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.

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

Energy you need to figure out how to implant (embed) as rich vibrant images the following terms and/or concepts. Joule Force Newton Power Watt Instrument Systems Coulomb 1 proton’s charge Charge 1 electron’s charge 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. 1 Watt is the rate when 1 Joule is used in 1 second. 1 watt=1 Joule/ second Charge What is it? Your guess is as good as mine. BUT here are some of its properties!!! is associated with objects. is packaged in electrons and protons. is measured in Coulombs. 1 electron carries 1.6 x 10 Coulombs proton carries 1.6 x 10 Coulombs. -19 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 12 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 Difference If one charged particle moves toward or away from a second charge particle, the first charge particle is now in a different energy state. Voltage is the potential difference per unit 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. V = Amount of charge on the first particle Energy difference Note: The second charged particle is not involved in the calculation but is the reason the first charged moved in the first place. 1 volt = 1 Joule/ 1 Coulomb The second charged particle is sometimes known as the reference or test charge. 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. = V(Amount of charge on the particle) Energy 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. (amount of charge) / ( change in time)current Current is measured in amperes. = Current is not the rate of movement of the charged particle. 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 ab v is just the voltage between two points in a static electric field But by drawing convention; +v b - v a 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 ab is positive Energy is absorbed when 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” + - i(t) v(t) If the actual current and voltage agree with their passive assignment the element is absorbing energy a b Terminal “a” is assigned to be point at higher voltage 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.) + - i(t) v(t) If the actual current and voltage agree with there passive assignment the element is absorbing energy a b Terminal “a” is assigned to be point at higher voltage Instrument Systems Electrical Engineering Fundamentals

Other conventions that might seem confusing. Practice Problem-- Assume the “passive labeling” system is being practiced. + - i(t) v(t) a b Terminal “a” is assigned to be point at higher voltage + - i(t) v(t) a bIf the actual current and voltages are in the orientations as shown, which two- terminal element device is delivering power? The one on the left True; terminal a is high energy point 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 A 3V a b + - 4A 3V a b 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)2) p(t) = v(t) i(t) p(t) = (3V)(2A) = 6 Watts 6W of power absorbed by this element p(t) = [-v(t)] i(t) p(t) = (-3V)(4A) =-12 Watts 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 A 5V a b A 3V a b 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)4) p(t) = v(t) [-i(t)] p(t) = (5V)(-2A) =-10 Watts 10W of power delivered by this element p(t) = [-v(t)] i(t) p(t) = (-3V)(-3A) =9 Watts 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? 1)2) v(t) = [-i(t)]R v(t) is negative indicating a voltage gain This two terminal electric circuit element is the resistor. The resistance of material used to make a resistor is always a positive number. b - + a v(t) b - + a v(t) = i(t)R v(t) 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? 1)2) - i(t) is negative indicating the charge in the capacitor is dropping. This two terminal electric circuit element is the capacitor. The capacitance of a capacitor is always a positive number b a + - i(t) a b - + i(t) = [dv/dt)]C i(t) is positive indicating a gain in the charge of the capacitor. i(t) = [ dv/dt)]C Energy being stored Energy being released Instrument Systems Electrical Engineering Fundamentals

Other conventions that might seem confusing. Are the following inductors absorbing or releasing energy? 1)2) - v(t) is negative indicating the inductor is releasing energy. This two terminal electric circuit element is the inductor. The inductance of practical inductors is always a positive number a b - + v(t) v(t) = [di/dt)]L v(t) is positive indicating the inductor is absorbing energy. v(t) = [ di/dt)]L Energy being stored Energy being released b a + - v(t) Instrument Systems Electrical Engineering Fundamentals

Other concepts that might seem confusing. The perfect world. The ideal voltage supply The ideal current source v(t) i(t) 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 car battery an ideal voltage 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. Is a lighting bolt, (like the one Ben Franklin felt when he was flying his kite) an ideal current supply? Instrument Systems Electrical Engineering Fundamentals

