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{
"metadata": {
"name": "",
"signature": "sha256:b7a82ebd5adf08f1d2cb1363a68626f83f8a9fdc0fb722a735bf408f1a7aed03"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter01:Introduction to Electronics"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E1 - Pg 8"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Find the range of tolerance\n",
"#soltion\n",
"#given\n",
"#color coding\n",
"orange=3.#\n",
"gold=5.#\n",
"yellow=4.#\n",
"violet=7.#\n",
"#band pattern\n",
"band1=yellow#\n",
"band2=violet#\n",
"band3=orange#\n",
"band4=gold#\n",
"#resistor color coding\n",
"r=(band1*10.+band2)*10.**(band3)#\n",
"tol=r*(band4/100.)#tolerance\n",
"ulr=r+tol##upper limit of resistance\n",
"llr=r-tol##lower limit of resistance\n",
"print 'The range of resistance =',llr/1000. ,'kOhm','to',ulr/1000,'kOhm'\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The range of resistance = 44.65 kOhm to 49.35 kOhm\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E2 - Pg 8"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Find the range of tolerance\n",
"#color coding\n",
"blue=6.#\n",
"gold=-1.#\n",
"gray=8.#\n",
"silver=10.#\n",
"#band pattern\n",
"band1=gray#\n",
"band2=blue#\n",
"band3=gold#\n",
"band4=silver#\n",
"#resistor color coding\n",
"r=(band1*10.+band2)*10.**(band3)#\n",
"tol=r*(band4/100.)#tolerance\n",
"ulr=r+tol##upper limit of resistance\n",
"llr=r-tol##lower limit of resistance\n",
"print 'The Range of resistance is',llr,'ohm','to',ulr,'ohm'\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Range of resistance is 7.74 ohm to 9.46 ohm\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E3 - Pg 19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Find the equivalent current source\n",
"#given\n",
"Vs=2.;#Volts #dc voltage source\n",
"Rs=1.;#ohm #internal resistance\n",
"Rl=1.;#ohm #load resistance\n",
"Ise=Vs/Rs;#ampere #equivalent current source\n",
"\n",
"# In accordance to figure 1.23a\n",
"Il1=Ise*(Rs/(Rs+Rl));#using current divider concept\n",
"Vl1=Il1*Rl;\n",
"print \"In accordance to figure 1.23a\\n\"\n",
"print \"The Load current (current source Il=\",Il1,'A'\n",
"print \"The Load voltage (current source Vl=\",Vl1,'V','\\n'\n",
"\n",
"# In accordance to figure 1.23b\n",
"Vl2=Vs*(Rs/(Rs+Rl));#using voltage divider concept\n",
"Il2=Vl2/Rl;\n",
"print \"\\nIn accordance to figure 1.23b\"\n",
"print \"\\nThe Load voltage (voltage source) Vl=\",Vl2,'V'\n",
"print \"The Load current (voltage source) Il=\",Il2,'A'\n",
"print \"\\nTherefore they both provide same voltage and current to load\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"In accordance to figure 1.23a\n",
"\n",
"The Load current (current source Il= 1.0 A\n",
"The Load voltage (current source Vl= 1.0 V \n",
"\n",
"\n",
"In accordance to figure 1.23b\n",
"\n",
"The Load voltage (voltage source) Vl= 1.0 V\n",
"The Load current (voltage source) Il= 1.0 A\n",
"\n",
"Therefore they both provide same voltage and current to load\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E4 - Pg 19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Find percentage variation in load current and load voltage\n",
"#given\n",
"Vs=10.;#volt#Supply voltage\n",
"Rs=100.;#ohm#internal resistance\n",
"\n",
"# In accordance to figure 1.24a\n",
"#For 1ohm - 10 ohm\n",
"Rl11=1.;#ohm#min extreme value of Rl\n",
"Rl12=10.;#ohm#max extreme value of Rl\n",
"Il11=Vs/(Rs+Rl11);\n",
"Il12=Vs/(Rs+Rl12);\n",
"Pi1=(Il11-Il12)*100./Il11;#Percentage variation in current\n",
"Vl11=Il11*Rl11;\n",
"Vl12=Il12*Rl12;\n",
"Pv1=(Vl12-Vl11)*100./Vl12;#Percentage variation in voltage\n",
"print '%s' %(\"In accordance to figure 1.24a \\n\");\n",
"print '%s %.2f %s' %(\"Percentage variation in Current(1-10 ohm)=\",Pi1,'percent');\n",
"print '%s %.1f %s ' %(\"Percentage variation in Voltage(1-10 ohm)=\",Pv1,'percent\\n\\n');\n",
"\n",
"# In accordance to figure 1.24b\n",
"#For 1kohm - 10kohm\n",
"Rl21=1000.;#ohm#min extreme value of Rl\n",
"Rl22=10000.;#ohm#max extreme value of Rl\n",
"Il21=Vs/(Rs+Rl21);\n",
"Il22=Vs/(Rs+Rl22);\n",
"Pi2=(Il21-Il22)*100./Il21;#Percentage variation in current\n",
"Vl21=Il21*Rl21;\n",
"Vl22=Il22*Rl22;\n",
"Pv2=(Vl22-Vl21)*100./Vl22;#Percentage variation in voltage\n",
"print '%s' %(\"In accordance to figure 1.24b \\n\");\n",
"print '%s %.f %s' %(\"Percentage variation in Current(1-10 ohm)=\",Pi2,'percent');\n",
"print '%s %.f %s ' %(\"Percentage variation in Voltage(1-10 ohm)=\",Pv2,'percent \\n');\n",
"print 'In book the percentage variation in voltage(1kohm-10kohm) is 9 percent due to' \n",
"print 'the incorrect value of Il22 i.e. 0.000999 Amp correct value is 0.0009901 Amp'\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"In accordance to figure 1.24a \n",
"\n",
"Percentage variation in Current(1-10 ohm)= 8.18 percent\n",
"Percentage variation in Voltage(1-10 ohm)= 89.1 percent\n",
"\n",
" \n",
"In accordance to figure 1.24b \n",
"\n",
"Percentage variation in Current(1-10 ohm)= 89 percent\n",
"Percentage variation in Voltage(1-10 ohm)= 8 percent \n",
" \n",
"In book the percentage variation in voltage(1kohm-10kohm) is 9 percent due to\n",
"the incorrect value of Il22 i.e. 0.000999 Amp correct value is 0.0009901 Amp\n"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}
|
