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Elements of Electrical Engr II

by: Dr. Reina Hane

Elements of Electrical Engr II ECE 207

Dr. Reina Hane
GPA 3.79

Clifford Grigg

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Clifford Grigg
Class Notes
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This 19 page Class Notes was uploaded by Dr. Reina Hane on Monday October 19, 2015. The Class Notes belongs to ECE 207 at Rose-Hulman Institute of Technology taught by Clifford Grigg in Fall. Since its upload, it has received 18 views. For similar materials see /class/225081/ece-207-rose-hulman-institute-of-technology in Electrical Engineering & Computer Science at Rose-Hulman Institute of Technology.

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Date Created: 10/19/15
11 Circuit Analysis Circuits are formed when circuit elements resistors etc are connected together Two common ways for elements to be connected are parallel and series Elements in parallel share both nodes and so have a common voltage Elements in series have share one node that is shared by no other element and so have a common current Is parallel and series the only ways in which elements can be connected lecture I outline Resistor Combinations Ohm s Law Power Senes Parallel Ohm s Law Power Examples Consider the circuit shown below The resistors could be lights heaters motors etc A B 1 Suppose identical 100 Q resistors A B and C are connected to the DC voltage supply as 100 Q 100 9 shown c i Which resistors dissipates the most power 100 9 ii Which resistors draws the most current I l iii What happens to the power dissipated in A and B when C is removed that is when C is replaced by an open circuit iv v What happens to the power dissipated in B and C ifA is removed lecture I outline N 0quot 5 Suppose identical bulbs A B and C are connected to the DC voltage supply as shown i How do the of the three bulbs compare in brightness C ii Which bulbs draws the most current iii What happens to the brightness of A and B when C is removed that is when C is I replaced by an open circuit I I iv What happens to the brightness of B and V C ifA is removed Consider two light bulbs each designed to have 110V across them One has a power rating of 40 W and the other has a power rating of 75 W i Which bulb has the higher resistance ii Which carries the greater current i In the circuit below how does the power dissipated in the resistors compare with one ii Suppose resistor A is removed How does the power dissipated in the remaining resistors change V 100 1000 lecture I uutlme 5 i In the circuit below how does the brightness ofthe bulbs compare with one another Suppose bulb A is removed How do the brightness of the remaining bulbs change v A B 4L confirm your predictions in 1i 1iii and 1iv adjust V5V confirm your predictions in 4D and 4H adjust V6V see below for wiring guides 06 V Bench Power Sl gply D I S P L A Y BE pawquot EV mi 25 mm 25 D pgbeee Ot06V i Press power ii Select 6V selects the O to 6V adjustable supply iii Press Output OnOff turns 0 to 6V supply on iv Turn dial to adjust output voltage 0 to 6V lecture I outline A 3911 Breadbording Circuit 1 adjust voltage to 5V r5V l r Er m E B u u amass E m EIEIE M n n aaaaa a a I m E E Using digital multimeter DMM to measure voltages and currents measuring current through A and B HIOOHI measuring voltage across C set DMM for DC Lo Lo III Q I H39 set DMM for DCV LO LO 5v mfg 39 l l El m lmlllll EEEEEEEEEEE aaaaaaaaaaa 5 EE a EH a EEEE lecture I outline A c39n Breadbording Circuit 4 adjust voltage to 6V l6V 1 Mr H E 9 m n F n F E m lecture I outline 51 Power Factor Correction The power factor pf is the ratio of real power to apparent power It is dimensionless and between 0 and 1 i a pure resistance has unity pf ii a pure inductance has zero pf lagging iii a pure capacitance has zero pf leading Almost all industrial loads are lagging Low pf results in unnecessarily high currents requiring larger conductors and resulting in unnecessary losses and in too much voltage being dropped across the feeders the power lines feeding the load lfa power source is required to deliver power average power to a load which has a low pf pf cos e the required current will be higherthan it would be ifthe power factor were larger This excess current causes excessive losses in the power lines and in the entire power system The power company is then required to devote capacity to this excess current that is present only because the load power factor is small that is because the load appears too strongly inductive I RL M 4 VS V load What is physically going on is that energy is being temporarily stored at the load during one part ofthe cycle and then sent back to