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# Circuit Analysis I EGR 214

GVSU

GPA 3.68

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This 14 page Class Notes was uploaded by Tyree Funk on Saturday September 26, 2015. The Class Notes belongs to EGR 214 at Grand Valley State University taught by Staff in Fall. Since its upload, it has received 73 views. For similar materials see /class/214380/egr-214-grand-valley-state-university in Engineering and Tech at Grand Valley State University.

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Date Created: 09/26/15

TMY W02 GRAND VALLEY STATE UNIVERSITY Padnos School of Engineering EGR 214 DIFFERENTIAL EQUATION SOLUTION FOR SINGLE TIME CONSTANT CIRCUITS The voltage and current responses to the single time constant C R and L R circuits illustrated in gure 1 is commonly referred to as the quotforcedquot as opposed to the quotnaturalquot response of first order systems FIG 1 By applying Kirchhoff39s Voltage Law to the circuits in gure 1 the following differential equations are derived di t dv t vs RitL and vs RC vt for tgt0 dt dt The first step in solving these differential equations is to separate the variables and find a form of the resulting equation that can be integrated directly This means grouping the variables with their differentials ie is with the dis and v s with the dv s 1 VsdtRidtLdi and VsdtRCdvvdt Notice however that in equation 1 the variables i or v and t cannot be separated Therefore a different strategy for finding the solution will have to be found Observe from equation 1 that the right side of the equation quotlooks likequot the derivative of a product That is let f pt then taking the derivative of both sides with respect to X p t orindifferentialform dfpdttdpdpt bi bi bi which can be integrated directly as Idltp t p t f A differential in a form that can be integrated directly is referred to as an quotexactquot differential A strategy for solving differential equations is to try and force the differential equation to be quotexac quot by multiplying both sides of the equation by an quotintegrating factorquot and then integrating In order to do this a definition for an quotexact differentialquot must be derived first EGR 214 First Order Differential Equations W02 Assume a function fx behaved well enough so that the derivative exists everywhere Let f f39x or in differential form df f39x dx x If f is a function of two variables p and t then the total differential is df 3 fdp dt The derivatives are constants at speci c values for p and t Let these P constants M and N be de ned as f and N Observe that if P dM dzf dN dzf dzf dzf dM dN d i that 7 an then so dt dp dt dp dt dp dp dt dt dp dt dp This implies that for a differential equation to be an exact differential then dMdN dt dp This criteria can be applied to the set of equations labeled 1 as piorv MLorRC NRiorv so that 11 14 at R and dE tc 1 Since the criteria for an exact differential fails these equations are not exact Therefore another approach needs to be found to make equations 1 exact differentials A common approach is to multiply both sides of the equation by another function y called an integrating factor and solve for y by applying the criteria for an exact differential The function y will then make the differential equation integrable hence the name quotintegrating factorquot An important thing to note is that this integrating factor is applicable to all physical systems Muliplying equation 1 by y results in the equations 2 yVs dt yRidtyLdi and yVs dt yRCdvyvdt By applying the criteria for an exact differential the following differential equations result dy dy RL and RC y dt y dt These equations are easily solved since the variables can be separated and the resulting expressions integrated directly EGR 214 First Order Differential Equations W02 1 y dt y R C Q Q yEdt and y L d R d l 1Ifdt and 1 39dt lny t and lny t These equations are usually expressed in exponential form as RLt y e lRCt and y e There are constants associated with the inde nite integration that are zero based on the initial conditions for these differential equations If these integrating factors are now substituted into equation set 2 then elmt VS dt elmt R i dt elmt L di and ellRCt VS dt ellRCt R C dV ellRCt v dt Observe that both equations are now exact differentials and can be written as V V se1RCtdt dV e1RCt RC eRLgt dt dieRL39 and L Both equations are now integrated to yield eRLt K1 i eRLt K2 and Vs e1RCt K1 V e1RCt K2 it 1 K1 K2e39RLt and Vt Vs K1 K2e391RCt The constants K1 and K2 are determined from the boundary conditions at t 0 and t 3 00 For the series RL circuit i0 0 and ioo VS R so that 1 Therefore K1 i0 and K2 ioo i01 K139K2 and i The complete solution for the current through a series RL circuit is then it 151 e39lRth fort gt 0 Veri cation of Ohm s Law Kirchoff s Voltage Law and Kircho s Current Law Brad Peirson 22405 EGR 214 7 Circuit Analysis I Laboratory Section 04 Prof Blauch Abstract The purpose of this report is to verify Ohm s law Kirchoff s Current Law and Kirchoff s Voltage Law Ohm s law relates voltage to resistance and current Kirchoff s laws deal solely with current and voltage A circuit was built using a given schematic This circuit was also drawn in PSpice The circuit was then analyzed using three separate methods the three laws the PSpice simulation and a digital multimeter Ohm s law and Kirchoff s laws were found to be valid after comparing the results from all three