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!'.Z"#!%LG)1)0J,!!1@!1!KLG&K%y"<)PIr!#G,!1"*LG&K""Is!$&J%9!&LG#e(>Is!"DG5\)0!1!LLG#e2rIs!!.Z!A!SLH9Q""Ir!#>z!A!>!(:&LG#e(.Ir!'<6!A!$LG&K)pJ,!!1@!1!KLG&K%y"<*"Ir!#6j!1"-LG&K!PIs!$.Z%9!&LG#e(>Is!"9Q5\1@!1!FLG#e3DIv!$1@%9!&LG#e(>Is!"DG5\4&!1!DLG#e3TIs!!+t !A!TLH9Q""Ir!#>z!A!>!(:*LG#e&jIr!'G,!A!%LG&K)`J,!!1@!1!KLG&K%y"<*bIs!"G,!A"1LG&K!pIs!$)0%9!&LG#e(>Is!"DG5\>z!A!:LG&K46Is!!.Z!A!SLH9Q""Ir!#>z!A!>!(:/LG+u$&Iu!()0!A!$LG&K)pJ,!!1@!1!KLG&K%9"<,&Iv!!6j!q"7LG6k*"J,!!1@!1!KLG&K#u"<,vIx!,1@%9!&LG#e(>Is!"#e5]9P !1"{LH9Q""Ir!#>z!A!2!(:;LG#e@NJ,!!1@!1!KLG&K#u"<-HIr!,9P%9!&LG#e(>Is!"#e5]9P!1"{LH9Q""Ir!#>z!A!2!(:;LG#e@NJ,!!1@!1!KLG&K#u"<-HIr!,9P%9!&LG#e(>Is!!DG5]9P!1#)LG#e+DIs!!DG5]9P!1#)LG#e+DIs!!DG5]9P!1#)LG#e+DIs!!DG5]9P!1#)LG#e+DIs!!DG5]9P!1#)LG#e+DIs!"#e5\4& !1!2LG#eBRIr!$DF!A!2!(:)LG#e$&Ir!-.Z!1!]LG&K#u"<*BIr!"&J!1#)LG#e+DIs!"#e5\4&!1!2LG#eBRIr!$DF!A!2!(:)LG#e$&Ir!-.Z!1!]LG&K#u"<*BIr!"&J!1#)LG#e+DIs!"DG5\#d#%!-LG#e)0Ir!"6j!1!OLJDG$&Ir!$DF!A!B!(:)LG#e$&Ir!#LV!Q!6LG#e(~Ir!$A`!1!2LG#e+DIs!#.[5\4&!1!2LG#e)0Ir !!)0!A!4LG#e(~Ir!$A`!1!2LG#e+DIs!#.[5\4&!1!2LG#e)0Ir!!.Z!A!2LG#e(~Ir!$A`!1!2LG#e+DIs!#.[5\4&!1!2LG#e)0Ir!!4&!A!0LG#e(~Ir!$A`!1!2LG#e+DIs!#.[5\4&!1!2LG#e)0Ir!!9P!A!.LG#e(~Ir!$A`!1!2LG#e+DIs!##e5]9P!1!PLG#e"rIs!!A`!1!OLG#e+4Ir!"&J!1!]LG&K&["<-HIr!#LV!1!- LG&K"bIr!#Ip!1!\LG#e$&Ir!$DF!A!B!(:;LG#e)0Ir!!Ip!A!(LG#e(~Ir!$A`!1!2LG#e+DIs!##e5]9P!1!PLG#e#tIs!!1@!1!OLG#e+4Ir!"&J!1!]LG&K&["<-HIr!#LV!1!3LG&K!`Ir!#Ip!1!\LG#e$&Ir!$DF!A!B!(:;LG#e)0Ir!".Z!A!"LG#e(~Ir!$A`!1!2LG#e+DIs!"#e5]9P)A!7LJ#e+4J&!$DF!A!6!(:kLG#e $fIt!#Ip!1!\LG#e.JIs!"9Q5`9P!1!4LG&K!PIr!#Ip!1!\LG#e.JIs!"9Q5`9P!1!2LG&K!pIr!#Ip!1!\LG#e.JIs!"9Q5`9P!1!0LG&K"2Ir!#Ip!1!\LG#e.JIs!"9Q5`9P!1!.LG&K"RIr!#Ip!1!\LG#e.JIs!"9Q5`9P!1!,LG&K"rIr!#Ip!1!\LG#e.JIs!"9Q5`9P!1!*LG&K#4Ir!#Ip!1!\LG#e.JIs!"9Q5`9P!1!(LG&K #TIr!#Ip!1!\LG#e.JIs!"9Q5`9P!1!&LG&K#tIr!#Ip!1!\LG#e.JIs!"9Q5`9P!1!$LG&K$6Ir!#Ip!1!\LG#e.JIs!"9Q5`9P!1!"LG&K$VIr!#Ip!1!\LG#e.JIs!!DG5`9P!A!fLJDG.JIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV -xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!!4'5hLV-xIs!$!Q  .EQN 31 -1 290 0 0 {0:E}NAME:340 .TXT 0 9 291 0 0 Cg a64.000000,64.000000,2 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard { \cf2 V.}} .EQN 7 -9 298 0 0 {0:R}NAME:10 .TXT 0 8 299 0 0 Cg a64.000000,64.000000,1 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}} \plain\cf1\fs28 \pard {\cf2\f1 W}} .EQN 8 -8 296 0 0 {0:wL}NAME:2*{0:\p}NAME*50*31.8*(10)^(-3) .TXT 0 25 297 0 0 Cg a63.000000,63.000000,2 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}} \plain\cf1\fs28 \pard {\cf2 }{\cf2\f1 W}} .TXT 7 -25 300 0 0 Cg a73.800000,73.800000,71 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard The value of L is set to be 31.8 mH. The frequency is set to be 50 Hz.