Other concepts that might seem confusing. 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. Hmmmmmm! Life does seems to be full of difficult questions! Circuit configurations Voltage as function of time that is available for use. Is the energizer bunnies battery an ideal voltage supply? v(t) b + - a s s = 1) 2) 3) Voltage value profile as a function of time remains the same. Instrument Systems Electrical Engineering Fundamentals

Other concepts that might seem confusing. Hmmmmmm! Life does seems to be full of difficult questions! Circuit configurations Is the energizer bunnies battery an ideal voltage supply? Ideal voltage supplies connected in series. v(t) = v (t) s1 v (t) s2 + v(t) b a s1 v (t) s2 v(t) b + - a s Instrument Systems Electrical Engineering Fundamentals

Other concepts that might seem confusing. Hmmmmmm! Life does seems to be full of difficult questions! Circuit configurations Is the energizer bunnies battery an ideal voltage supply? v(t) b + - a s b a s1 v (t) s2 v(t) = v (t) s1 v (t) s2 = Ideal voltage supplies connected in parallel. v(t) b a s1 v (t) s2 Ideal voltage supplies connected in series. v(t) = v (t) s1 v (t) s2 + Instrument Systems Electrical Engineering Fundamentals

Other concepts that might seem confusing. Circuit configurations Ideal current sources connected in series. i(t) = i (t) s1 i (t) s2 + Ideal current source i(t) b a s1 i (t) s2 Ideal current sources connected in parallel. b a i(t) s1 i (t) s2 i(t) = i (t) s1 i (t) s2 = Instrument Systems Electrical Engineering Fundamentals

Other conventions that might seem confusing. 2) 3) a b - + c - d + A current i (t) entering a dotted terminal in one inductor produces an induced voltage which is positive at the other dotted terminal. 1 i 1 If a second current i (t) enters the other inductor at its dotted terminal there is also a voltage induced in the first inductor. (Mutual induction) 2 i 2 v (t) ) A current i (t) entering a dotted terminal in one inductor produces a voltage across the terminals of that inductor. 1,v (t), 1 (Mutual inductance) M 4) If a second current i (t) enters the other inductor at its dotted terminal there is a voltage drop produced across that inductor This four terminal electric circuit element is a coupled inductor (transformer). Instrument Systems Electrical Engineering Fundamentals Things to remember about transformer passive diagram symbolism.

Other conventions that might seem confusing. 2) 3) a b - + c - d + A current i (t) entering a dotted terminal in one inductor produces an induced voltage which is positive at the other dotted terminal. 1 i 1 If a second current i (t) enters the other inductor at its dotted terminal there is also a voltage induced in the first inductor. (Mutual induction) 2 i 2 v (t) ) A current i (t) entering a dotted terminal in one inductor produces a voltage across the terminals of that inductor. 1,v (t), 1 (Mutual inductance) M 4) If a second current i (t) enters the other inductor at its dotted terminal there is a voltage drop produced across that inductor a b - + i 1 v (t) = 5)5) This four terminal electric circuit element is a coupled inductor (transformer). v (t) c - + i 2 d Instrument Systems Electrical Engineering Fundamentals Things to remember about transformer passive diagram symbolism. v (t) = 2

Other conventions that might seem confusing. v (t) 2 = M (d /dt ) i 1 i 2 a b - + c - d + i 1 M v (t) 1 = L (d /dt ) i 1 1 v (t) 2 = L (d /dt) i 2 2 v (t) 1 = M (d /dt ) i 2 v (t) v (t) = 1 v (t) v (t) = 2 The voltage difference v 1 (t) between point b and point a is the sum of the voltage, v 1 induced by the change in current i 1 and the mutually induced voltage v 1 produced by the change in the flow of current i 2. The voltage difference v 2 (t) between point d and point c is the sum of the voltage, v 2 induced by the change in current i 2 and the mutually induced voltage v 2 produced by the change in the flow of current i 1. Instrument Systems Electrical Engineering Fundamentals Things to remember about transformer passive diagram symbolism. M 12 M 1 2 = M = M 12 M 21 =

