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{
"metadata": {
"name": "",
"signature": "sha256:542e92bd62de523a8cd9abe5adfd1eec8e64f2ebaa6b2ac5b00d208e9be115e3"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 5 : Analysis Methods"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.8 Page No : 76"
]
},
{
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"input": [
"import math \n",
"#From figure 5.13(a)\n",
"\n",
"#Applying KVL equation to the loop \n",
"I = (20.+10)/(3+6)\n",
"\n",
"# Calculation and Results\n",
"#As current will not flow in upper 3 ohm resistor so Thevenin voltage is equal to either of the two parallel branches\n",
"V1 = 20.-I*3\n",
"print \"Thevenin voltage = %dV\"%(V1)\n",
"\n",
"# Left 3 ohm and 6 ohm resistor are in parallel and their equivalent is in series with 3 ohm\n",
"R1 = 3+(3.*6)/(3+6)\n",
"print \"Thevenin resistance = %dohm\"%(R1)\n",
"\n",
"#Now to find Norton's equivalent\n",
"I1 = V1/R1\n",
"print \" Norton current = %dA\"%(I1)\n",
"#The value of resistance in Norton equivalent will not change but will come in parallel with current source\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thevenin voltage = 10V\n",
"Thevenin resistance = 5ohm\n",
" Norton current = 2A\n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}
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