diff options
Diffstat (limited to 'Basic_Electrical_Engineering/Chapter18.ipynb')
| -rwxr-xr-x | Basic_Electrical_Engineering/Chapter18.ipynb | 542 |
1 files changed, 542 insertions, 0 deletions
diff --git a/Basic_Electrical_Engineering/Chapter18.ipynb b/Basic_Electrical_Engineering/Chapter18.ipynb new file mode 100755 index 00000000..99017984 --- /dev/null +++ b/Basic_Electrical_Engineering/Chapter18.ipynb @@ -0,0 +1,542 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 18: ELECTRICAL MEASURING INSTRUMENTS\n",
+ "\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.1,Page number: 598\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the deflecting torque in Newton-metres for a PMMC instrument.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "I=15e-03 #Current flowing through the coil(in Amperes) \n",
+ "B=0.2 #Flux density in the air gap(in Tesla)\n",
+ "l=2e-02 #Length of the magnetic field(in m)\n",
+ "d=2.5e-02 #Mean width of the coil(in m) \n",
+ "r=d/2 #Radius of the coil(in cm) \n",
+ "n1=42 #Number of turns(lower limit) \n",
+ "n2=43 #Number of turns(upper limit)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "F1=I*B*l*n1\n",
+ "F2=I*B*l*n2\n",
+ "net_torque=(F1+F2)*r\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The deflecting torque is %e Nm.\" %(net_torque)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The deflecting torque is 6.375000e-05 Nm.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.2,Page Number: 604\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the shunt resistance for measuring a maximum current of 10 mA.\"\"\" \n",
+ "\n",
+ "#Variable Declaration:\n",
+ "Ifsd=10e-03 #Maximum current(in Amperes)\n",
+ "Im=100e-06 #Full-scale deflection current(in Amperes) \n",
+ "Rm=100 #Meter Resistance(in Ohms) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Ish=Ifsd-Im\n",
+ "Rsh=(Im*Rm)/Ish\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The shunt resistance needed is %.6f Ohms.\" %(Rsh)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The shunt resistance needed is 1.010101 Ohms.\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.3,Page number: 605"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Designing an universal shunt for a multi-range ammeter.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "Im=100e-06 #Full-scale deflection current(in Amperes)\n",
+ "Rm=100.0 #Internal resistance(in Ohms)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "\"\"\"For 1-mA range,the required shunt can be calculated as follows.\"\"\"\n",
+ "Ifsd=1e-03\n",
+ "Rsh=(Im*Rm)/(Ifsd-Im)\n",
+ "\n",
+ "Rm=900.0\n",
+ "R=100.0\n",
+ "\"\"\"(a)Range-switch at 1 mA:\"\"\"\n",
+ "Rm1=Rm\n",
+ "Ish1=(1e-03)-(0.1e-03)\n",
+ "Rsh1=(Rm1*Im)/(Ish1)\n",
+ "\n",
+ "\"\"\"(b)Range-switch at 10 mA:\"\"\"\n",
+ "\"\"\"Rm2=Rm+R1\"\"\"\n",
+ "Ish2=(10e-03)-(0.1e-03)\n",
+ "\"\"\"Rsh2=R2+R3+R4+R5=R-R1=100-R1;\"\"\"\n",
+ "\"\"\"Rsh2=(Rm2*Im)/Ish2;\"\"\"\n",
+ "R1=(9900.0-900.0)/100.0\n",
+ "\n",
+ "\"\"\"(c)Range-switch at 100 mA:\"\"\"\n",
+ "\"\"\"Rm3=Rm+R1+R2\"\"\"\n",
+ "Ish3=(100e-03)-(0.1e-03)\n",
+ "\"\"\"Rsh3=R3+R4+R5=R-R1-R2=100-90-R2=90-R2;\"\"\"\n",
+ "\"\"\"Rsh3=(Rm3*Im)/Ish3;\"\"\"\n",
+ "R2=(9990.0-990.0)/1000.0\n",
+ "\n",
+ "\"\"\"(d)Range-switch at 500 mA:\"\"\"\n",
+ "\"\"\"Rm4=Rm+R1+R2+R3\"\"\"\n",
+ "Ish3=(500e-03)-(0.1e-03)\n",
+ "\"\"\"Rsh4=R4+R5=R-R1-R2-R3=100-90-9-R3=1-R3;\"\"\"\n",
+ "\"\"\"Rsh4=(Rm4*Im)/Ish4;\"\"\"\n",
+ "R3=(4999.0-999.0)/5000.0\n",
+ "\n",
+ "\n",
+ "\"\"\"(e)Range-switch at 1 A:\"\"\"\n",
+ "\"\"\"Rm5=Rm+R1+R2+R3+R4\"\"\"\n",
+ "Ish3=(1000e-03)-(0.1e-03)\n",
+ "\"\"\"Rsh5=R5=R-R1-R2-R3-R4=100-90-9-0.8-R4=0.2-R4;\"\"\"\n",
+ "\"\"\"Rsh5=(Rm5*Im)/Ish5;\"\"\"\n",
+ "R4=(1999.8-999.8)/10000.0\n",
+ "R5=R-R1-R2-R3-R4\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The resistor values are: \"\n",
+ "print \"R1=%.2f Ohms.\" %(R1)\n",
+ "print \"R2=%.2f Ohms.\" %(R2)\n",
+ "print \"R3=%.2f Ohms.\" %(R3)\n",
+ "print \"R4=%.2f Ohms.\" %(R4)\n",
+ "print \"R5=%.2f Ohms.\" %(R5)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistor values are: \n",
+ "R1=90.00 Ohms.\n",
+ "R2=9.00 Ohms.\n",
+ "R3=0.80 Ohms.\n",
+ "R4=0.10 Ohms.\n",