the source during another part A low pf indicates a significant transfer ofstored energy with respect to energy dissipation the transfer ofstored energy does no useful work Energy is just transferred between the source and load lecture 5 outline 52 Almost all industrial load are inductive because of motors and other electric machines This provides a practical way to correct a low power factor nearer to one By adding a capacitor a power factorcorrecting capacitor in parallel with the inductive load we may improve ef ciency and save a great deal of money Power companies encourage power factor correction so that taking care to maintain a high power factor can easily result in savings in the million of dollars per year for a large plant lt s usual to correct the power factorto 09095 Adding too much capacitance is wasteful and ifwe overcorrect the power factor becomes leading and its magnitude to decrease with additional capacitance load VS 23 jmc s P jQL General Analysis lecture 5 outline 53 Example i Given 132 kV 60 Hz source nd the pfc capacitor required to correct a load of 36 MW at pf 0707 lag to pf 095 lag Compare the line current just magnitude before and after installing the pfc capacitor iii Compare the apparent power before and after installing the pfc capacitor Suppose the power company charges 9 KVAmonth in demand charges How much money do we save by installing pfc capacitors v iv v What is going on physically with power factor correction is that less energy is being swapped between the load and source Much of the stored energy is now being swapped between the inductive elements in the original load and the power factor correction capacitor lecture 5 outline A very common way to measure power is via an ammeter a voltmeter and a wattmeter as shown below 0 0 Id VS 0 oa sIjQL The scheme is this with the voltmeter measuring voltage rms the ammeter measuring current rms and the wattmeter measuring power average power in W We can quickly determine all quantities of interest for power measurement Example Suppose I 20 A V 120 V and P 2 kW Find i apparent power i39 power factor assume lagging iii complex power iv reactive power v vt and it assume 60 Hz with vt taken as reference v v lecture 5 outline 55 More sophisticated power meters like those in the power lab measure phasor current and phasor voltage These meters measure current voltage power factor power apparent power and reactive power The meter assumes current and voltage polarities as shown below that is the polarities that result in power flowing from the source to the load 80 for example ifthe load is inductive the reactive power will be positive wattmeter ammeter Would the readings be affected if the meter voltage polarity were reversed How lecture 5 outline Stress and Strain Measurement Introduction Stress and strain can be measured in many ways typically based in some fashion upon the fact that a structure is deformed or strained when it experiences stress Piezoelectric transducers use the fact that in piezoelectric materials electric charge is separated when the materials is strained and vice versa Other transducers are often based on the resultant changes that the strain produces in capacitance inductance or resistance We are going to focus on perhaps the most widely used transducer the metallic strain gage Its fundamental principles of operation use and application necessary signal conditioning and its limitations will be explored Metallic electric resistance strain gages Metallic ER strain gages use the fact that a metal is deformed or is strained when it experiences stress 39n The strain results in a change in its resistance R z pLA R R AR The shape is usually not a simple F bar39 unstressed stressed lecture 16 uutlme The most common form for metallic strain gages is forthe wire grid to be formed from constantan a CuNi alloy and the backing to be polyimide a hightemperature polymer This metal foil is bonded to the 162 4 direction of strain foil trace solder pads plastic backing element to be strained so that it will experience the same strain as that of the element While this one type is most common metallicfoil strain gages are manufactured from a variety of different metals and alloys with a range of resistance values a range of sensitivity of resistance to strain and varying resistancetotemperature characteristics Common nominal resistance values the resistance of the unstrained strain gage are 120 Q 350 Q and 1000 Q Applications Strain gages can be related directly to stress through the elastic modulus Young39s modulus E Othertypical measurements are force torque and pressure N 0quot If the relationship between applied force and strain can be determined for a given structure then