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Electrical engineering background concepts This presentation is partially animated. Only use the control panel at the bottom of screen to review what you have seen. When using your mouse, make sure you click only when it is within the light blue frame that surrounds each slide. Instrument Systems Electrical Engineering Fundamentals An instrument system is the correct combination of sensor, controller, and final control element that will allow a process to operate in an automatic mode. The vary first instrument systems were hydraulic and for the most part major examples of public works. The Roman aquifers in addition to supplying drinking water had a system of weirs and wheels designed to do specific jobs. Instrument Systems Electrical Engineering Fundamentals you need to figure out how to implant (embed) as rich vibrant images the following terms and/or concepts. Force Newton Energy Joule Power Watt Charge Coulomb 1 electron’s charge 1 proton’s charge Instrument Systems Electrical Engineering Fundamentals Force is the “push or pull” that alters the energy of an entity (system). 1 Newton of force will be needed to just stop a 1 kg mass object that is accelerating at 1 meter/ second squared. 1 Newton = 1 kg . 1 meter / sec 2 Instrument Systems Electrical Engineering Fundamentals Energy is the characteristic of an entity (system) that is associated with its motion, its potential motion and/or its lack of motion. Objects (systems) that are in an identical environment are grouped together and said to be in the same Energy State. ( The unit of energy is Joules which are defined as: The energy used when 1 Newton of force managed to move the object 1 meter. ) 1 Joule = 1 Newton . meter Instrument Systems Electrical Engineering Fundamentals Power is a rate concept. Power is the rate energy is used. Your guess is as good as mine. BUT here are some of its properties!!! 1 Watt is the rate when 1 Joule is used in 1 second. -19 1 electron 1.6 xsecond 10 Coulombs. 1 watt carries = 1 Joule/ -19 1 proton carries 1.6 x 10 Coulombs. Charge is associated with objects. What is it? is packaged in electrons and protons. is measured in Coulombs. made Coulomb’s law famous! Instrument Systems Electrical Engineering Fundamentals Coulombs Law sort of says that two charge particles in the same elevator (or any where else for that matter) will not stay still. Coulomb determined that the force needed to keep them from moving was equal to: F (in Newtons) = k ( Q ) (Q )/ r 2 1 2 Practice Problem Show that two 1 Coulomb each charged particles in free space 1 meter apart need to have 1 million tons of force ( give or take a few tons) applied in the correct direction to keep them from moving. Instrument Systems Electrical Engineering Fundamentals Potential Note: Difference The second charged particle is not involved in If one charged particle moves toward or away from a the calculation but is the reason the first second charge particle, the first charge particle is now in a charged moved in the first place. different energy state. 1 volt = 1 Joule/ 1 Coulomb The second charged particle is sometimes known as the reference or test charge. The difference between these two energy states (the one the first particle is in now and the one the first particle use to be in) is the potential difference of the first particle. Voltage is the potential difference per unit charge. V = Energy difference Amount of charge on the first particle Instrument Systems Electrical Engineering Fundamentals Electric Field If one charged particle moves because of the presence of a test charge, the first charge is said to move in (through) an electric force field. The charged particle can move toward or away from the test charge but the field is still the region where the influence of the test charge is felt by the first charge. The electric field lines define the electric field with respect to the test charge. The direction the first charge travels across the electric field lines determines if the change in energy can be used by us to do work. Instrument Systems Electrical Engineering Fundamentals Electric Field Once a charge particle is in an electric field, the only thing of interest is the voltage changes because the charged particle moved to different places in the electric field ( Once you have purchased that stock, you only care about the changes in the price, not what the stock costs.) Knowledge of the voltage difference will let you compute the energy difference between the two points in the electric field if the charge on the particle is know. Energy = V (Amount of charge on the particle) difference Use a volt meter for this value. Many times the charge particle is an electron and the charge is know. Instrument Systems Electrical Engineering Fundamentals Current Current is the rate of movement of electric charge. Current is not the rate of movement of the charged particle. Current is measured in amperes. current = (amount of charge) / ( change in time) 1 ampere = 1 Coulomb / second Decision