tests 10 Introduction Ohm s law Kirchoff s Voltage Law and Kirchoff s Current Law are essential in the analysis of linear circuitry Kirchoff s laws deal with the voltage and current in the circuit Ohm s law relates voltage current and resistance to one another These three laws apply to resistive circuits where the only elements are voltage andor current sources and resistors Using the three laws any resistance of current through or voltage across a resistor can be found if any two are already known The purpose of this report is to provide veri cation of these laws Section 2 outlines the three laws and gives simple examples and conventions Section 3 analyzes the circuit based solely on the schematic and generates the three equations needed for further analysis Section 4 contains a circuit simulation run in P Spice Section 5 contains the data and calculations from the measurement of the physical circuit Section 5 also contains the error analysis between the methods Section 6 is the results discussion section Section 7 is the experimental results and conclusions section 20 Circuit Analysis Techniques The following sections describe the three basic laws that are used in analyzing linear circuits 21 Ohm s Law Ohm s law is used to relate voltage to current and resistance It states that voltage is directly proportional to current and resistance This is stated mathematically as V1R 1 where V ls the voltage aeross an element othe erremtm volts Ils the eurreht passmg through the element m amps andR ls the resrstahee of the element m ohms Gwen any two ofthese quantltles Ohm39s law can be usedto solve for the thlrd 22 Kirchn santzge Law Klrchoff s Voltage Law KV39L states that the sum of all voltages m a eloseolloop mustbe zero A eloseolloop ls apath m a erreurt that doesn39t eohtam any other eloseol loops Loops land 2 m Flgure l are examples ofclosedloops Flgure 1 An example ofKV39L The pehmeter of the erreurt ls also a eloseolloop but smee rtmeluoles loops 1 and Ifloop the KVL equahoh ls mVZeVso 1 Thls equahoh holols true only lfthe passlve slgn eohvehhoh ls sahsfreol 1n the ease of v 71 an M u Wheneverposslble 23 Kirehm39s Curmsz Kunhu s Currant LawKCL dais wnh the cunmls uwmg mm and em Ufa gveh nude KCL states that the sum ufall eurrems 212 nude must equal zem Tms rs illustrated m Frgure 2 Frgure 2 An Example ufKCL The equaueh ub39amed hy KCL fur the nude shuwn m Frg 215 I 7 I1 7 I n 2 tn the use ufKCL the passrve srgh eehvehueh dais wnh the direcnm ereurrems wnh awed m the nude Currems amenng the hederhug have eppesne 5x315 as Lhuse eehuhg the nude The passrve 5131 eehvehueh wnh rewect m KVL an 315 he apphed m KCL On many sehemaues the pulanues ufreslsturs are alrady assrgred 5 the direchuns er eurrems shuuld he assigued sueh that the man 15 emenhg the pusmve lammal Tms wru srrhphry later ultimatum 3n Amlyxis The mrmltanalyzedmhs hhmmryrs shuwn m Frgure 3 I1 A V1 V2 I3 Mi j A R3 V3 R4 V4 Vs 7 Figure 3 Resistive Circuit The currents voltages and polarities were labeled as shown on the given schematic The current directions and voltage polarities have all been assigned such that the passive sign convention has been satis ed wherever possible The four resistors were chosen at random Their resistance was then measured with a multimeter The nominal and measured resistance values are given in Table 1 Table 1 Nominal and Measured Re i lance Values Resistor Nominal Value Measured Value k9 k9 R1 68 675 R 22 2193 R 10 985 R4 1 994 KVL was applied to the two closed loops ofthe circuit using the symbolic labeling in Figure 3 The KVL equations are shown in 3 and 4 KVL ABD A V V2 7 V 0 3 KVL BC D B V V4 7 V3 0 lt4 Current 12 is the same as 14 so only node B generates a KCL equation This equation is given in 5 17127130 5 n L ems analysis mm mum39 39 39 vuha c and currents 39 PSpice 40 Simulation TL r 4 PSpice rm schematic is given in Figure 4 l1 R1 B IZ R2 0 A v1 675k V2 9 85k 156m 3 97 11uA 4 v3 v4 R3 2193k R4 994k V5 10v 39 39 D 7 Figure 4 PSpice simulation diagram with results The measured resistance values were used in the simulation in order to obtain accurate results The ground was arbitrarily placed to give PSpice a reference node The currents were then input into 5 to verify the equation 116637196117970110 6 Given that all three terms in 6 are in microamps the PSpice simulation does verify KCL In order to use the KVL equations Ohm s law had to be applied to the resistors and the current owing through them This gave the voltage across each resistor These voltages were then inserted into 3 and 4 78750 212745 10 00205 7 1932470195013 212745 000033 8 The sum of the voltages in a closed loop should be zero if KVL is true Neither 7 nor 8 produced zero volts The sum of voltages produced while not exactly zero were extremely close to zero This was strong evidence that KVL is true 50 Experimental Results The circuit was constructed on a breadboard using the four random resistors Ten volts was measured on the CADET trainer s power supply and applied to the circuit The voltages across and currents through each of the resistors was then measured and recorded in Table 2 Table 2 Measured values of resistance and current Voltage Measured Current Measured V mA V1 737 11 108 V2 1803 12 081 V3 2 621 13 025 V4 0871 The measured voltage values were inserted into 3 and 4 to verify KVL 7371803 10 827 9 18030871 2621 0053 10 The measured currents were inserted into 5 to verify KCL at node B 106 081 0250 11 There was no error