Lecture 2. Engineering

Transcription

1 Lecture 2 Fundamentals of Electrical Engineering Introduction 1. Define current, voltage, and power, including their units. 2. Calculate power and energy, as well as determine whether energy is supplied or absorbed by a circuit element. 3. State t and apply basic circuit it laws. 4. Solve for currents, voltages, and powers in simple circuits. 1

2 Electrical Current Electrical current is the time rate of flow of electrical charge through a conductor or circuit element. The units are amperes (A), which are equivalent to coulombs per second (C/s). Electrical Current i ( t ) q( t) dq ( t ) = dt t = i( t) dt + q( t t 0 0 ) 2

3 Dissipates energy in the form of heat One example is a battery that stores energy in the form of an electrochemical reaction Stores energy in an electric field Stores energy in a magnetic field 3

4 Direct Current Alternating Current When a current is constant with time, we say that we have direct current, abbreviated as dc. On the other hand, a current that varies with time, reversing direction periodically, is called alternating current, abbreviated as ac.. 4

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6 Voltages The voltage associated with a circuit element is the energy transferred per unit of charge that flows through the element. The units of voltage are volts (V), which are equivalent to joules per coulomb (J/C). uphill: battery downhill: resistor 6

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8 Power and Energy p ( t ) = v ( t ) i ( t ) w = t 2 t 1 p( t) dt Watts Joules Current is flowing in the passive configuration. Energy is being absorbed by the element (e.g. a resistor) If the current flows opposite to the passive configuration, the power is given by p = -vi Energy is being provided by the element (e.g. a battery) 8

9 P a = i a v a = (2A)*(12V) = 24W, energy is being absorbed P b = -i b v b = -(1A)*(12V) = -12W, energy is being supplied P c = i c v c = (-3A)*(12V) = -36W, energy is being supplied Power and Energy v(t) = 12 V i(t) = 2e -t A p(t) = v(t)i(t) = 24e -t = t w p ( t ) dt = 24 e dt = 24 e ( 24 e ) = 24 J

10 An example would be a battery. An example would be a transformer.? You want don t to do this! An ideal voltage source has a voltage v x =12 V independent of the load An ideal conductor requires that v x = 0 10

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12 Resistors and Ohm s Law a b v v = ab = ir i ab R The units of resistance are Volts/Amp which are called ohms. The symbol for ohms is omega: Ω 12

13 George Simon Ohm, I = V R In 1805 Ohm entered the University of Erlangen but he became rather carried away with student life. Rather than concentrate on his studies he spent much time dancing, ice skating and playing billiards. Conductance G = 1 R i = Gv The units of conductance are 1/Ω or Ω -1. The units are called siemens 13

14 Resistance Related to Physical Parameters R = ρl A ρ is the resistivity of the material used to fabricate the resistor. The units of resitivity are ohm-meters (Ω-m) Power dissipation in a resistor p = vi = Ri 2 = v 2 R 14

15 KIRCHHOFF S CURRENT LAW The net current entering a node is zero. Alternatively, the sum of the currents entering a node equals the sum of the currents leaving a node. Gustav Kirchhoff ( ) Kirchhoff's Current Law The principle of conservation of electric charge implies that: At any point in an electrical circuit where charge density is not changing in time, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point. A charge density changing in time would mean the accumulation of a net positive or negative charge, which typically cannot happen to any significant degree because of the strength of electrostatic forces: the charge buildup would cause repulsive forces to disperse the charges. 15

16 (a) () Currents into the node = 1A+3A = 4A current out of the node i a (b) Currents into the node = 3A+1A+i b = current out of the node = 2A so i b =2A-4A=-2A (c) Currents into the node = 1A+i c +3A+4A = current out of the node= 0 amps so i c =-8A Series Connection 16

17 Which elements are in series? KIRCHHOFF S VOLTAGE LAW The algebraic sum of the voltages equals zero for any closed path (loop) in an electrical circuit. 17

18 Loop 1: -v a +v b +v c = 0 Loop 2: -v c -v d +v e = 0 Loop 3: v a -v e +v d -v b = 0 18

19 Parallel Connection KVL through A and B: -v a +v b = 0 v a = v b KVL through ha and dc: -v a -v c = 0 v a = -v c 19

20 -3V - 5V + v c = 0 v c = 8V -8V - (-10V) + v e = 0 v e = -2V Which elements are in series? Which elements are in parallel? Find the current, voltage and power for each element: 20

21 Power P R = i R v R = (2A)(10V) = 20W (power dissipated) = i R2 R = V R2 /R P S = -v S i S = -(10V)(2A) = -20W (power supplied) P R + P S = 0 Example What is the current i R flowing through the resistor? What is the power for each element in the circuit? Which elements are absorbing power? 21

22 Since all of the elements are in series, the same current i R = 2A runs through each of them The voltage drop across the resistor: v R = ir = (2A)(5Ω) = 10V Apply KVL: v c =v R + 10 = 20V The power dissipated in each element: p R = i R2 R = (2A) 2 (5Ω) = 20W (absorbing) p vs = iv = (2A)(10V) = 20W (absorbing) p cs = -iv = -(2A)(20V) = 40W (supplying) Example Use Ohm s law, KVL and KCL to find V x 22

23 I T I R2 I R3 R 1 R 4 R 2 R 3 Ohm s Law: V R4 = (1A)(5Ω) = 5V = V R2 = V R3 I R2 = 5V/10Ω = 0.5A I R3 = 5V/5Ω = 1A KCL: I T = I R2 + I R3 + I R4 = 0.5A + 1A + 1A = 2.5A Ohm s Law: V R1 = I T R 1 = (2.5A)(5Ω) = 12.5V KVL: V x = V R1 + V R4 = 12.5V+5V = 17.5V 23