1 = i 2 = i 3 = i 4 = i 5 = i 6 = i n + 1.5 V _ -i + 6 i n = 0 6 i n = i i 6(1A) = i i = 6A Discussion #2 – Kirchhoff’s Laws ECEN 301 12
Kirchhoff’s Current Law (KCL) Example1 : find i 0 and i 4 i s = 5A, i 1 = 2A, i 2 = -3A, i 3 = 1.5A i 0 i 1 i 2 i s + i 3 i 4 V s _ Discussion #2 – Kirchhoff’s Laws ECEN 301 13
Kirchhoff’s Current Law (KCL) Example1 : find i 0 and i 4 i s = 5A, i 1 = 2A, i 2 = -3A, i 3 = 1.5A Node a NB : First thing to do – decide on unknown current i 0 i 1 i 2 directions. • If you select the wrong direction it won’t Node b matter • a negative current indicates current is i s + flowing in the opposite direction. i 3 i 4 V s • Must be consistent _ • Once a current direction is chosen must keep it Node c Discussion #2 – Kirchhoff’s Laws ECEN 301 14
Kirchhoff’s Current Law (KCL) Example1 : find i 0 and i 4 i s = 5A, i 1 = 2A, i 2 = -3A, i 3 = 1.5A Node a Find i at Node a : Find i at Node c : 0 4 i 0 i 1 i 2 i i i 0 i i i 0 0 1 2 4 s 3 i i i Node b i i i 0 1 2 4 3 s 2 3 i s 1 . 5 5 + 1 A i 3 3 . 5 A i 4 V s _ Node c Discussion #2 – Kirchhoff’s Laws ECEN 301 15
Kirchhoff’s Current Law (KCL) Example2 : using KCL find i s1 and i s2 i 3 = 2A, i 5 = 0A, i 2 = 3A, i 4 = 1A i s2 i s1 i 2 + R 2 _ V s2 R 4 + _ V s1 i 4 R 3 i 5 i 3 R 5 Discussion #2 – Kirchhoff’s Laws ECEN 301 16
Kirchhoff’s Current Law (KCL) Example2 : using KCL find i s1 and i s2 i 3 = 2A, i 5 = 0A, i 2 = 3A, i 4 = 1A i s2 Supernode KCL at supe rnode : i s1 i 2 + i i i 0 R 2 _ V s2 1 3 5 s R 4 i i i + _ V s1 s 1 3 5 2 0 i 4 R 3 i 5 i 3 R 5 2 A Discussion #2 – Kirchhoff’s Laws ECEN 301 17
Kirchhoff’s Current Law (KCL) Example2 : using KCL find i s1 and i s2 i 3 = 2A, i 5 = 0A, i 2 = 3A, i 4 = 1A Node a i s2 KCL at Nod e a : i s1 i 2 + R 2 _ V s2 i i i 0 s 2 s 1 2 R 4 + i i i _ V s1 s 2 2 s 1 i 4 3 2 R 3 i 5 i 3 R 5 1 A Discussion #2 – Kirchhoff’s Laws ECEN 301 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) Discussion #2 – Kirchhoff’s Laws ECEN 301 19
Voltage Voltage is also called potential difference Very similar to gravitational potential energy Voltages 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 ba _ + v ab => the work required to move a positive a b charge from terminal a to terminal b _ + v ba => the work required to move a positive v ab charge from terminal b to terminal a v ab = va - vb v ba = - v ab Discussion #2 – Kirchhoff’s Laws ECEN 301 20
Voltage Polarity of voltage direction (for a given current direction) indicates whether energy is being absorbed or supplied i i a a b b _ _ + + v ab v ba • Since i is going from + to – energy • Since i is going from – to + energy is is being absorbed by the element being supplied by the element ( active ( passive element ) element ) Discussion #2 – Kirchhoff’s Laws ECEN 301 21
Voltage Polarity of voltage direction (for a given current direction) indicates whether energy is being absorbed or supplied i i a a b b _ _ + + v ab v ba i a + + + Absorbing energy v ab v 1 Supplying energy 1.5 V ( load ) _ ( source ) _ _ ( passive element) ( active element) i POSITIVE voltage NEGATIVE voltage b Discussion #2 – Kirchhoff’s Laws ECEN 301 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 Another ground symbol (earth ground) (chasis ground) Discussion #2 – Kirchhoff’s Laws ECEN 301 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 0 n i n 1 a + + v v 0 v v v + ab 1 ab a b v ab v 1 1.5 V v v v v 0 _ _ ab 1 a ab _ i 1 . 5 V 1 . 5 0 b 1 . 5 V Use Node b as the reference voltage (ground): v b = 0 Discussion #2 – Kirchhoff’s Laws ECEN 301 25
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