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' }E 2002 University of North Carolina at Charlotte, ALL RIGHTS RESERVEDFF&    `' ??"NnR ' T Click to edit Master title style! !f  C 0Abottom logo"\  C &Acrown2"RB  s *޽h ? 3f3fff̙3ff̙y___PPT10Y+D=' = @B + 0,(  , , 0(y P   y X*  , 0-y    y Z*  , 681y _P  y X*  , 65y _  y Z* H , 0޽h ? 3380___PPT10.jipl @#(  x  c $' NnR ' x  c $d' 0p   '   T' ?? 0   g)University of North Carolina at Charlotte"*(2*f3H  0޽h ? 3f3fff̙3ff̙y___PPT10Y+D=' = @B +Cf  jb X(  X X  fA+ ? ??` 8 $D 0x X c $@   V X Z ??p ,$ 0 <Kirchoff s current law states,& 2 t X T< ??p@,$ 0 ( Ientering node = 0\ 2%  X Z( ??p $ ,$ 0 Dor& 2x X Z ??p`^,$ 0 & Iexiting node = 0\ 2%  B X  fD8c?? @ @ ,$D   0B  X  fD8c?? P @ ,$D  0B  X  fD8c?? ,$D  0  X Z' ??  ,$ 0 JI1, 2   X Z@' ?? ,$ 0 JI2, 2  X Th' ?? ,$ 0 JI3, 2  X Z ' ?? P D ,$  0 JI4, 2  X Z' ?? pD ,$ 0 QI1+ (-I2) +(- I3)+ I4 = 0 2          X Z\"' ?? ,$ 0 &(-I1) + I2 + I3 + (- I4) = 0 2        X Z/' ??p 0,$ 0 Dor& 2  X Z5' ?? @ ,$ 0 O KCL tells us,& 2  X Z:' ?? 0 $ ,$  0 XI4: 2  X Z3' ??  ,$ 0 JR1, 2  X ZC' ?? / ,$  0 JV2, 2  X ZXH' ?? P  ,$  0 JR3, 2  X TL' ??` `,$ 0 G Same equation(2B X  fDp??  ,$D  0B X # lDp??0   ,$D  0B X@  fDp??  ,$D 0B X # lDp??0 ,$D  0 X TQ' ?? p` @,$ 0 IGiven this diagram, 2B  X  `D8c?? ,$D  0H X 0޽h ? 3f3fff̙3ff̙GyG___PPT10YG+XxDMC' = @B DC' = @BA?%,( < +O%,( < +D] ' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-6B'blinds(horizontal)*<3<*XD' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-6B'blinds(horizontal)*<3<*XD' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-6B'blinds(horizontal)*<3<*XD ' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =%(DI' =4@BB5BB%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*XD' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*XD' =-g6B fade*<3<*XD' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =%( D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =%( D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =%( D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-s6Bwipe(down)*<3<*XD' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* X%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* X%(D' =-s6Bwipe(left)*<3<* XD' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* X%(D' =%(|D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* X%(D' =-s6Bwipe(down)*<3<* XD' =%(pD' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =%(dD7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* X%(D' =-s6Bwipe(left)*<3<* XD] ' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-6B'blinds(horizontal)*<3<*XD' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-6B'blinds(horizontal)*<3<*XD' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-6B'blinds(horizontal)*<3<*XD ' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-s6Bwipe(left)*<3<*XD' =%(D9' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-u6Bwipe(right)*<3<*XD' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-s6Bwipe(down)*<3<*XD' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*X%(D' =-s6Bwipe(down)*<3<*X++0+X' ++0+X' ++0+X' ++0+X' ++0+ X' ++0+ X' ++0+X' ++0+X' ++0+X' ++0+X' ++0+X' ++0+X' ++0+X' ++0+X' ++0+X' ++0+X' ++0+X' ++0+X' +E  Y(     fA+ ? ??` 8 $D 0x  c $p;@  ;   Z; ??pC,$ 0 Kirchoff s current law can also be observed in an alternate form.