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{
"metadata": {
"name": "",
"signature": "sha256:b71c5763a44a53c43becf4b38bc4dc6ca15fcaba00869ac6f1172cbb9803b804"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 1 - ELECTRONIC MATERIALS AND COMPONENTS"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E1 - Pg 43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the value of resistor\n",
"print '%s' %(\"Refer to the figure 1.52\")\n",
"print '%s' %(\"Hold the resistor as shown in the figure such that tolerance is on your extreme right.\")\n",
"print '%s' %(\"Now the value of the resistor is equal to\")\n",
"print '%s' %(\" Red Black Blue Gold\")\n",
"print '%s' %(\" 2 0 6 (+/-)5%\")\n",
"red=2. #red value\n",
"blk=0 #black value\n",
"blu=6. #blue value\n",
"gld=5. #gold value\n",
"value_res=(red*10.+blk)*10.**blu #value of resistor\n",
"print '%s %.0f %s %.0f' %(\"\\n The value of resistor is\",value_res,\"ohm (+/-)\",gld)\n",
"per_val=0.05*value_res\n",
"pos_value_res=value_res+per_val #positive range of resistor\n",
"neg_value_res=value_res-per_val #negative range of resistor\n",
"print '%s %.0f %s %.0f %s ' %(\"\\n The value of resistor is\",neg_value_res*1e-6,\" Mohm and\",pos_value_res*1e-6,\"Mohm\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Refer to the figure 1.52\n",
"Hold the resistor as shown in the figure such that tolerance is on your extreme right.\n",
"Now the value of the resistor is equal to\n",
" Red Black Blue Gold\n",
" 2 0 6 (+/-)5%\n",
"\n",
" The value of resistor is 20000000 ohm (+/-) 5\n",
"\n",
" The value of resistor is 19 Mohm and 21 Mohm \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E2 - Pg 43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the value of the resistor\n",
"print '%s' %(\"With the help of colour coding table, one finds\")\n",
"print '%s' %(\" 1st_Band 2nd_Band 3rd_Band 4th_Band\")\n",
"print(\" Yellow Violet Orange Gold\")\n",
"print '%s' %(\" 4 7 10**3 (+/-)5%\")\n",
"yel=4. #yellow value\n",
"vio=7. #violet value\n",
"org=1e3 #orange value\n",
"gld=5. #gold value in %\n",
"val_res=(yel*10.+vio)*org\n",
"print '%s %.2f %s %.0f' %(\"\\n The value of resistor is\",val_res*1e-3,\"kohm (+/-)\",gld)\n",
"gld_ab=0.05 #absolute gold value\n",
"per_val=gld_ab*val_res\n",
"print '%s %.0f %s'%(\"\\n Now, 5%% of 47k_ohm =\",per_val, \"ohm\")\n",
"range_high=val_res+per_val #higher range\n",
"range_low=val_res-per_val #lower range\n",
"print'%s %.2f %s %.2f %s %.2f %s %.2f %s'%(\"\\n\\n Thus resistance should be within the range\",val_res*1e-3, \"kohm(+/-)\",per_val*1e-3 ,\"Kohm\\n or between\",range_low*1e-3 ,\"kohm and\",range_high*1e-3, \"kohm\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"With the help of colour coding table, one finds\n",
" 1st_Band 2nd_Band 3rd_Band 4th_Band\n",
" Yellow Violet Orange Gold\n",
" 4 7 10**3 (+/-)5%\n",
"\n",
" The value of resistor is 47.00 kohm (+/-) 5\n",
"\n",
" Now, 5%% of 47k_ohm = 2350 ohm\n",
"\n",
"\n",
" Thus resistance should be within the range 47.00 kohm(+/-) 2.35 Kohm\n",
" or between 44.65 kohm and 49.35 kohm\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E3 - Pg 44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the value of the resistor\n",
"print '%s' %(\"The specification of the resistor from the color coding table is as follows\")\n",
"print '%s' %(\" 1st_Band 2nd_Band 3rd_Band 4th_Band\")\n",
"print '%s' %(\" Gray Blue Gold Silver\")\n",
"print '%s' %(\" 8 6 10**(-1) (+/-)10%\")\n",
"gray=8. #gray value\n",
"blu=6. #blue value\n",
"gld=10.**-1 #gold value\n",
"sil=10. #silver value in %\n",
"val_res=(gray*10.+blu)*gld\n",
"print '%s %.2f %s %.2f' %(\"\\n The value of resistor is\",val_res, \"ohm (+/-)\",sil)\n",
"sil_ab=0.1 #absolute gold value\n",
"per_val=sil_ab*val_res\n",
"print '%s %.2f %s' %(\"\\n Now, 10%% of 8.6 ohm =\",per_val,\" ohm\")\n",
"range_high=val_res+per_val #higher range\n",
"range_low=val_res-per_val #lower range\n",
"print '%s %.2f %s %.2f %s %.2f %s %.2f %s' %(\"\\n\\n Obviously resistance should lie within the range\",val_res,\"ohm(+/-)\",per_val,\"ohm\\n or between\",range_high,\"ohm and\",range_low,\"ohm\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The specification of the resistor from the color coding table is as follows\n",