Other conventions that might seem confusing. An example of a three coupled coil system that is complaint with the passive diagram symbolism. i (t) 3 M 12 M 23 M 13 i (t) v (t) A) how many equations are needed to describe this three coupled inductor system? Passive diagram example v (t) = = = three v (t) = = = But it is more convenient to arrange them this way! Instrument Systems Electrical Engineering Fundamentals

An example of a three coupled coil system that is complaint with the passive diagram symbolism. i (t) 3 M 12 M 23 M 13 i (t) v (t) Passive diagram example v (t) = = = Other conventions that might seem confusing. L (d /dt) i 1 1 i 2 2 i 3 3 After a bit more manipulation and substitution we get. + v (t) 1 = L (d /dt) i 1 1 M (d /dt ) i 2 12 Instrument Systems Electrical Engineering Fundamentals v (t) 1 = M (d /dt ) i 2

An example of a three coupled coil system that is complaint with the passive diagram symbolism. i (t) 3 M 12 M 23 M 13 i (t) v (t) Passive diagram example v (t) = = = Other conventions that might seem confusing. L (d /dt) i 1 1 i 2 2 i 3 3 After a bit more manipulation and substitution we get. + v (t) 1 + = 2 = = L (d /dt) i 1 1 M (d /dt ) i 2 12 M (d /dt ) i 3 13 M (d /dt ) i 1 31 L (d /dt) i 2 2 M (d /dt ) i L (d /dt) i 3 3 M (d /dt ) i 2 32 M (d /dt ) i 1 21 Note: M 31 = M 13 M 23 = M 32 M 21 = M 12 Instrument Systems Electrical Engineering Fundamentals v (t) 1 = M (d /dt ) i 2

An example of a three coupled coil system that is complaint with the passive diagram symbolism. i (t) 3 M 12 M 23 M 13 i (t) v (t) Passive diagram example Other conventions that might seem confusing. + v (t) 1 + = 2 = = L (d /dt) i 1 1 M (d /dt ) i 2 12 M (d /dt ) i 3 13 M (d /dt ) i 1 31 L (d /dt) i 2 2 M (d /dt ) i L (d /dt) i 3 3 M (d /dt ) i 2 32 M (d /dt ) i 1 21 Instrument Systems Electrical Engineering Fundamentals 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. Try the next practice problem to see how this works.

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? A) the defined voltages and corresponding currents do not all conform to the passive sign convention. Note that either: B) the positive voltage terminal is not at the dot. or 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. 1) Reverse direction of i 1 i (t) 3 M 12 M 23 M 13 i (t) v (t) ) Reverse direction of i 2 3) Reverse direction of v 2 4) Reverse direction of v 3 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? 1) Reverse direction of i 1 i (t) 3 M 13 M 23 i (t) v (t) 2 1 2) Reverse direction of i 2 3) Reverse direction of v 2 4) Reverse direction of v 3 + v (t) 1 + = 2 = = L (d /dt) i 1 1 M (d /dt ) i 2 12 M (d /dt ) i 3 13 M (d /dt ) i 1 31 L (d /dt) i 2 2 M (d /dt ) i L (d /dt) i 3 3M (d /dt ) i 2 32 M (d /dt ) i i (t) 1 M v (t) 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? + v (t) 1 + = 2 = = L (d /dt) i 1 1 M (d /dt ) i 2 12 M (d /dt ) i 3 13 M (d /dt ) i 1 31 L (d /dt) i 2 2 M (d /dt ) i L (d /dt) i 3 3M (d /dt ) i 2 32 M (d /dt ) i v (t) 1 + = L (d /dt) i 1 1 M (d /dt ) i 2 12 M (d /dt ) i v (t) 2 = L (d /dt) i 2 2 M (d /dt ) i M (d /dt ) i v (t) 3 + = M (d /dt ) i 1 31 L (d /dt) i 3 3M (d /dt ) i i (t) 3 M 13 M 23 i (t) v (t) 2 1 i (t) 1 M v (t) Thus the voltages for the three coupled coils are modeled with the following three equations. i (t) 3 M 12 M 23 M 13 i (t) v (t) Instrument Systems Electrical Engineering Fundamentals