} .EQN 8 0 301 0 0 {0:\a}NAME:{0:atan}NAME(({0:wL}NAME)/({0:R}NAME)) .EQN 11 0 319 0 0 {0:\a}NAME={0}?_n_u_l_l_ .TXT 0 16 318 0 0 Cg a61.000000,61.000000,4 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard { \cf2 rad.}} .TXT 7 -16 305 0 0 Cg a73.800000,73.800000,68 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard The load angle is calculated for the specified values of R, L and f.} .EQN 9 0 306 0 0 {0:Z}NAME:\(({0:R}NAME)^(2)+({0:wL}NAME)^(2)) .EQN 17 0 307 0 0 {0:A}NAME:({0:E}NAME)/({0:Z}NAME)*{0:sin}NAME({0:\a}NAME) .EQN 12 0 308 0 0 {0:Z}NAME={0}?_n_u_l_l_ .TXT 0 12 309 0 0 Cg a60.000000,60.000000,2 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}} \plain\cf1\fs28 \pard {\cf2 }{\cf2\f1 W}} .EQN 0 21 276 0 0 {0:A}NAME={0}?_n_u_l_l_ .TXT 0 8 277 0 0 Cg a64.000000,64.000000,4 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard { \cf2 Amp.}} .TXT 9 -41 270 0 0 Cg a73.800000,73.800000,322 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard The value of the coefficient A has been computed. Now the plots of the \par source voltage, the current thr ough the diode and the voltage at the \par cathode of diode are obtained. A range variable, called {\f1 q}, is\par created first and it is varied from 0 deg to 360 deg. At each degree,\par the current through the diode is computed. } .EQN 19 0 269 0 0 {0:\q}NAME:0;360 .EQN 10 0 268 0 0 ({0:Cur}NAME)[({0:\q}NAME):({0:E}NAME)/({0:Z}NAME)*{0:sin}NAME(({0:\q}NAME)/(180)*{0:\p}NAME-{0:\a}NAME)+{0:A}NAME*({0:e}NAME)^(-(({0:R}NAME*{0:\q}NAME*{0:\p}NAME)/(180*{0:wL}NAME))) .TXT 10 0 266 0 0 Cg a74.800000,74.800000,643 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard The expression for current, Cur{\f1\dn q }, shown above would yield positive values \par if {\f1 q} is less than the conduction angle and would yield negative values if \par {\f1 q} is higher than the conduction angle. The statement below creates an \par array called i{\f1\dn q } and it is set equal to Cur{\f1\dn q }{\f1 ,}{ \f1\dn }if Cur{\f1\dn q } is positive. If Cur{\f1\dn q } \par is negative, i{\f1\dn q } is set equal to zero. The expression within the brackets is\par similar to arithmetic {\b IF }operator in C language. Given\par x :=if (expr1,expr2, expr3 ),\par \par x is assigned the value of expr2, if expr1 evalautes to Boolean TRUE value \par and x is assigned the value of expr3, if expr1 evalautes to Boolean FALSE value. } .TXT 41 -2 265 0 0 Cg a72.000000,72.000000,213 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard This arithmetic IF operation is carried out for each value of array\par \pard i{\f1\dn q } taking into account the corresponding value of Cur{\f1\dn q }{\f1 . }That is, both i{\f1\dn q } \par and Cur{\f1\dn q } have the same index, {\f1 q} when evaluation is performed.