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2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 2 8 Basic Components and Electric Circuits Base Quantity length mass time electric current thermodynamic temperature amount of substance luminous intensity Name meter kilogram second ampere kelvin mole candela Symbol m kg s A K mol cd Table 1: SI (International System of Units) Base Units 2.0.1 Examples • Force unit: newton (N): force required to accelerate 1kg mass by 1m/s/s. • Work/ Energy: – Joule=1 Newton.Meter=kg.m.m/s/s – Cal=4.187 J – kwh=3.6e6 J – BTU=1055.1 J • Power=rate of change of work (w.r. to time) – watt=J/s – hp=745.7 watts – Note: J=watt.s ⇒ KJ=Kw.s ⇒ 3600KJ=kw.h ⇒ 3.6 Meg.J=kwh • Find the cost of heating 50 liters of water from 10 to 75 degrees kelvin if the cost of 1 KWH=0.08 JD. If the power consumed by the heating element is 2 KW, find the time needed for heating the water. (The small calorie or gram calorie approximates the energy needed to increase the temperature of 1 gram of water by 1 C. This is about 4.184 joules=4.184 Watt.second.) Answer: Energy required = 50*1000*(75-10)*1 Calories = 50*1000*(75-10)*1*4.187 Joules (or Watt.second) = 13607750 Watt.second = 13607750/1000/60/60 KWH = 3.7799 KHW. Cost of heating = 3.7799*0.08 = 0.3024 JD. If 2 KW heater is used, then time needed to heat the water = 3.7799/2 = 1.89 Hours. 2.1 Introduction An electric circuit is an interconnection of electrical elements. 2.1.1 SI System of Units Meter (m), Kilogram (kg), second (s), Ampere (A), Kelvin (K), Luminous intensity Candela (cd), amount of Substance Mole (mol) 2.1.2 SI Prefixes E=exa= 1018 , P=peta= 1015 , T=tera= 1012 G=giga= 109 , M=mega= 106 , k=kilo= 103 h=hecto= 102 , da=deca= 10, d=deci= 10−1 c=centi= 10−2 , m=milli= 10−3 , µ=micro= 10−6 n=nano= 10−9 , p=pico= 10−12 , f=femto= 10−15 a=atto= 10−18 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 2.2 9 Charge and Current Charge is an electrical property of the atomic particles of which matter consists, measured in coulombs (C). 1. In a coulomb there are 1 1.602×10−19 = 6.24 × 1018 electrons. 2. charges occur in multiples of electron charge, i.e. multiples of 1.602 × 10−19 C. 3. law of conservation of charge: charge can neither be created nor be destroyed, only transferred. ⇒ algebraic sum of electric charges in a system does not change Universally accepted convention is that current is the net flow of positive charges. Electric current is the time rate of change of charge, measured in ampere (A)- flow of +ve charges A point can not hold or accumulate charge and hence the algebraic sum of currents entering (+ve) and leaving (-ve) any node (point) equals zero. ∆ dq dt i= , q(t) instantaneous value of charge. ∆ and q(t) = R t t0 idt + q(t0 ) A direct current (dc) is a current that remains constant with time An alternating current (ac) is a current that varies sinusoidally with time current: value and direction 2.3 Voltage Electromotive force (emf)=voltage=potential difference. ∆ dw =work dq vab = needed to move a unit of charge from a to b w in joules, q in coulombs, v in volts. 1 volt=1 joule/coulomb=1 newton meter/coulomb Voltage (or potential difference) is the energy required to move a unit charge through an element, measured in volts vab = −vba Voltage: value plus polarity (+ and - signs) 2.4 Power and Energy Power is the energy (-ve) if supplied (delivered) or (+ve) if absorbed (consumed) per unit of time. It is also the product of voltage and current dw = vi watt or J/s p= dt According to the passive sign convention power assumes a positive sign when the current enters the positive polarity of the voltage across an element. Absorbed is +ve, delivered is -ve. 2.4.1 Law of conservation of Energy The algebraic sum of power in a circuit, at any instant of time, must be zero (power (+ve) absorbed + power (-ve) delivered=0) : Pp = 0 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 2.5 10 Circuit Elements • Circuit element = mathematical model. • Simple circuit element = mathematical model of two terminal electrical device. Completely characterized by its v − i relationship. 