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| author | Jovina Dsouza | 2014-06-17 19:17:50 +0530 |
|---|---|---|
| committer | Jovina Dsouza | 2014-06-17 19:17:50 +0530 |
| commit | c6f0d6aeb95beaf41e4b679e78bb42c4ffe45a40 (patch) | |
| tree | 8a9939717a888c4b79115cb5e30a7100179148f8 /Basic_Electronics_and_Linear_Circuits/ch10.ipynb | |
| parent | 3df19050b78c9fad037dd5f464f05ccf2b43c40e (diff) | |
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diff --git a/Basic_Electronics_and_Linear_Circuits/ch10.ipynb b/Basic_Electronics_and_Linear_Circuits/ch10.ipynb new file mode 100644 index 00000000..eed36852 --- /dev/null +++ b/Basic_Electronics_and_Linear_Circuits/ch10.ipynb @@ -0,0 +1,141 @@ +{
+ "metadata": {
+ "name": "Ch 10"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10: Power Amplifiers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.1 Page No.345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 10.1\n",
+ "#Program to Determine the Transformer Turns Ratio\n",
+ "\n",
+ "#Given Circuit Data\n",
+ "RL=16 # Ohms, load resistance\n",
+ "RLd=10000.0 # Ohms ,effective load resistance\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "N12=math.sqrt(RLd/RL) #N12=N1/N2\n",
+ "\n",
+ "# Result\n",
+ "print \" The Transformer Turns Ratio is N1/N2\",N12,\":1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The Transformer Turns Ratio is N1/N2 25.0 :1\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.2 Page No.345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 10.2\n",
+ "# Determine the Effective Resistance seen\n",
+ "# looking into the Primary\n",
+ "\n",
+ "#Given Circuit Data\n",
+ "Rl=8.0 #Ohms, load resistance\n",
+ "N12=15.0 #N12=N1/N2, transformer turns ratio\n",
+ "\n",
+ "#Calculation\n",
+ "Rld=(N12)**2*Rl #effective resistance\n",
+ "\n",
+ "# Result\n",
+ "print \" The Effective Resistance seen looking into the Primary, Rld = \",Rld/10**3,\"k ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.3 Page No.353"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 10.3\n",
+ "#(a)\n",
+ "# Determine the Second, Third & Fourth Harmonic Distortions \n",
+ "\n",
+ "#Given Circuit Data\n",
+ "#io=15*sin(600*t)+1.5*sin(1200*t)+1.2*sin(1800*t)+0.5*sin(2400*t)\n",
+ "#current in components 1,2,3,4\n",
+ "I1=15 #A\n",
+ "I2=1.5 #A\n",
+ "I3=1.2 #A\n",
+ "I4=0.5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "D2=(I2/I1)*100 #percentage harmonic distribution of component 2\n",
+ "D3=(I3/I1)*100 #percentage harmonic distribution of component 3\n",
+ "D4=(I4/I1)*100 #percentage harmonic distribution of component 4\n",
+ "\n",
+ "#Result\n",
+ "print \" The Second Harmonic Distortion is, D2 = percent .\",D2\n",
+ "print \" The Third Harmonic Distortion is, D3 = percent .\",D3\n",
+ "print \" The Fourth Harmonic Distortion is, D4 = percent .\",round(D4,2)\n",
+ "\n",
+ "#(b)\n",
+ "import math\n",
+ "P1=1 #say\n",
+ "\n",
+ "#Calculation\n",
+ "D=math.sqrt(D2**2+D3**2+D4**2) #Distortion Factor\n",
+ "P=(1+(D/100)**2)*P1\n",
+ "Pi=((P-P1)/P1)*100\n",
+ "\n",
+ "#Result\n",
+ "print \"The Percentage Increase in Power because of Distortion is, Pi (in percent)= \",round(Pi,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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