1 files changed, 0 insertions, 36 deletions
diff --git a/Electronic_Principles_/Chapter_3_New.ipynb b/Electronic_Principles_/Chapter_3_New.ipynb
index e2552e78..e4397d8b 100644
--- a/Electronic_Principles_/Chapter_3_New.ipynb
+++ b/Electronic_Principles_/Chapter_3_New.ipynb
@@ -27,19 +27,13 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Example 3.2.py\n",
-      "#A diode has a power rating of 5W. If diode voltage is 1.2V and diode current is 1.75A,\n",
-      "#What is the power dissipation? Will the diode be destroyed?\n",
       "\n",
-      "#Variable declaration\n",
       "Pr=5     #Power rating(W)\n",
       "Vd=1.2   #diode voltage(V)\n",
       "Id=1.75  #diode current(A)\n",
       "\n",
-      "#Calculation\n",
       "PD=Vd*Id #Power dissipaion(W)\n",
       "\n",
-      "#Result\n",
       "print 'Power Dissipation =',PD,'W'\n",
       "print 'PD(',PD,'W) < ''Pr(',Pr,'W), So diode will not be destroyed.'"
      ],
@@ -69,19 +63,14 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Example 3.3.py\n",
-      "#Use the ideal diode to calculate the VL & IL in Fig. 3-6a.\n",
       "\n",
       "\n",
-      "#Variable declaration\n",
       "Vs=10     #Source voltage(V)\n",
       "RL=1      #Load resistance(KOhm)\n",
       "VL=Vs     #LOad voltage(V)\n",
       "\n",
-      "#Calculation\n",
       "IL=VL/RL  #Load current(mA)\n",
       "\n",
-      "#Result\n",
       "print 'Vs will be appearing across RL'\n",
       "print 'Load voltage VL =',Vs,'V'\n",
       "print 'Load current IL =',IL,'mA'"
@@ -113,24 +102,19 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Example 3.4.py\n",
-      "#Calculate the VL & IL in Fig. 3-6b using an ideal diode.\n",
       "\n",
-      "#Variable Declaration\n",
       "print 'As per figure 3-6b, Thevenize the circuit to the left of the diode'\n",
       "R1=6   #Resistance(KOhm)\n",
       "R2=3   #Resistance(KOhm)\n",
       "RL=1   #Load Resistance(KOhm)\n",
       "Vs=36  #Supply voltage(V)\n",
       "\n",
-      "#Calculation\n",
       "Vth=R2*Vs/(R1+R2)   #ThCevenin voltage(V)\n",
       "Rth=(R1*R2)/(R1+R2) #Thevenin resistance(KOhm)\n",
       "Rt=Rth+RL           #total resistance(KOhm)\n",
       "IL=Vth/Rt           #Load current(mA)\n",
       "VL=IL*RL            #Load voltage(V)\n",
       "\n",
-      "#Result\n",
       "print 'Vth =',Vth,'V & Rth =',Rth,'KOhm'\n",
       "print 'Visualize diode as closed switch,'\n",
       "print 'IL =',IL,'mA'\n",
@@ -165,21 +149,16 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Example 3.5.py\n",
-      "#Use the second approximation to calculate the VL, IL & PD in Fig 3-8. \n",
       "\n",
-      "#Variable Declaration\n",
       "print 'As per Second approximation in Fig.3-8,'\n",
       "Vd=0.7    #diode voltgage(V)\n",
       "Vs=10     #supply voltage(V)\n",
       "RL=1      #Load resistance(KOhm)\n",
       "\n",
-      "#Calculation\n",
       "VL=Vs-Vd  #Load voltage(v)\n",
       "IL=VL/RL  #Load current(mA)\n",
       "PD=Vd*IL  #diode power(mW)\n",
       "\n",
-      "#Result\n",
       "print 'IL =',IL,'mA & VL =',VL,'V'\n",
       "print 'Diode power PD =',PD,'mW'"
      ],
@@ -210,10 +189,7 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Example 3.6.py\n",
-      "#Calculate the VL, IL & PD in fig. 3-9a using the second approximation.\n",
       "\n",
-      "#Variable Declaration\n",
       "print 'As per Second approximation in Fig.3-9a,'\n",
       "Vd=0.7  #diode voltgage(V)\n",
       "Vs=36   #supply voltage(V)\n",
@@ -221,14 +197,12 @@
       "R2=3    #Resistance(KOhm)\n",
       "RL=1    #Load resistance(KOhm)\n",
       "\n",
-      "#Calculation\n",
       "Vth=R2*Vs/(R1+R2)#Thevenin Voltage(V)\n",
       "Rth=(R1*R2)/(R1+R2)#Thevenin resistance(KOhm)\n",
       "IL=(Vth-Vd)/R2#Load current(mA)\n",
       "VL=IL*RL#Load voltage(V)\n",
       "PD=Vd*IL#diode power(mW)\n",
       "\n",
-      "#Result\n",
       "print 'Thevenize the circuit to the left of the diode'\n",
       "print 'Vth =',Vth,'V & Rth =',Rth,'KOhm'\n",
       "print 'VL =',round(VL,2),'V & IL =',round(IL,2),'mA'\n",
@@ -263,17 +237,13 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Example 3.7.py\n",
-      "#The IN4001 has a bulk resistance of 0.23 Ohm. What is VL, IL & PD?\n",
       "\n",
-      "#Variable Declaration\n",
       "print 'In Fig.3-11a,'\n",
       "Vd=0.7    #diode voltgage(V)\n",
       "Vs=10     #supply voltage(V)\n",
       "RL=1000L  #Load resistance(Ohm)\n",
       "Rb=0.23   #bulk resistance\n",
       "\n",
-      "#Calculation\n",
       "print 'As per third approximation, we get fig.3-11b'\n",
       "if Rb<(RL/100):\n",
       "    print'If Rb < 0.01RL than ignore Rb & use second approximation.'\n",
@@ -281,7 +251,6 @@
       "    IL=(VL/RL)*1000   #Load current(mA)\n",
       "    PD=Vd*IL          #diode power(mW)\n",
       "\n",
-      "#Result    \n",
       "    print 'IL =',IL,'mA & VL =',VL,'V'\n",
       "    print 'Diode power PD =',PD,'mW'"
      ],
@@ -314,10 +283,7 @@
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "#Example 3.8.py\n",
-      "#Repeat the preceding example for RL of 10 Ohm.\n",
       "\n",
-      "#Variable Declaration\n",
       "print 'In Fig.3-11a, take RL = 10 Ohm'\n",
       "Vd=0.7    #diode voltgage(V)\n",
       "Vs=10     #supply voltage(V)\n",
@@ -326,14 +292,12 @@
       "RT=Rb+RL  #Total reistance(Ohm)\n",
       "VT=Vs-Vd  #total voltage(V)\n",
       "\n",
-      "#Calculation\n",
       "print 'RT =',RT,'Ohm & VT =',VT,'V'\n",
       "IL=VT/RT      #Load current(mA)\n",
       "VL=IL*RL      #Load voltage(V)\n",
       "VD=Vd+(IL*Rb) \n",
       "PD=VD*IL      #diode power(W)\n",
       "\n",
-      "#Result\n",
       "print 'IL =',round(IL,2),'mA & VL =',round(VL,2),'V'\n",
       "print 'VD =',round(VD,2),'V'\n",
       "print 'Diode power PD =',round(PD,2),'W'"