&B 2A 8  T; ??p  ,$ 0 *J Ientering node =  Iexiting node& 2% %     (B   fD8c?? @ @ ,$D   0B    fD8c?? P @ ,$D   0B    fD8c?? ,$D  0   Z|$; ??  ,$  0 JI1, 2    Z@); ?? ,$ 0 JI2, 2   TT.; ?? ,$ 0 JI3, 2   Z2; ?? P D ,$  0 JI4, 2   Z6; ?? x,$ 0 I1 + I4 = I3 + I2 2        Z@; ?? @ ,$ 0 O KCL tells us,& 2   ZtE; ?? 0 $ ,$  0 XI4: 2   ZJ; ??  ,$ 0 JR1, 2   ZdO; ?? / ,$ 0 JV2, 2   ZS; ?? P  ,$ 0 JR3, 2   TdX; ?? p` @,$ 0 IGiven this diagram, 2  T[; ??@: ZContinued& *  2 f3B   `D8c?? ,$D  0H  0޽h ? 3f3fff̙3ff̙..___PPT10.+?T^D+' = @B D+' = @BA?%,( < +O%,( < +D' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*D ' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(DI' =4@BB5BB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*D' =-g6B fade*<3<*D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%( D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%( D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%( D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-s6Bwipe(down)*<3<*D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-s6Bwipe(left)*<3<* D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =%(|D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-s6Bwipe(down)*<3<*D' =%(pD' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(dD7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-s6Bwipe(left)*<3<* D' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*++0+; ++0+; ++0+ ; ++0+ ; ++0+; ++0+; ++0+; ++0+; ++0+; ++0+; ++0+; ++0+; ++0+; +d  VN %(   0 # 0e0eAf  ?A@ A5%8c8c     ?1d0u0@Ty2 NP'p<'p@A)BCD|E?pPPH $D 0 ;r  S p;@  ;   T* ??@: ^Continued& . 2 f3f3  T; ??@,$ 0 \&As previously show&  2  T; ??,$ 0 X Ientering a node =  Iexiting that nodeJ- 2  (|  T ??0@ ,$ 0 However, in many circuits we encounter it becomes useful to picture a current entering or exiting a network as apposed to a node. & 2   T; ?? @p ,$ 0 b,For example, consider the following circuit:- 2-   Zp; ?? 0 ,$ 0 JR1, 2    T; ??@ p` ,$ 0 JR2, 2   T; ?? ,$  0 JR3, 2   Z; ??  ,$  0 JIs, 2 E  T; ?? @0& ,$  0 9Suppose we wanted to find the voltage across resistor R3.8: 27   Tܺ; ??0 @06,$ 0 d.1) We could easily write a KCL equation at Vo./ 2/  Zľ; ?? @,$  0 JVo, 2    Z; ?? p5 ,$  0 M+0 2  T; ??@p`,$ 0 I-, 2 o  T; ??@F,$ 0 EBut through observation we can see that Is flows directly through R3.VF 2)  B   fD8c?? ,$D  0  Z$; ?? ,$ 0 JIs, 2    T; ?? @p& ,$ 0 hTo find Vo, Ohm s Law can be directly applied at R3.h5 2  E'M. ! T; ??0 @6,$ 0 v@Noticing this fundamental concept, makes this problem very easy.A 2A| " T; ??@@F,$ 0 And in addition to the original KCL technique we ve learned we can say& H 2H $ T; ?? @@ ,$ 0 d Ientering a network =  Iexiting that networkJ3 2  (  % T4; ??@@ `,$ 0 TThis concept is very useful!!! 2H  0޽h ? 3f3fff̙3ff̙SDKD___PPT10+D.@.+ޚDO>' = @B D >' = @BA?%,( < +O%,( < +D2' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =%(DI' =4@BB5BB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*D' =-g6B fade*<3<*D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D3' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-o6Bwipe(up)*<3<*D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D^' =%(D' =%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* 5%(D2' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*!%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*"%(D ' =%(D' =%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*"%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*!