{\f1\dn }} .EQN 17 2 256 0 0 ({0:i}NAME)[({0:\q}NAME):{0:if}NAME(({0:Cur}NAME)[({0:\q}NAME)>0.0,({0:Cur}NAME)[({0:\q}NAME),0.0) .TXT 7 0 257 0 0 Cg a73.800000,73.800000,63 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard The voltage across the resistor can be computed as shown below.} .EQN 7 0 258 0 0 ({0:vR}NAME)[({0:\q}NAME):{0:R}NAME*({0:i}NAME)[({0:\q}NAME) .TXT 7 0 259 0 0 Cg a73.800000,73.800000,330 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard Next the voltage across the inductor is computed. Again the arithemtic IF\par operation is carried out. When the diode current is positve, the voltage\par across the inductor is the difference between the source voltage and the\par voltage across the resistor. When the diode current is zero, the voltage\par across the inductor is also zero.} .EQN 22 0 254 0 0 ({0:vL}NAME)[({0:\q}NAME):{0:if}NAME(({0:i}NAME)[({0:\q}NAME)>0.0,{0:E}NAME*{0:sin}NAME(({0:\q}NAME)/(180)*{0:\p}NAME)-({0:vR}NAME)[({0:\q}NAME),0.0) .TXT 10 0 255 0 0 Cg a74.800000,74.800000,458 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard Next the conduction angle is determined. MathCad allows a particular construct. Assign a guess value for conduction angle. Here it is called\par {\f1 b} and is assigned a value of { \f1 p} radians. Next form a block staring\par with {\b Given} statement. The equality that should exist is presented below.\par The program then evaluates the correct value of {\f1 b} for which the \par equality is true. Then that value of {\f1 b} is assigned to another \par variable called {\f1 d} in this program. } .EQN 22 0 252 0 0 {0:\b}NAME:{0:\p}NAME .EQN 12 -1 253 0 0 {0:Given}NAME .EQN 9 0 249 0 0 ({0:E}NAME)/({0:Z}NAME)*{0:sin}NAME({0:\b}NAME-{0:\a}NAME)+{0:A}NAME*({0:e}NAME)^(-(({0:R}NAME*{0:\b}NAME)/({0:wL}NAME)))÷0 .EQN 9 0 244 0 0 {0:\d}NAME:{0:Find}NAME({0:\b}NAME) .EQN 0 28 245 0 0 {0:\d}NAME={0}?_n_u_l_l_ .EQN 0 10 246 0 0 {0:radians}NAME .TXT 6 -38 247 0 0 Cg a71.000000,71.000000,229 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}} \plain\cf1\fs28 \pard {\cf2 When wt = }{\cf2\f1 d}{\cf2 , the current through the load becomes zero. The dc value of this current is found out as shown below. Here this current is \par expressed as a function of x and this function is integrated over the\par period of conduction.