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 3. St t and d apply l b basic i circuit i it llaws. 4. Solve for currents, voltages, and powers in simple circuits. 1 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 ) = dqq (t ) dt t q (t ) = ∫ i (t )dt + q (t0 ) t0 2 Dissipates energy in the form of heat One example is a battery that stores i the th form f off energy in an electrochemical reaction Stores energy in a magnetic field Stores energy gy in an electric field 3 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 5 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 7 Power and Energy p(t ) = v (t )i (t ) Watts t2 w = ∫ p(t )dt Joules t1 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 Pa = iava = (2A)*(12V) = 24W, energy is being absorbed Pb = -ibvb = -(1A)*(12V) = -12W, energy is being supplied Pc = icvc = (-3A)*(12V) = -36W, energy is being supplied Power and Energy v(t) = 12 V i( ) = 2e i(t) 2 -tt A p(t) = v(t)i(t) = 24e-t ∞ w= ∫ 0 ∞ ∫ p (t )dt = 24e −t dt = −24e −∞ − (−24e 0 ) = 24 J 0 9 An example would be a battery. An example would be a transformer. ? You don’t want to do d this! • An ideal voltage source has a voltage vx =12 V independent of the load • An ideal conductor requires that vx = 0 10 11 Resistors and Ohm’s Law a b v = iR vab = iab R The units of resistance are Volts/Amp which are called “ohms”. The symbol for ohms is omega: Ω 12 George Simon Ohm, 1789-1854 V I= 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. http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Ohm.html Conductance 1 G= R i = Gv The units of conductance are 1/Ω or Ω-1. The units are called “siemens” 13 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 2 v p = vi = Ri 2 = R 14 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 t leaving l i a node. d Gustav Kirchhoff (1824-1887) Kirchhoff's Current Law The principle of conservation of electric charge i li that: implies h 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 charge, significant degree because of the strength of electrostatic forces: the charge buildup would cause repulsive forces to disperse the charges. http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Kirchhoff.html 15 ((a)) Currents into the node = 1A+3A = 4A current out of the node ia (b) Currents into the node = 3A+1A+ib = current out of the node = 2A so ib=2A-4A=-2A (c) Currents into the node = 1A+ic+3A+4A = current out of the node= 0 amps so ic =-8A Series Connection 16 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 Loop 1: -va+vb+vc = 0 Loop 2: -vc-vd+ve = 0 Loop 3: va-ve+vd-vb = 0 18 Parallel Connection KVL through A and B: -va+vb = 0 → va = vb KVL through h h A and dC C: -va - vc = 0 → va = -vc 19 -3V - 5V + vc = 0 → vc = 8V -8V - (-10V) + ve = 0 → ve = -2V Which elements are in series? Which elements are in parallel? Find the current, voltage and power for each element: 20 Power PR = iRvR = (2A)(10V) = 20W (power dissipated) = iR2R = VR2/R PS = -vSiS = -(10V)(2A) = -20W (power supplied) PR+ PS = 0 Example • What is the current iR flowing through the resistor? • What is the power for each element in the circuit? • Which elements are absorbing power? 21 • Since all of the elements are in series, the same current iR= 2A runs through each of them • The voltage drop across the resistor: vR= iR = (2A)(5Ω) = 10V • Apply KVL: vc = vR + 10 = 20V • The power dissipated in each element: pR = iR2R = (2A)2(5Ω) = 20W (absorbing) pvs= iv = (2A)(10V) = 20W (absorbing) pcs= -iv = -(2A)(20V) = 40W (supplying) Example Use Ohm Ohm’ss law law, KVL and KCL to find Vx 22 IT R1 IR2 IR3 R2 R3 R4 Ohm’s Law: VR4 = (1A)(5Ω) = 5V = VR2 = VR3 IR2 = 5V/10Ω = 0.5A IR3 = 5V/5Ω = 1A KCL: IT = IR2 + IR3 + IR4 = 0.5A + 1A + 1A = 2.5A Ohm’s Law: VR1= ITR1 = (2.5A)(5Ω) = 12.5V KVL: Vx = VR1 + VR4 = 12.5V+5V = 17.5V 23