%(D' =4@BBBB%(D' =1:Bhidden*o3>+B#style.visibility<* 5%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*%(D' =%(DF' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*$%(D' =-u6Bdiamond(in)*<3<*$D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%%(++0+t ++0+t ++0+t ++0+ t ++0+ t ++0+ t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+ t ++0+!t ++0+!t ++0+"t ++0+"t ++0+$t ++0+%t +T  [S0(   0 # 0e0eA;  ?@ 5%8c8c     ?1d0u0@Ty2 NP'p<'p@A)BCD|E? H $D 0 tx  c $2t@  t '  Z3t ??@ ,$D 0 i!Given: The circuit pictured below," 2B  Z9t ??@ 0,$  0 Find: The voltage Vo using KCLJ 2    Z|?t ?? /@,$ 0 VVo8 2    Z@Et ?? @t@,$ 0 VR18 2    ZJt ?? ,$  0 VR28 2     ZOt ??  @,$  0 JR3, 2    Z\Tt ?? pV ,$ 0 VIs8 2     Z`Yt ??` ,$ 0 Q Solution:,  2   Z]t ?? ,$ 0 0 Ientering =  Iexiting 2% %     (  B    fD8c??   ,$D   0B  # lD8c?? ,$@   0  Z(it ?? $@  > 2 Z  Zkt ??  ,$ 0  Is+I2 = I3v  2     Zst ??P ^p ,$ 0 Is + (-Vo/R2) = Vo/R3 2       Z~t ??  ,$ 0 Is = - (-Vo/R2) + Vo/R3 2       Zt ??P 3p,$ 0 Is = Vo (1/R2 + 1/R3) 2       Tt ?? ,$ 0  Vo = Is/(1/R2 + 1/R3) 2       Tt ?? 0P ,$ 0 DI2& 2   T̋t ??0 ,$ 0 DI3& 2   Ţt ??  ,$ 0 .pNotice, R1 has no effect on the value of Vo, since Is flows directly into the parallel combination of R2 and R3.q 2      2  B   `D8c?? P ,$@  0B   `D8c?? `P ,$D  0H  0޽h ? 3f3fff̙3ff̙5 5___PPT104+WBDO1' = @B D 1' = @BA?%,( < +O%,( < +D:' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*"%(D' =%(DI' =4@BB5BB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*D' =-g6B fade*<3<*D ' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-s6Bwipe(left)*<3<*D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-s6Bwipe(down)*<3<* DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D2' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =%(D3' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-o6Bwipe(up)*<3<*D3' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-o6Bwipe(up)*<3<*D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(++0+t ++0+t ++0+t ++0+t ++0+ t ++0+ t ++0+ t ++0+ t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+t ++0+t +ڬ  11`%/S1(   %0 # 0e0eAg ' ?A@ A5%8c8c     ?1d0u0@Ty2 NP'p<'p@A)BCD|E? H ' $D 0 t  T4t ?? p,$1 0 However if multiple resistors appear in parallel, current division can be equated by placing the conductance you want over the sum of conductances in parallel. 2   fAN %? ??P @H 8 %$D, 0@  Tt ?? p,$- 0 Notice with current division, it s the resistor you don t want to find the current through divided by the sum of resistors in parallel.^ 2'3x  c $@t@  t r  TTt ??j@p,$ 0 Current division provides a relationship for dividing current among parallel resistors in inverse proportion to their resistances. 2  ZDv ??p@,$ 0 `(For example, consider the circuit below:) 2)  Zv ?? =,$) 0 PCurrent Division states, 2    Zv ?? ( ,$ 0 KIin, 2    Tt ?? ,$ 0 JR1, 2    Zv ??| ,$  0 JR2, 2 B    fD;?? 00 ,$D 0B   fD;?? ,$D  0  Zv ??0 pP  > 2   Zv ?? @4 ,$ 0 JI1, 2   Tv ??  ,$  0 JI2, 2   Zv ?? 4 ,$+ 0 Dor& 22   ` ??P P ,$D.  02   ` ?? p ,$D/  0  Z%v ??J,$2 0 &I1 = Iin [G1/ G s] 2   %    fAM $? ??Pp3 8 $$D* 0t  Td/v ?? @,$  0 >Using KCL, we can derive a current relationship for I1 and I2.b? 25  o  Zd7v ??p | ,$ 0  Iin=I1+I2r  2     TP@v ?? 0 ,$ 0 Iin=Vo/R1+Vo/R2 2        TKv ??  ,$ 0 Iin=Vo(1/R1+1/R2) 2      TUv ?? @,$ 0 Vo = Iin/(1/R1+1/R2) (1)  2    B    `D?? pp ,$@  0B !  `D??   ,$D  0 " T`v ??p ,$ 0 Vo=I1R1 or I2R2 (2) 2     0 ' Zplv ??p n ,$  0 r+ Vo -P(  ( Trv ?? ,$ 0 Vo = Iin/(1/R1+1/R2) (1)  2     ) T|v ??  ,$ 0 Vo=I1R1 or I2R2 (2) 2       * T`v ??0 P ,$ 0 USubstituting equ.2 into equ.1 2 + Tv ??P 0 ,$  0 -;Vo = I1R1 = Iin/(1/R1+1/R2) Or Vo = I2R2 = Iin/(1/R1+1/R2) 0n8n              , Tv ?? 6 ,$% 0 9I1= Iin(1/R1) /(1/R1+1/R2) Or I2 = Iin(1/R2)/(1/R1+1/R2) 0n8n            - Tv ?? ,$& 0 aMultiplying these two equations by (R1R2)/(R1R2) provides a useful equation for current division.b0n%    2H  0޽h ? 3f3fff̙3ff̙zz___PPT10z.+Dr' = @B Dq' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D:' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DI' =4@BB5BB%(D' =1:Bvisible*o3>+B#style.visibility<*%%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*%D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*%D' =-g6B fade*<3<*%D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =%( D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%( D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%( D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*'%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(Do' =%(D&' =%(Dc' =4@BBBB%()))D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-o6Bwipe(up)*<3<* Dc' =4@BBBB%()))D' =1:Bvisible*o3>+B#style.visibility<*!%(D' =-o6Bwipe(up)*<3<*!D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*"%(DF ' =%(D ' =%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bhidden*o3>+B#style.visibility<*!%(D' =4@BBBB%(D' =1:Bhidden*o3>+B#style.visibility<* %(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*"%(D' =%(D' =%(Dp' =A@BB BB0B%()))D' =1:Bvisible*o3>+B#style.visibility<*(%(D' =-o6Bdissolve*<3<*(D' =%(Dp' =A@BB BB0B%()))D' =1:Bvisible*o3>+B#style.visibility<*)%(D' =-o6Bdissolve*<3<*)D' =%(D' =%(Dp' =A@BB BB0B%()))D' =1:Bvisible*o3>+B#style.visibility<**%(D' =-o6Bdissolve*<3<**D' =%(Dp' =A@BB BB0B%()))D' =1:Bvisible*o3>+B#style.visibility<*+%(D' =-o6Bdissolve*<3<*+D ' =%(D' =%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*+%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<**%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*(%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*)%(D' =%(Dp' =A@BB BB0B%()))D' =1:Bvisible*o3>+B#style.visibility<*,%(D' =-o6Bdissolve*<3<*,D' =%(Dp' =A@BB BB0B%()))D' =1:Bvisible*o3>+B#style.visibility<*-%(D' =-o6Bdissolve*<3<*-D!' =%(D' =%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*-%(D' =%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*,%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(DG' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'checkerboard(down)*<3<*D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*D' =%(DG' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'checkerboard(down)*<3<*D_' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*Dv' =%(D' =%(D' =4@BBB B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<*D' =%(D' =%(D' =A@BBBB0B%(D' =1:Bhidden*o3>+B#style.visibility<*%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =%(DT' =A@BBB B0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =-6B'blinds(horizontal)*<3<*+P+0+v ++0+v ++0+v ++0+v ++0+v ++0+v ++0+ v ++0+ v ++0+ v ++0+v ++0+v ++0+v ++0+v ++0+v ++0+v ++0+v ++0+v ++0+v ++0+v ++0+v ++0+v ++0+y ++0+y ++0+"y ++0+"y ++0+'y ++0+(y ++0+(y ++0+)y ++0+)y 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