00020001010008main.ACT0002020012eActivity Save.EAC010000002edf  electric.EAC main.ACT ,0 $ ] '[5`E` al Networks Q|by Ravinder Kumar Alcorn State University !! , MS 39096?rkM@aF.edur" ObjX:T$Solution of elementary modeling probs # ctrical networks involves solv1syst0f Aear equacs and1terpret^the9 . Purposethis e-Activity to demonstrate howwe.  #$WhatZaې?.8Electric networks are used to establish functioning of e7#al devices. They contain power sour whicDo  current_flow through the . Resistorregulate,requirF b. Ckis measu.in terms amperes ( ),(volts (v$andan&Ohmo see a diagram.an1 , click onsp Problem be(. \%Picture Definitions M] 80Mp 1a߫4 JyD`$",#$($"9`K'1=r٦(Pe( +P&9L_r 0@!N{Z 'sZq !b (b O!A 'By(Q!@ B9@/qB{- I\oE  [R0 @<8Dx`P$ED@D"( $}$D $A@ydžP@dq`tϴ7Q@@K@{9oN8< x9EPD E QE }PǀKzq9؊!(J( ,$( !J/z(=B'(E( >'zx(|z#E؄>jL   T@*@ <  @ @  @ 0 dQ`@ <? TƆ R;_pŢ֢  Up 9 yg8L (DP@QA` GCAQ DDQE yȜsFOperation of electric networks follows Ohm's Law and Kirchoffs. V\_ .(tOhm's law, named after German -physicist, Georg'E (1789-1851), _gives aBrelation between thempower<resk ance, andcurrent. It states  V = RI where R is d in s, I '!u%ampere!VE (charge)T volts.am[r\{ Kirchoff Laws (tap?)+MU 66 1aΠ< ` l d0? ``s=,8 d90HH`M"\0@0 `0 @H 0 \؊0m`t P  x <6Y< 0;lPk8!Jkl<ppDؘtH0#  #| 8 R H~a0'0I U x`I`_3U8_4mHc1"Hz:-*  @p0["|H `  %0 @ ~H 1@x  0` (0rs!H$S!D"0H`#Pp0AH —` A0 ` xi  o  U@8`Y` lx000 鞐ؠFؠThO`h 0H`x0 3 @،0C``a1P0xi (0000~0>0BH`x!xp<-bΠ0ĠH@r$ž >@H ;uQ 1a]˜]˜ 1a6`g_UGustav Kirchoff (1824-1887) was a German physicist who developed the following laws. ] egBefore jt%3se+, we ask you to observeUat'5a9three cuits: CABC, DABD, and CBDwVTtwo: (loops)A:BD shecommo oints calljunctions or nodeލ: FG?'s Lawr N-+h sum of currekf<inanyh is equald6 cur+rents flowing out of that node.   Kirchoff's Law for Cuits$-rThe sumJ2e voltage drops (caused by resistors) around any cO equalsBtotalH,[,(provid^power sources). l[\ ProblemMU 66P 1aΠ< ` l d0? ``s=,8 d90HH`M"\0@0 `0 @H 0 \؊0m`t P  x <6Y< 0;lPk8!Jkl<ppDؘtH0#  #| 8 R H~a0'0I U x`I`_3U8_4mHc1"Hz:-*  @p0["|H `  %0 @ ~H 1@x  0` (0rs!H$S!D"0H`#Pp0AH —` A0 ` xi  o  U@8`Y` lx000 鞐ؠFؠThO`h 0H`x0 3 @،0C``a1P0xi (0000~0>0BH`x!xp<-bΠ0ĠH@r$ž >@H ;uQ 1a]˜]˜ 1a6`[  Solution of the Problem Click picture strip C3ZJ. SpS is to determineo currents I,, and. *+1. We will apply Kirchoff's Law for Nodes< A: (sum flowing inn: 0=I+I *sum of the currents flowing outnode A 1L I(Therefore, by Kirchoff's Law  NGs G = K OR..