" 1st_Band 2nd_Band 3rd_Band 4th_Band\n",
" Gray Blue Gold Silver\n",
" 8 6 10**(-1) (+/-)10%\n",
"\n",
" The value of resistor is 8.60 ohm (+/-) 10.00\n",
"\n",
" Now, 10%% of 8.6 ohm = 0.86 ohm\n",
"\n",
"\n",
" Obviously resistance should lie within the range 8.60 ohm(+/-) 0.86 ohm\n",
" or between 9.46 ohm and 7.74 ohm\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E4 - Pg 44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the load voltage and load current\n",
"print '%s' %(\"Refer to the figure 1.53\")\n",
"Vs=2. #supply voltage in V\n",
"Rs=1. #resistance in ohm\n",
"Is=Vs/Rs\n",
"print '%s %.2f %s' %(\"\\n Current Is =\",Is,\" A \\n\")\n",
"print '%s' %(\" Internal resistance remains the same but is now connected in parralel with the current sourceIS,as shown in Figure 1.51(a)\")\n",
"print '%s' %(\" Now,we connect a load resistance R_L=1 ohm across the terminals of two representations ,and find I_L and V_L. From Figure 1.54(b) and using the current-divider concept,one obtains\")\n",
"RL=1. #load resistance in ohm\n",
"IL=Is*(Rs/(Rs+RL)) #load current using current-divider\n",
"VL=IL*RL #load voltage\n",
"print '%s %.0f %s' %(\"\\n Load voltage =\",VL,\"V\")\n",
"print '%s %.0f %s' %(\"\\n Load current =\",IL,\"A \\n\")\n",
"print '%s' %(\"From equation 53(b),using the voltage-divider concept,one obtains\")\n",
"VD_vl=Vs*(RL/(RL+Rs)) #load voltage using voltage divider \n",
"VD_il=VL/RL #load current\n",
"print '%s %.0f %s' %(\"\\n Load voltage =\",VD_vl,\"V\")\n",
"print '%s %.0f %s' %(\"\\n Load current =\",VD_il,\"A \\n\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Refer to the figure 1.53\n",
"\n",
" Current Is = 2.00 A \n",
"\n",
" Internal resistance remains the same but is now connected in parralel with the current sourceIS,as shown in Figure 1.51(a)\n",
" Now,we connect a load resistance R_L=1 ohm across the terminals of two representations ,and find I_L and V_L. From Figure 1.54(b) and using the current-divider concept,one obtains\n",
"\n",
" Load voltage = 1 V\n",
"\n",
" Load current = 1 A \n",
"\n",
"From equation 53(b),using the voltage-divider concept,one obtains\n",
"\n",
" Load voltage = 1 V\n",
"\n",
" Load current = 1 A \n",
"\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E5 - Pg 45"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the percentage variation in current, current for two extreme values R_L\n",
"print '%s' %(\"Refer to the figure 1.55\")\n",
"print '%s' %(\"(a) R_L varies from 1 ohm to 10 ohm.\")\n",
"print '%s' %(\"Currents for two extreme values of R_L are\")\n",
"Vs=10. #supply voltage\n",
"RL1=1. #resistance RL1\n",
"Rs=100. #source resistance\n",
"IL1=(Vs/(RL1+Rs))\n",
"RL2=10.\n",
"IL2=(Vs/(RL2+Rs))\n",
"per_var_cur=((IL1-IL2)/IL1)*100.\n",
"print '%s %.2f %s' %(\"\\n Percentage variation in current =\",per_var_cur,\"\\n\")#answer in the text book took a .3 decimal round off value\n",
"print '%s' %(\" Now,load voltage for the two extreme values of R_L are\")\n",
"VL1=IL1*RL1\n",
"VL2=IL2*RL2\n",
"per_var_vol=((VL2-VL1)/VL2)*100.\n",
"print '%s %.2f %s' %(\"\\n Percentage variation in current =\",per_var_vol,\"\\n\")\n",
"\n",
"print '%s' %(\"(b) R_L varies from 1 k-ohm to 10 k-ohm(Figure 1.55(b))\")\n",
"print '%s' %(\"Currents for the two extreme values R_L are\")\n",
"RL11=1000.\n",
"IL11=(Vs/(RL11+Rs))\n",
"RL22=10000.\n",
"IL22=(Vs/(RL22+Rs)) #mistake in book value\n",
"per_var_cur11=((IL11-IL22)/IL11)*100.\n",
"print '%s %.2f %s' %(\"\\n Percentage variation in current =\",per_var_cur11,\"\\n\") #mistake in book value\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Refer to the figure 1.55\n",
"(a) R_L varies from 1 ohm to 10 ohm.\n",
"Currents for two extreme values of R_L are\n",
"\n",
" Percentage variation in current = 8.18 \n",
"\n",
" Now,load voltage for the two extreme values of R_L are\n",
"\n",
" Percentage variation in current = 89.11 \n",
"\n",
"(b) R_L varies from 1 k-ohm to 10 k-ohm(Figure 1.55(b))\n",
"Currents for the two extreme values R_L are\n",
"\n",
" Percentage variation in current = 89.11 \n",
"\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