calculation performed on the KVL and KCL results because the expected value was zero Ohm s law was used to calculate the resistance values using the values of the current through and voltage across each A percent error analysis was then performed between the measured values and the Ohm s law values A sample calculation from the analysis is shown in 12 error W X 100 gtlt100 103 12 675 682 measured These errors would ultimately be used to determine the validity of Ohm s law The calculated resistances as well as the errors are given in Table 3 Table 3 Measured calculated and percent error for resistances Resistor Measured Value Calculated Value Error 19 19 R1 675 682 103 R2 2193 223 166 R3 985 105 619 R4 994 108 796 60 Discussion The simulation of the circuit veri ed KCL This was shown by 6 PSpice found the sum of the two currents exiting the node to be equal to the current entering the node The experimental results of the KCL test were not so close There was less than two percent error between the measured and calculated 11 s This error is acceptable in stating that KCL is true The simulated KVL test was not as expected Neither 7 nor 8 produced zero volts However both of the equations added up to voltages small enough to be negligible in stating that KVL is valid As with the KCL tests the experimental KVL results in 9 and 10 did not meet the predicted values Both of the equations produced values within TMY W02 GRAND VALLEY STATE UNIVERSITY Padnos School of Engineering EGR 214 L amp C EQUIVALENT CIRCUITS AT t 0 and t 00 FOR STC SYSTEMS Since the form of the response for all single time constant STC systems is known yt yoo y0 yoo em for t gt 0 it is only necessary to determine the boundary values y0 and yoo in order to nd the complete response Both of these boundary values can be determined by analyzing an equivalent circuit valid at each boundary Inductance The inductance tries to keep the current owing through a system An impedance equivalent circuit at t 0 and at t 00 can be found by analyzing the system response when the current is interrupted at t 0 after having been connected for a long period of time as in gure 1 Io L l FIG 1 When the switch is opened at time t 0 the current is interrupted The equivalent inductance however tries to keep it owing in the same direction by increasing the voltage across the inductor in a direction opposite to the cause of the initial current ow Therefore for an inductor alone the voltage increases toward in nity while the current through the inductor drops to zero The impedance which is the ratio of voltage to current then goes to in nity That is di v 3 oo ZIW T T 33 0 The equivalent circuit model with the same behavior characteristics possessed by an inductance as described above is shown in gure 2 The initial current ow is represented by the current source owing in the same direction as the initial conditions and the equivalent inductance becomes an open circuit EGR 214 L amp C Eq th W02 At t 0 a V open Io b FIG 2 Therefore at t 0 the energy accumulated at the initial condition t 0 is represented by the current source 10 and the 39 r J ofthe 39 J t 39 t t 39J goes to in nity or an open circuit The Norton39s equivalent circuit a current source in parallel with the impedance of the inductance is an equivalent circuit model that represents the behavior of an inductance which is to maintain the ow of the current in the system As time progresses the rate of change of the current goes to zero Therefore the voltage across the equivalent inductance goes to zero and the equivalent impedance becomes a short circuit The t gt 00 equivalent circuit is shown in gure 3 All 00 a V short I0 b FIG 3 Capacitance The capacitance tries to keep the voltage constant across a system In this respect it is the dual of the inductance parameter An impedance equivalent circuit at t 0 and t 00 can be found by analyzing the system response when the voltage source is removed at t 0 after having been connected for a long period of time as in figure 4 EGR 214 L amp C Eq th W02 T o gt FIG 4 When the switch is opened at time t 0 the capacitance will try to keep the voltage constant at V0 In order to do this a current must ow in such a direction that the polarity of the voltage drop is in the same direction as the source As the voltage drops towards zero the current must increase to in nity The impedance ofthe capacitance then goes to zero or a short circuit at t 0 That is V0 7 V0 7 V0 Z39W CLV w 3 dt t0 The equivalent circuit model with the same behavior characteristics possessed by a capacitance as described above is shown in gure 5 The initial voltage is represented by a voltage source with the same polarity direction as the source while the equivalent capacitance becomes a short circuit Att0 a gt short V0 b FIG 5 Therefore at t 0 the energy accumulated at the initial condition t 0 is represented by the voltage source V0 and the 39 r J of the J 39 39J goes to zero or a short circuit The Thevenin39s equivalent circuit a voltage source in series with the impedance of the capacitance is an equivalent circuit model that represents the behavior of a capacitance which is to maintain the value of the voltage in the system 39 As time progresses the rate of change of voltage goes to zero Therefore the current through the equivalent capacitance goes to zero and the equivalent impedance becomes an open circuit The t gt 00 equivalent circuit is shown in gure 6

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