-I20 |?wlw applied to B also givsame equation.OP#Cui&F:4S ! CABC:%% areq ree resistorH?FBy Ohm's Law voltage drop due to left side resistor is 2I, k9middl616, andlthe righq. s{MCurrent directionpositive (fromterminalnega).p 1Therefore,t8-is = 2+I+  = 4_I There is only'e battery (power source) and its voltag3*8v. Hence by Kirchoff's law for cu= # , 4I+I = 8#G#RS C? DABDBmatwo resistors:bCBy OhmLdrop due to left side>4I,/=middl:1:.,je, total=    9J There is only'e battery (power source) and its voltag3-16v. Hence by Kirchoff's law for cu@ $ 4I+I =N A%Now we have three equationsp-P1I-+ V0To+08=֜Hwhich a6sufficient to de7miBth unknowns I,,H.#Matric ofsystem 1- 40(GI =.L816 P \RREFˈ R rref(1- 0448(I016u)Ru"u0(ǀKzq9؊!(J( ,$( !J/z(=B'(E( >'zx(|z#E؄>jL   T@*@ <  @ @  @ 0 dQ`@ <? TƆ R;_pŢ֢  Up 9 yg8L (DP@QA` GCAQ DDQE yȜsFOperation of electric networks follows Ohm's Law and Kirchoffs. V\_ .(tOhm's law, named after German -physicist, Georg'E (1789-1851), _gives aBrelation between thempower<resk ance, andcurrent. It states  V = RI where R is d in s, I '!u%ampere!VE (charge)T volts.am[r\{ Kirchoff Laws (tap?)+MU 66 1aΠ< ` l d0? ``s=,8 d90HH`M"\0@0 `0 @H 0 \؊0m`t P  x <6Y< 0;lPk8!Jkl<ppDؘtH0#  #| 8 R H~a0'0I U x`I`_3U8_4mHc1"Hz:-*  @p0["|H `  %0 @ ~H 1@x  0` (0rs!H$S!D"0H`#Pp0AH —` A0 ` xi  o  U@8`Y` lx000 鞐ؠFؠThO`h 0H`x0 3 @،0C``a1P0xi (0000~0>0BH`x!xp<-bΠ0ĠH@r$ž >@H ;uQ 1a]˜]˜ 1a6`g_UGustav Kirchoff (1824-1887) was a German physicist who developed the following laws. ] egBefore jt%3se+, we ask you to observeUat'5a9three cuits: CABC, DABD, and CBDwVTtwo: (loops)A:BD shecommo oints calljunctions or nodeލ: FG?'s Lawr N-+h sum of currekf<inanyh is equald6 cur+rents flowing out of that node.   Kirchoff's Law for Cuits$-rThe sumJ2e voltage drops (caused by resistors) around any cO equalsBtotalH,[,(provid^power sources). l[\ ProblemMU 66 1aΠ< ` l d0? ``s=,8 d90HH`M"\0@0 `0 @H 0 \؊0m`t P  x <6Y< 0;lPk8!Jkl<ppDؘtH0#  #| 8 R H~a0'0I U x`I`_3U8_4mHc1"Hz:-*  @p0["|H `  %0 @ ~H 1@x  0` (0rs!H$S!D"0H`#Pp0AH —` A0 ` xi  o  U@8`Y` lx000 鞐ؠFؠThO`h 0H`x0 3 @،0C``a1P0xi (0000~0>0BH`x!xp<-bΠ0ĠH@r$ž >@H ;uQ 1a]˜]˜ 1a6`[  Solution of the Problem Click picture strip C3ZJ. SpS is to determineo currents I,, and. *+1. We will apply Kirchoff's Law for Nodes< A: (sum flowing inn: 0=I+I *sum of the currents flowing outnode A 1L I(Therefore, by Kirchoff's Law  NGs G = K OR..-I20 |?wlw applied to B also givsame equation.OP#Cui&F:4S ! CABC:%% areq ree resistorH?FBy Ohm's Law voltage drop due to left side resistor is 2I, k9middl616, andlthe righq. s{MCurrent directionpositive (fromterminalnega).p 1Therefore,t8-is = 2+I+  = 4_I There is only'e battery (power source) and its voltag3*8v. Hence by Kirchoff's law for cu= # , 4I+I = 8#G#RS C? DABDBmatwo resistors:bCBy OhmLdrop due to left side>4I,/=middl:1:.,je, total=    9J There is only'e battery (power source) and its voltag3-16v. 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