+ "R5=0.10 Ohms.\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.4,Page number: 609"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"To convert a d'Arsonval meter movement into a voltmeter.\"\"\" \n",
+ "\n",
+ "#Variable Declaration:\n",
+ "Im=100e-06 #Current sensitivity(in Amperes)\n",
+ "Rm=100 #Resistance of the coil(in Ohms)\n",
+ "Vfsd=100 #Full-scale deflection of voltmeter(in Volts) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Rs=(Vfsd/Im)-Rm\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"To convert the d'Arsonval meter movement into a voltmeter of range 100V, we connect a resistor Rs in series.\"\n",
+ "print \"Rs = %.3f kilo Ohms. \" %(Rs/1000)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "To convert the d'Arsonval meter movement into a voltmeter of range 100V, we connect a resistor Rs in series.\n",
+ "Rs = 999.900 kilo Ohms. \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.5,Page number: 609\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the multipiler resistance and the voltage mutiplyiing factor of a dc voltmeter.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "Im=50e-06 #Full-scale deflection current(in Amperes) \n",
+ "Rm=1e03 #Meter Resistance(in Ohms) \n",
+ "Vfsd=50 #Full-scale deflection of voltmeter(in Volts) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Rs=(Vfsd/Im)-Rm\n",
+ "n=Vfsd/(Im*Rm)\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The multiplier resistance needed is %.2f kilo Ohms.\" %(Rs/1000)\n",
+ "print \"(b)The voltage multiplying factor is %d.\" %(round(n,0))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The multiplier resistance needed is 999.00 kilo Ohms.\n",
+ "(b)The voltage multiplying factor is 1000.\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.6,Page number: 612"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the reading and error in measurement of voltmeter.\"\"\"\n",
+ " \n",
+ "\"\"\" NOTE: All resistances expressed in kilo Ohms.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "range_A=50 #Range of voltmeter-A(in Volts)\n",
+ "range_B=50 #Range of voltmeter-B(in Volts)\n",
+ "sens_A=1000 #Sensitivity of voltmeter-A(in Ohm/Volts) \n",
+ "sens_B=20000 #Sensitivity of voltmeter-B(in Ohm/Volts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Vt=150.0*(50.0/(100.0+50.0))\n",
+ "R_i1=(range_A*sens_A)/1000.0\n",
+ "Req=1/((1/R_i1)+(1.0/50))\n",
+ "V1=(150.0)*(Req/(100+Req))\n",
+ "R_i2=(range_B*sens_B)/1000.0\n",
+ "Req=1/((1/R_i2)+(1.0/50))\n",
+ "V2=(150.0)*(Req/(Req+100))\n",
+ "err_A=((Vt-V1)/Vt)*100.0\n",
+ "err_B=((Vt-V2)/Vt)*100.0\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The reading of voltmeter-A is %.2f V\\nThe reading of voltmeter-B is %.2f V.\" %(V1,V2)\n",
+ "print \"\\nThe error in the reading of voltmeter-A is %.2f percent.\" %(err_A)\n",
+ "print \"The error in the reading of voltmeter-B is %.2f percent.\" %(err_B)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reading of voltmeter-A is 30.00 V\n",
+ "The reading of voltmeter-B is 48.39 V.\n",
+ "\n",
+ "The error in the reading of voltmeter-A is 40.00 percent.\n",
+ "The error in the reading of voltmeter-B is 3.23 percent.\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.7,Page number: 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the deflection in an ammeter.\"\"\"\n",
+ "\n",
+ "from math import sin,asin,degrees,pow\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "I1=20.0 #Initial current(in Amperes) \n",
+ "I2=12.0 #Final current(in Amperes)\n",
+ "angle1=60 #Initial deflection(in degrees)\n",
+ "\"\"\" Given: Deflecting torque is directly proportional to the current.\"\"\"\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "\"\"\"For spring control: Controlling torque is directly proportional to deflection.\n",
+ " For steady state deflection, controlling torque=deflecting torque.\n",
+ " Therefore,deflection is directly proportional to current. \"\"\"\n",
+ "angle2_a=(I2/I1)*angle1\n",
+ "\"\"\"For gravity control: Controlling torque is directly proportional to sine of the deflection angle.\n",
+ " For steady state deflection, controlling torque=deflecting torque.\n",
