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# Circuit Analysis II EGR 314

GVSU

GPA 3.68

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This 3 page Class Notes was uploaded by Tyree Funk on Saturday September 26, 2015. The Class Notes belongs to EGR 314 at Grand Valley State University taught by Staff in Fall. Since its upload, it has received 58 views. For similar materials see /class/214377/egr-314-grand-valley-state-university in Engineering and Tech at Grand Valley State University.

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Date Created: 09/26/15

TMY F99 GRAND VALLEY STATE UNIVERSITY Padnos School of Engineering EGR 314 NOTE 2 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 td p 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 quotexactquot 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 Assume a function fX behaved well enough so that the derivative eXists everywhere Let 1 EGR 314 NOTE 2 F99 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 gdp d fdt The derivatives are constants at speci c values for p and t Let these P df df constants M and N be defined as M d and N Observe that if P dM dzf dN dzf dzf dzf M dN d at and 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 This criteria can be applied to the set of equations labeled 1 as piorv MLorRC NRiorv dRC dt 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 dL so that E at R and at 1 Since the cr1ter1a for an exact d1fferent1al fails these equatlons 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 314 NOTE 2 F99 d y3dt and d ydt y L y RC dy R dy l d d d Iy J t an I y IR C t lny t and lnyR iC t These equations are usually expressed in exponential form as RLt ye and ye lRCt 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 5e1Rctdt dV e1RCt RC EARWat dieRL39 and L Both equations are now integrated to yield eRLt K1 ieRLt1lt2 and Vse1Rct K1 Ve1Rct1lt2 V it Es K1 K2 e39lRLt 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 i0ISK1K2 and moo The complete solution for the current through a series RL circuit is then V it elkL for t gt 0 For the series RC circuit V0 0 and Voo VS so that 3

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