1. An ideal voltage source produces a specific potential difference across it terminals regardless of what is connected to it. 2. An ideal current source produces a specific current through its terminals regardless of what is connected to it. 3. Voltage and current sources can be independent or dependent on either current or voltage. 1.2ix + − H 9.2iy F 2.7v6 + − E 2.1v2 G • VS (V), Independent Voltage Source • IS (I), Independent Current Source • VCVS (E), (Voltage Controlled Voltage Source) • CCCS (F), (Current Controlled Current Source) • VCCS (G), (Voltage Controlled Current Source) • CCVS (H), (Current Controlled Voltage Source) Voltage sources’ symbols are HEV Current sources’ symbols are FIG As an example, (voltage) transformer can be viewed as a VCVS, vp N = Nps vs As an example, (current) transformer can be viewed as a CCCS, ip Np = is Ns =Ampere.turns 12V + − V 1.5A I sources.m4 Symbols used for Electric Sources 2.5.1 G= i v Conductance = 1 R measured in siemens (S). Notes: 1. power dissipated in a resistor is always positive. (can not be negative) 2. short circuit means R = 0 3. open circuit means R = ∞ 2.5.2 Networks and Circuits Circuits must be closed while networks need not be. Any circuit is a network. 2.5.3 Ohm’s Law v = iR. R=resistance (bidirectional, constant value, linear element). 2.5.4 Power Absorption p = vi = i2 R = v 2 /R 2.5.5 PSPICE Syntax First line is comment line. Next comes the description lines, one line per element, and then comes the (dot) command lines, and finally (.) END line. 2 BASIC COMPONENTS AND ELECTRIC CIRCUITS 2.6 11 Tutorial # 1 1. Practical Application: Wire Gauge R = ρ Al Conductor size (AWG) 28 24 18 14 12 6 4 2 2.6.1 Cross-sectional Area (mm2 ) 0.0804 0.205 0.823 2.08 3.31 13.3 21.1 33.6 Ohms per 1000 ft at 20o C 65.3 25.7 6.39 2.52 1.59 0.3952 0.2485 0.1563 Example 2.3, page 21, 6th ed. A wire of length 4000-ft, 4 AWG, has resistance R = (4000f t)(0.2485Ω)/(1000f t) = 0.992Ω. If i = 100A, then power dissipated in the wires=i2 R = 1002 × 0.994 = 9940W . 2. Practice Problem: 2V +− 8A p2 + 5V p1 − I = 5A 3A p3 + 0.6I − p4 + 3V − ch1pp7.m4 p1 = −40W, p2 = 16W, p3 = 9W, p4 = 15W Figure 3: HKD7eC1PP7 p1 = (+5)(−8) = −40W, p2 = (+2)(+8) = +16W, p3 = (0.6I)(+3) = (0.6 ∗ 5)(3) = +9W, p4 = (+3)(+5) = +15W 3. Find the cost of heating 50 liters of water from 10 to 75 degrees kelvin if the cost of 1 KWH=0.08 JD. If the power consumed by the heating element is 2 KW, find the time needed for heating the water. (The small calorie or gram calorie approximates the energy needed to increase the temperature of 1 gram of water by 1 C. This is about 4.184 joules=4.184 Watt.second.) Answer: Energy required = 50*1000*(75-10)*1 Calories = 50*1000*(75-10)*1*4.187 Joules (or Watt.second) = 13607750 Watt.second = 13607750/1000/60/60 KWH = 3.7799 KHW. Cost of heating = 3.7799*0.08 = 0.3024 JD. If 2 KW heater is used, then time needed to heat the water = 3.7799/2 = 1.89 Hours. 4. How do you provide good earth for a building? Practically, what is considered as an acceptable resistance in Ω’s for a proper earth? Answer: Dig a hole for approximately half a meter and insert a copper rod of approximately 1 meter height inside the hole. Connect a reasonably thick wire to the rod and this point will be your EARTH for the building. For a normal house, as recommended by IEEE, good earth should be less than 5 Ω. Picture to the right from Wikipedia - The free encyclopedia.