strain gages can be used to measure the force If the relationship between applied torque and strain can be determined for a given structure then strain gages can be used to measure the torque If the relationship between applied pressure and strain examples include a container or a membrane can be determined for a give structure then strain gages can be used to measure the pressure lecture 16 uutlme 163 Theoretical background For a resistor of uniform cross section A of length L and of resistivity p 1 dR dp dL dA Assuming a circular cross section same end result is obtained the boxed equation below if other cross sections are used 02 2 439 A D The axial strain is related to the transverse strain via v Poisson39s ratio a material property This permits us to write v T v 8 Substituting these relationships into 1 zdlg2vg R p a 3 Strain gages are characterized by the strain gage factor 8 s dRR 12v dpp 8 8 a a lecture I 6 outline 164 Overview of their characteristics For metallic strain gages the gage factor is typically around 26 usually closer to 2 Semiconductor strain gages can have gage factors higher than 100 For metallic gages strains as high as 004 4 elongation are measured routinely For semiconductor gages the strain is limited to about 0003 03 elongation For a constant temperature S is fairly constant with strain Depending on the material S can be sensitive to temperature changes Specific examples Measuring force Strain gages can be used to measure force Given a structure39s geometry and composition the relationship between applied force and resulting stress is calculated Knowing material properties the resulting stress is related to the strain a For a given gage factor S the resulting change in resistance is easily calculated AR m 88R Measuring torque Strain gages can be used to measure torque Given a structure39s geometry and composition the relationship between applied torque and resulting stress can be calculated Knowing material properties the resulting stress is related to the strain a For a given gage factor S the resulting change in resistance is easily calculated lecture I 6 outline 165 Using strain gages in a measurement system In a system designed to measure for example stress the strain gage provides a change in resistance This is the signal available from the strain gage This change in resistance is typically transformed to a change in voltage often by using a Wheatstone bridge Then since the change in voltage is usually too small to be useful an amplifying circuit is used to raise the voltage changes to a useful level Signal conditioning for metallic strain gages Wheatstone Bridge The Wheatstone Bridge is used to convert a change of resistance into a change of voltage V 9 With Vb measured across an open circuit the current through R1 is the same as that through R4 the current through R2 is the same as that through R3 By voltage division we have Vb VS R2 RS R R The bridge is quotbalancedquot that is Vb 0 for lecture 16 uutlme 166 Now let39s replace one of the resistances on the bridge with a strain gage resistor and make the other three resistances the resistance of the strain gage39s nominal resistance Note that when the strain gage experiences no strain all four resistances in the bridge will have the same value Analysis to nd Vb By voltage division V RFAR R V 2R22RAR 2R2 RAR b 4R2 2RAR AR R V for AR ltlt R almost always the case 4 4 5i 5 4R AR 4R This configuration is called the oneactivearm bridge since only one of the arms responds to the measurand Vb E V5 for oneactivearm bridge Often bene ts are available by placing more than one strain gage resistance into the bridge Consider a cantilever beam Place R1 and R3 in tension and R2 and R4 in compression Bene ts are elimination of torsion loading effects 2 more sensitivity 3 temperature compensation R2 and R4 on underside of beam 39 39 V lecture I 6 outline This configuration is called the four activearm bridge since all of the arms respond to the measurand Vs Analysis to find Vb Again by voltage division Vb RARRAR Vs RARRAR Vs RR RR 2R and we obtain for the fouractivearm bridge be sure to note the assumptions that this relation rests on identical strain gages common temperature variations Typical voltages What voltages should we typically expect for Vb That is fortypical strain gages with typical strains is Vb going to be 10 V7 1 V 1mV or 1uV Fouractivearm Wheatstone bridge typical R metallic strain gage 120 Q typical gage factor 8 2 typical strain 2 0001 typical VS 10 V Vb SeVg 20mV lecture 16 uutlme 168 This is a fairly small voltage amplification is often required Differential Amplifier whereRmR1mR2Rng4 lecture I 6 outline


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