}} .EQN 17 0 248 0 0 {0:cur}NAME({0:x}NAME):({0:E}NAME)/({0:Z}NAME)*{0:sin}NAME({0:x}NAME-{0:\a}NAME)+{0:A}NAME*({0:e}NAME)^(-(({0:R}NAME*{0:x}NAME)/({0:wL}NAME))) .EQN 11 0 241 0 0 {0:I.DC}NAME:(1)/(2*{0:\p}NAME)*(0&{0:\d}NAME`{0:cur}NAME({0:x}NAME)&{0:x}NAME) .EQN 0 39 242 0 0 {0:I.DC}NAME={0}?_n_u_l_l_ .TXT 0 16 243 0 0 Cg a55.000000,55.000000,3 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard { \cf2 Amp}} .EQN 8 -52 236 0 0 &&(_n_u_l_l_&_n_u_l_l_)&({0:i}NAME)[({0:\q}NAME)@&&(_n_u_l_l_&_n_u_l_l_)&{0:\q}NAME 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 0 0 1 1 NO-TRACE-STRING 0 2 1 0 1 1 NO-TRACE-STRING 0 3 2 0 1 1 NO-TRACE-STRING 0 4 3 0 1 1 NO-TRACE-STRING 0 1 4 0 1 1 NO-TRACE-STRING 0 2 5 0 1 1 NO-TRACE-STRING 0 3 6 0 1 1 NO-TRACE-STRING 0 4 0 0 1 1 NO-TRACE-STRING 0 1 1 0 1 1 NO-TRACE-STRING 0 2 2 0 1 1 NO-TRACE-STRING 0 3 3 0 1 1 NO-TRACE-STRING 0 4 4 0 1 1 NO-TRACE-STRING 0 1 5 0 1 1 NO-TRACE-STRING 0 2 6 0 1 1 NO-TRACE-STRING 0 3 0 0 1 1 NO-TRACE-STRING 0 4 1 0 1 1 NO-TRACE-STRING 0 1 1 42 26 18 0 3 .TXT 44 -1 215 0 0 Cg a71.000000,71.000000,311 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard { \cf2 The ac source shown in Fig. 1.1 has to supply this dc current. An ac source should not normally be required to supply a dc current. The \par mains ac supply is distributed using transformers and a transformer is \par not designed to supply a dc current. It is preferable to avoid using half-wave rectifier circuits. }} .EQN 15 5 213 0 0 &&(_n_u_l_l_&_n_u_l_l_)&({0:vL}NAME)[({0:\q}NAME)@&&(_n_u_l_l_&_n_u_l_l_)&{0:\q}NAME 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 0 0 1 1 NO-TRACE-STRING 0 2 1 0 1 1 NO-TRACE-STRING 0 3 2 0 1 1 NO-TRACE-STRING 0 4 3 0 1 1 NO-TRACE-STRING 0 1 4 0 1 1 NO-TRACE-STRING 0 2 5 0 1 1 NO-TRACE-STRING 0 3 6 0 1 1 NO-TRACE-STRING 0 4 0 0 1 1 NO-TRACE-STRING 0 1 1 0 1 1 NO-TRACE-STRING 0 2 2 0 1 1 NO-TRACE-STRING 0 3 3 0 1 1 NO-TRACE-STRING 0 4 4 0 1 1 NO-TRACE-STRING 0 1 5 0 1 1 NO-TRACE-STRING 0 2 6 0 1 1 NO-TRACE-STRING 0 3 0 0 1 1 NO-TRACE-STRING 0 4 1 0 1 1 NO-TRACE-STRING 0 1 1 39 24 18 0 3 .EQN 41 -4 183 0 0 &&(_n_u_l_l_&_n_u_l_l_)&{0:E}NAME*{0:sin}NAME(({0:\q}NAME)/(180)*{0:\p}NAME)@&&(_n_u_l_l_&_n_u_l_l_)&{0:\q}NAME 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 0 0 1 1 NO-TRACE-STRING 0 2 1 0 1 1 NO-TRACE-STRING 0 3 2 0 1 1 NO-TRACE-STRING 0 4 3 0 1 1 NO-TRACE-STRING 0 1 4 0 1 1 NO-TRACE-STRING 0 2 5 0 1 1 NO-TRACE-STRING 0 3 6 0 1 1 NO-TRACE-STRING 0 4 0 0 1 1 NO-TRACE-STRING 0 1 1 0 1 1 NO-TRACE-STRING 0 2 2 0 1 1 NO-TRACE-STRING 0 3 3 0 1 1 NO-TRACE-STRING 0 4 4 0 1 1 NO-TRACE-STRING 0 1 5 0 1 1 NO-TRACE-STRING 0 2 6 0 1 1 NO-TRACE-STRING 0 3 0 0 1 1 NO-TRACE-STRING 0 4 1 0 1 1 NO-TRACE-STRING 0 1 1 37 26 18 0 3