time ! We have to pick the sign of this 1 Coulomb charge. Positive? Or negative? Yep, Ben Franklin wins, the reference charge is the positive charge, even though the electron is the most common carrier of charge in systems we will deal with. Instrument Systems Electrical Engineering Fundamentals Concepts that might seem confusing. A charge moving from point “b” to point “a” (lower voltage to higher voltage) requires energy from the outside world and the charge is moving up the field A charge moving from point “b” to point “a” produces an electrical current that itself produces a net force. This force produced by an electric current is referred to as magnetic force, and it possess many of the properties of the magnetic force associated with an ordinary bar magnet. A charged particle has a static electric field around it. A moving charged particle creates a current and has a magnetic field around the path, usually a wire, it is traveling on. Instrument Systems Electrical Engineering Fundamentals Conventions that might seem confusing. + v v - ab is just the voltage between two points in a static electric field ab But by drawing convention; When a charge moves from point “b” to point “a” (lower voltage to higher voltage) energy is required from the outside world and the charge is moving across the field lines as it goes up the field Terminal “a” is assigned to be point at higher voltage + - v v b Energy is absorbed when v ab is positive a Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. The passive sign convention for labeling the voltage and current of a two-terminal electric circuit element. When the element current and voltage are labeled with the assumed current to enter the terminal of assumed higher voltage the element is labeled using the “passive sign convention” If the actual current and voltage agree with their passive assignment the element is absorbing energy Terminal “a” is assigned to be point at higher voltage + b i(t) v(t) a - Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. If either (but not both) the current or voltage differ from the assumed passive assignment, then that negative component is opposite in direction of its assumed orientation and the energy is flowing in the opposite direction. (The two port element is delivering power.) If the actual current and voltage agree with there passive assignment the element is absorbing energy Terminal “a” is assigned to be point at higher voltage + b i(t) v(t) a - Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Practice Problem-- Assume the “passive labeling” system is being practiced. + b i(t) If the actual current and voltages are in the orientations as shown, which twoterminal element device is delivering power? - b i(t) v(t) v(t) The one on the left a - Terminal “a” is assigned to be point at higher voltage True; terminal a is high energy point a False; terminal a not high energy point + Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Practice Problem-- Assume the “passive labeling” system is being practiced. If the actual current and voltages are in the orientations as shown, what is the power and direction of power for each of the following? 1) + b 2A p(t) = v(t) i(t) 2) - b 4A p(t) = (-3V)(4A) =-12 Watts p(t) = (3V)(2A) = 6 Watts 3V a - p(t) = [-v(t)] i(t) 3V 6W of power absorbed by this element a + 12W of power delivered by this element Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Practice Problem-- Assume the “passive labeling” system is being practiced. If the actual current and voltages are in the orientations as shown, what is the power and direction of power for each of the following? 3) + b 2A p(t) = v(t) [-i(t)] 4) - b -3A p(t) = (-3V)(-3A) =9 Watts p(t) = (5V)(-2A) =-10 Watts 5V a - p(t) = [-v(t)] i(t) 3V 10W of power delivered by this element a + 9W of power absorbed by this element Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Is there a voltage gain or drop across the following resistors? This two terminal electric circuit element is the resistor. The resistance of material used to make a resistor is always a positive number. 1) 2) b v(t) = [-i(t)]R + - b v(t) v(t) + a v(t) is negative indicating a voltage gain v(t) = i(t)R -a v(t) is positive indicating a voltage drop Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Is there a charge gain or drop in each of the following capacitor? This two terminal electric circuit element is the capacitor. The capacitance of a capacitor is always a positive number 1) 2) - b i(t) = [- dv/dt)]C i(t) + a Energy being stored Energy being released i(t) is negative indicating the charge in the capacitor is dropping. + b i(t) = [dv/dt)]C i(t) - a i(t) is positive indicating a gain in the charge of the capacitor. Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Are the following inductors absorbing or releasing energy? This two terminal electric circuit element is the inductor. The inductance of practical inductors is always a positive number 1) 2) - b v(t) = [ - di/dt)]L v(t) + a Energy being stored Energy being released v(t) is negative indicating the inductor is releasing energy. + b v(t) = [di/dt)]L v(t) - a v(t) is positive indicating the inductor is absorbing energy. Instrument Systems Electrical Engineering Fundamentals Other concepts that might seem confusing. The perfect world. The ideal voltage supply v(t) The ideal current source i(t) Is a car battery an ideal voltage supply? The voltage developed in an ideal voltage supply is a dictated value that does not change with age, use (misuse), application, snow, sleet, rain, or etc, it just keeps providing the voltage it was advertised to provide. Is a lighting bolt, (like the one Ben Franklin felt when he was flying his kite) an ideal current supply? An ideal current source will provide the stated amount of current (a specific number of moving charges/unit time) no matter what the demand for that flow of charge becomes or how long that demand lasts. Instrument Systems Electrical Engineering Fundamentals Other concepts that might seem confusing. Is the energizer bunnies battery an ideal voltage supply? Hmmmmmm! Life does seems to be full of difficult questions! Circuit configurations Voltage as function of time that is available for use. b v (t) + s - v(t) a 1) 2) 3) v (t) = v(t) s Voltage value profile as a function of time remains the same. Voltage value profile can be fixed value, dc, shaped like a sine wave, ac, or any other desired shape. The main point is that the specific voltage value of the ideal supply at a specific instant in time is always that exact value desired no matter what the circumstances are. Instrument Systems Electrical Engineering Fundamentals Other concepts that might seem confusing. Is the energizer bunnies battery an ideal voltage supply? Hmmmmmm! Life does seems to be full of difficult questions! Circuit configurations b b v v (t) + s - v(t) v s1 s2 (t) (t) + - + - v(t) a a Ideal voltage supplies connected in series. v(t) = v s1(t) + v s2(t) Instrument Systems Electrical Engineering Fundamentals Other concepts that might seem confusing. Is the energizer bunnies battery an ideal voltage supply? Hmmmmmm! Life does seems to be full of difficult questions! Circuit configurations b b b v (t) vv s1 (t) s a + + - - v + (t)v(t) s2 - v(t) s2 (t) (t) + - + - v(t) a a Ideal voltage supplies connected in parallel. v(t) = v s1(t) v s1 = v s2(t) Ideal voltage supplies connected in series. v(t) = v s1(t) + v s2(t) Instrument Systems Electrical Engineering Fundamentals Other concepts that might seem confusing. Ideal current source Circuit configurations i(t) b i (t) s1 i (t) s2 a Ideal current sources connected in parallel. i(t) = i (t) + i s2(t) s1 b i (t) s2 i (t) s1 i(t) a Ideal current sources connected in series. i(t) = i (t) = i s2(t) s1 Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Things to remember about transformer passive diagram symbolism. This four terminal electric circuit element is a coupled inductor (transformer). 1) A current i1(t) entering a dotted terminal in one inductor produces a voltage,v1 (t), across the terminals of that inductor. 2) A current i1(t) entering a dotted terminal in one inductor produces an induced voltage which is positive at the other dotted terminal. M (Mutual inductance) 3) If a second current i 2 (t) enters the other inductor at its dotted terminal there is a voltage drop produced across that inductor. 4) If a second current i 2 (t) enters the other inductor at its dotted terminal there is also a voltage induced in the first inductor. (Mutual induction) i1 b + v (t) + v (t) 1 1 a d i2 + c v (t) + v2 (t) 2 Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Things to remember about transformer passive diagram symbolism. b d 5) + i1 i2 + This four terminal electric circuit v (t) + v (t) v (t) + v (t) = v (t) 1 1 1 a c v (t) element = 2 inductor 2 2 is a coupled (transformer). 1) A current i1(t) entering a dotted terminal in one inductor produces a voltage,v1 (t), across the terminals of that inductor. 2) A current i1(t) entering a dotted terminal in one inductor produces an induced voltage which is positive at the other dotted terminal. M (Mutual inductance) 3) If a second current i 2 (t) enters the other inductor at its dotted terminal there is a voltage drop produced across that inductor. 4) If a second current i 2 (t) enters the other inductor at its dotted terminal there is also a voltage induced in the first inductor. (Mutual induction) i1 b + v (t) + v (t) 1 1 a d i2 + c v (t) + v2 (t) 2 Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Things to remember about transformer passive diagram symbolism. M = MM 12 = M 21 i1 b d i2 + + v (t) = v (t) + v (t) 1 1 1 a c v (t) = L1 (d i1 /dt ) 1 v (t) = v2 (t) + v (t) 2 2 v (t) = L 2(d i2 /dt) 2 v (t) = M (d i2 /dt ) 1 M 12 = v (t) = M (d i1 /dt ) 2 M 21 The voltage difference v1(t) between point b and point a is the sum of the voltage, v in current i1 and the mutually induced voltage v 1 1 induced by the change produced by the change in the flow of current i2. The voltage difference v2(t) between point d and point c is the sum of the voltage, v 2 induced by the change in current i2 and the mutually induced voltage v 2 produced by the change in the flow of current i1. Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Passive diagram example i 1 (t) M 12 + + A) how many equations are needed to describe this three coupled inductor system? - M 13 - three v (t) = v (t) + v (t) + v (t) 2 2 2 2 v (t) = v (t) + v (t) + v (t) 3 3 3 3 M 23 An example of a three coupled coil system that is complaint with the passive diagram symbolism. + i 3 (t) v (t) = v (t) + v (t) + v (t) 1 1 1 1 i 2 (t) But it is more convenient to arrange them this way! v (t) = 1 v (t) + v (t) 1 + v (t) 1 v (t) = v (t) 2 2 + v (t) 2 + v (t) 2 v (t) = v (t) 3 3 + v (t) 3 + 1 v (t) 3 Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Passive diagram example i1 /dt) + v (t) = L1v(d(t) 1 + v (t) i /dt) + v (t) + L2v(d(t) 2 2 2 v (t) 1 v (t) = 2 v (t) = 3 v 3 (t) + v (t) i 1 (t) 1 1 2 i /dt) L3v(d(t) 3 3 v (t) + 3 M 12 + + + i 3 (t) After a bit more manipulation and substitution we get. v (t) = 1 L1 (d i1 /dt) + M 12 (d i2 /dt ) An example of a three coupled coil system that is complaint with the passive diagram symbolism. - M 13 i 2 (t) M 23 v (t) = M (d i2 /dt ) 1 Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Passive diagram example i1 /dt) + v (t) = L1v(d(t) 1 1 v (t) = 2 v (t) = 3 v 2 v (t) i 1 (t) v + 1 1 i /dt) + (t) + L2v(d(t) 2 2 v (t) + 3 v 3 (t) M 12 (t) v (t) 2 i /dt) L3v(d(t) 3 3 + + + i 2 (t) - - M 13 + i 3 (t) L1 (d i1 /dt) v (t) = 2 M v (t) = 3 M 21 + M (d i1 /dt ) + (d i1 /dt ) 31 + M (d i2 /dt ) 12 + L2 (d i2 /dt) + (d i2 /dt ) 32 + M 23 v (t) = M (d i2 /dt ) 1 Note: After a bit more manipulation and substitution we get. v (t) = 1 An example of a three coupled coil system that is complaint with the passive diagram symbolism. M M M (d i3 /dt ) 13 23 (d i3 /dt ) L3 (d i3 /dt) M M 31 23 21 = M = M = M 13 32 12 Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Passive diagram example M 12 i 1 (t) The equations below model the three inductor circuit to the right. If the circuit you have to deal with is not wired this way, you can manipulate the equations by multiplying specific components by (-1) to get the correct model. + + i 2 (t) - - M 13 + i 3 (t) Try the next practice problem to see how this works. v (t) = 1 L1 (d i1 /dt) v (t) = 2 M v (t) = 3 M 21 + M (d i1 /dt ) + (d i1 /dt ) 31 + M (d i2 /dt ) + M L2 (d i2 /dt) + M (d i2 /dt ) 32 + 12 An example of a three coupled coil system that is complaint with the passive diagram symbolism. 13 (d i3 /dt ) 23 (d i3 /dt ) L3 (d i3 /dt) M 23 Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Practice Problem-- Assume the “passive labeling” system is being practiced. If the actual current and voltages in the following diagram are in the orientations as shown, what are the three equations that determine the voltages across each of the coils, respectively? i 1 (t) M12 + Note that either: i 2 (t) + A) the defined voltages and corresponding currents do not all conform to the passive sign convention. or B) the positive voltage terminal is not at the dot. - - M M 13 + - 1) Reverse direction of i1 2) Reverse direction of i2 3) Reverse direction of v2 4) Reverse direction of v3 23 i 3 (t) To get to the correct set of equations, use negative signs in the model equation set everywhere such a sign corresponds to a reversal of an item in the diagram to return the diagram to its passive diagram form. 4 reversals must be accomplished. Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Practice Problem-- Assume the “passive labeling” system is being practiced. If the actual current and voltages in the following diagram are in the orientations as shown, what are the three equations that determine the voltages across each of the coils, respectively? i 1 (t) + - + + - - M M 13 - + v (t) = 1 - L1 (d i1 /dt) - v (t) = 2 - M - v (t) = 3 - M i 2 (t) M12 + - 23 i 3 (t) 21 - + M 12 (d i2 /dt ) - L2 (d i2 /dt) - M (d i1 /dt ) + (d i1 /dt ) + 31 + M 2) Reverse direction of i2 3) Reverse direction of v2 4) Reverse direction of v3 (d i3 /dt ) + M 23 (d i3 /dt ) (d i2 /dt ) + 32 1) Reverse direction of i1 13 L3 (d i3 /dt) Instrument Systems Electrical Engineering Fundamentals Other conventions that might seem confusing. Practice Problem-- Assume the “passive labeling” system is being practiced. If the actual current and voltages in the following diagram are in the orientations as shown, what are the three equations that determine the voltages across each of the coils, respectively? i i11(t) (t) M12 ++ - + - - - M M 13 - + v (t) = 1 i (t) i 22(t) + + - + - - L1 (d i1 /dt) - - M + 12 (d i2 /dt ) + M 13 (d i3 /dt ) - i1 /dt ) + forL2 (d voltages (d i2three /dt) coupled v (t) = + M 23 (d i3 /dt ) ThusM 21 the the 2 coils are modeled with the following three equations. v (t) = - M 31 (d i1 /dt ) +- M 32 (d i2 /dt ) + L3 (d i3 /dt) 3 23 v (t) = 1 i 3 (t) -+v2(t) - - L1 (d i1 /dt) - = -+ M 21 (d i1 /dt ) -+ - - + v (t) = + M (d i1 /dt ) + 31 3 M 12 (d i2 /dt ) L2 (d i2 /dt) M + M 13 (d i3 /dt ) - + M 23 (d i3 /dt ) - (d i2 /dt ) + 32 L3 (d i3 /dt)