+ " Therefore,sine of the angle of deflection is directly proportional to the current. \"\"\" \n",
+ "angle2_b=asin((I2/I1)*sin(radians(angle1)))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The deflection for a current of 12A when the instrument is spring controlled is %.2f degrees.\" %(angle2_a) \n",
+ "print \"(b)The deflection for a current of 12A when the instrument is gravity controlled is %.2f degrees.\" %(degrees(angle2_b)) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The deflection for a current of 12A when the instrument is spring controlled is 36.00 degrees.\n",
+ "(b)The deflection for a current of 12A when the instrument is gravity controlled is 31.31 degrees.\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.8,Page number: 618"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the deflection in degrees in a gravity-controlled instrument.\"\"\"\n",
+ "\n",
+ "from math import asin,degrees\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "W=0.005 #Controlling weight(in kilograms)\n",
+ "l=2.4e-02 #Distance of controlling weight from the axis(in metres)\n",
+ "torque=1.05e-04 #Deflecting torque(in kg-m)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "angle=asin(torque/(W*l))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The deflection corresponding to a deflecting torque of 1.05e-04 kg-m is %.2f degrees.\" %(degrees(angle))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The deflection corresponding to a deflecting torque of 1.05e-04 kg-m is 61.04 degrees.\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.9,Page number: 618"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the deflection in an ammeter.\"\"\"\n",
+ "\n",
+ "from math import sin,asin,degrees,pow\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "I1=10.0 #Initial current(in Amperes) \n",
+ "I2=5.0 #Final current(in Amperes)\n",
+ "angle1=90 #Initial deflection(in degrees)\n",
+ "\"\"\" Given: Deflecting torque is directly proportional to square of the current.\"\"\"\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "\"\"\"For spring control: Controlling torque is directly proportional to deflection.\n",
+ " For steady state deflection, controlling torque=deflecting torque.\n",
+ " Therefore,deflection is directly proportional to square of the current. \"\"\"\n",
+ "angle2_a=pow((I2/I1),2)*angle1\n",
+ "\"\"\"For gravity control: Controlling torque is directly proportional to sine of the deflection angle.\n",
+ " For steady state deflection, controlling torque=deflecting torque.\n",
+ " Therefore,sine of the angle of deflection is directly proportional to square of the current. \"\"\" \n",
+ "angle2_b=asin((pow((I2/I1),2))*sin(radians(angle1)))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The deflection for a current of 5A when the instrument is spring controlled is %.2f degrees.\" %(angle2_a) \n",
+ "print \"(b)The deflection for a current of 5A when the instrument is gravity controlled is %.2f degrees.\" %(degrees(angle2_b)) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The deflection for a current of 5A when the instrument is spring controlled is 22.50 degrees.\n",
+ "(b)The deflection for a current of 5A when the instrument is gravity controlled is 14.48 degrees.\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.10,Page number: 619 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the current required to produce a deflection of 60 degrees in a moving coil instrument.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "w=4e-02 #Width of the coil(in metres)\n",
+ "l=5e-02 #Length of the coil(in metres)\n",
+ "N=80.0 #Number of turns in the coil\n",
+ "torque_control=0.5e-07 #Controlling torque per degree deflection of the coil(in Nm)\n",
+ "B=0.1 #Magnetic flux density in the air gap(in Wb per square-metre)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "A=w*l\n",
+ "torque_c=torque_control*60.0\n",
+ "I=torque_c/(B*N*A)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The current required to give a deflection of 60 degrees is %e A.\" %(I)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current required to give a deflection of 60 degrees is 1.875000e-04 A.\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |