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| author | hardythe1 | 2014-07-25 12:37:07 +0530 |
|---|---|---|
| committer | hardythe1 | 2014-07-25 12:37:07 +0530 |
| commit | efb9ead5d9758d5d0bed7a22069320b14f972e40 (patch) | |
| tree | 88c95b7c5d4bccfbcb2bdf16bf2bef0b73184808 /Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint.ipynb | |
| parent | 6846da7f30aadc7b3812538d1b1c9ef2e465a922 (diff) | |
| download | Python-Textbook-Companions-efb9ead5d9758d5d0bed7a22069320b14f972e40.tar.gz Python-Textbook-Companions-efb9ead5d9758d5d0bed7a22069320b14f972e40.tar.bz2 Python-Textbook-Companions-efb9ead5d9758d5d0bed7a22069320b14f972e40.zip | |
adding books
Diffstat (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint.ipynb')
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diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint.ipynb new file mode 100755 index 00000000..5b0974af --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint.ipynb @@ -0,0 +1,1627 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 20: Transformers</h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1, page no. 317</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine the secondary voltage, assuming an ideal transformer\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "N1 = 500;# primary turns\n",
+ "N2 = 3000;# secondary turns\n",
+ "V1 = 240;# in Volts\n",
+ "\n",
+ "#calculation:\n",
+ " #For an ideal transformer, voltage ratio = turns ratio\n",
+ "V2 = V1*N2/N1\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n secondary voltage \",round(V2,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " secondary voltage 1440.0 V"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 2, page no. 317</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine its output voltage.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "tr = 2/7;# turns ratio\n",
+ "V1 = 240;# in Volts\n",
+ "\n",
+ "#calculation:\n",
+ " #A turns ratio of 2:7 means that the transformer has 2 turns on the primary \n",
+ " #for every 7 turns on the secondary\n",
+ "V2 = V1/tr\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n output voltage \",round(V2,2),\" V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " output voltage 840.0 V"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 3, page no. 317</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine secondary voltage and current.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "tr = 8/1;# turns ratio\n",
+ "I1 = 3;# in Amperes\n",
+ "V1 = 240;# in Volts\n",
+ "\n",
+ "#calculation:\n",
+ " #A turns ratio of 8:1 means that the transformer has 28 turns on the \n",
+ " #primary for every 1turns on the secondary\n",
+ "V2 = V1/tr\n",
+ " #secondary current\n",
+ "I2 = I1*tr\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n secondary voltage is \",round(V2,2),\" V and secondary current is \", round(I2,2),\" A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " secondary voltage is 30.0 V and secondary current is 24.0 A"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4, page no. 318</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the transformer turns ratio and the current taken from the supply.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "V1 = 240;# in Volts\n",
+ "V2 = 12;# in Volts\n",
+ "P = 150;# in Watts\n",
+ "\n",
+ "#calculation:\n",
+ "I2 = P/V2\n",
+ " #A turns ratio = Vp/Vs\n",
+ "tr = V1/V2# turn ratio\n",
+ " # V1/V2 = I2/I1\n",
+ " #current taken from the supply\n",
+ "I1 = I2*V2/V1\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n turn ratio is \",round(tr,2),\" and current taken from the supply is \",round(I1,2),\" A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " turn ratio is 20.0 and current taken from the supply is 0.63 A"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5, page no. 318</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine (a) the full-load secondary current,\n",
+ "#(b) the minimum load resistance which can be connected across the secondary winding to give full load kVA, \n",
+ "#(c) the primary current at full load kVA.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "S = 5000;# in VA\n",
+ "tr = 10;# turn ratio\n",
+ "V1 = 2500;# in Volts\n",
+ "\n",
+ "#calculation:\n",
+ " #A turns ratio of 8:1 means that the transformer has 28 turns on the primary for every 1turns on the secondary\n",
+ "V2 = V1/tr\n",
+ " #transformer rating in volt-amperes = Vs*Is\n",
+ "I2 = S/V2\n",
+ " #Minimum value of load resistance\n",
+ "RL = V2/I2\n",
+ " # tr = I2/I1\n",
+ "I1 = I2/tr\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)full-load secondary current is \",round(I2,2),\" A\"\n",
+ "print \"\\n (b)minimum load resistance is \",round(RL,2),\" ohm\"\n",
+ "print \"\\n (c) primary current is \",round(I1,2),\" A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)full-load secondary current is 20.0 A\n",
+ "\n",
+ " (b)minimum load resistance is 12.5 ohm\n",
+ "\n",
+ " (c) primary current is 2.0 A"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 6, page no. 319</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the values of the magnetizing and core loss components of the noload current.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "V1 = 2400;# in Volts\n",
+ "V2 = 400;# in Volts\n",
+ "I0 = 0.5;# in Amperes\n",
+ "Pc = 400;# in Watts\n",
+ "\n",
+ "#calculation:\n",
+ " #Core loss (i.e. iron loss) P = V1*I0*cos(phi0)\n",
+ "pf = Pc/(V1*I0)\n",
+ "phi0 = math.acos(pf)\n",
+ " #Magnetizing component\n",
+ "Im = I0*math.sin(phi0)\n",
+ " #Core loss component\n",
+ "Ic = I0*math.cos(phi0)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)magnetizing component is \",round(Im,3),\" A and Core loss component is \",round(Ic,3),\" A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)magnetizing component is 0.471 A and Core loss component is 0.167 A"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 7, page no. 320</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine (a) the iron loss current, \n",
+ "#(b) the power factor on no-load, and (c) the magnetizing current.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "V = 240;# in Volts\n",
+ "I0 = 0.8;# in Amperes\n",
+ "P = 72;# in Watts\n",
+ "f = 50;# in Hz\n",
+ "\n",
+ "#calculation:\n",
+ " #Power absorbed = total core loss, P = V*I0*cos(phi0)\n",
+ " #Ic = I0*cos(phi0)\n",
+ "Ic = P/V\n",
+ "pf = Ic/I0\n",
+ " #From the right-angled triangle in Figure 20.2(b) and using\n",
+ " #Pythagoras\u2019 theorem, \n",
+ "Im = (I0*I0 - Ic*Ic)**0.5\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a) Core loss component is \",round( Ic,2),\" A\"\n",
+ "print \"\\n (b) power factor is \",round( pf,2),\"\"\n",
+ "print \"\\n (c)magnetizing component is \",round(Im,2),\" A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a) Core loss component is 0.3 A\n",
+ "\n",
+ " (b) power factor is 0.37 \n",
+ "\n",
+ " (c)magnetizing component is 0.74 A"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 8, page no. 321</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "S = 100000;# in VA\n",
+ "V1 = 4000;# in Volts\n",
+ "V2 = 200;# in Volts\n",
+ "N2 = 100;# sec turns\n",
+ "f = 50;# in Hz\n",
+ "\n",
+ "#calculation:\n",
+ " #Transformer rating = V1*I1 = V2*I2\n",
+ " #primary current\n",
+ "I1 = S/V1\n",
+ " #secondary current\n",
+ "I2 = S/V2\n",
+ " #primary turns\n",
+ "N1 = N2*V1/V2\n",
+ " #maximum flux\n",
+ " #assuming E2 = V2\n",
+ "Phim = V2/(4.44*f*N2)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)primary current is \",round( I1,2),\" A and secondary current is \",round( I2,2),\" A\"\n",
+ "print \"\\n (b)number of primary turns is \",round( N1,2),\"\"\n",
+ "print \"\\n (c)maximum value of the flux is \",round(Phim*1000,2),\"mWb\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)primary current is 25.0 A and secondary current is 500.0 A\n",
+ "\n",
+ " (b)number of primary turns is 2000.0 \n",
+ "\n",
+ " (c)maximum value of the flux is 9.01 mWb"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 9, page no. 322</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine (a) the maximum value of the flux density in the core, and\n",
+ "#(b) the voltage induced in the secondary winding.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "V1 = 250;# in Volts\n",
+ "A = 0.03;# in m2\n",
+ "N2 = 300;# sec turns\n",
+ "N1 = 25;# prim turns\n",
+ "f = 50;# in Hz\n",
+ "\n",
+ "#calculation:\n",
+ " #e.m.f. E1 = 4.44*f*Phim*N1\n",
+ " #maximum flux density,\n",
+ "Phim = V1/(4.44*f*N1)\n",
+ " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
+ " #maximum core flux density\n",
+ "Bm = Phim/A\n",
+ " #voltage induced in the secondary winding,\n",
+ "V2 = V1*N2/N1\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)maximum core flux density \",round( Bm,2),\" T\"\n",
+ "print \"\\n (b)voltage induced in the secondary winding is \",round( V2,2),\" V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)maximum core flux density 1.5 T\n",
+ "\n",
+ " (b)voltage induced in the secondary winding is 3000.0 V"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 10, page no. 323</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the number of primary and secondary turns\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "V1 = 500;# in Volts\n",
+ "V2 = 100;# in Volts\n",
+ "Bm = 1.5;# in Tesla\n",
+ "A = 0.005;# in m2\n",
+ "f = 50;# in Hz\n",
+ "\n",
+ "#calculation:\n",
+ " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
+ " #maximum core flux density\n",
+ "Phim = Bm*A\n",
+ " #e.m.f. E1 = 4.44*f*Phim*N1\n",
+ " #primary turns,\n",
+ "N1 = V1/(4.44*f*Phim)\n",
+ " #secondary turns,\n",
+ "N2 = V2*N1/V1\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n no. of primary and secondary turns are \",round(N1,2),\" turns, and \",round(N2,2),\" turns respectively\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " no. of primary and secondary turns are 300.3 turns, and 60.06 turns respectively"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 11, page no. 323</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate (a) the number of primary and secondary turns and \n",
+ "#(b) the cross-sectional area of the core.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "emfpt = 15;# in Volts\n",
+ "V1 = 4500;# in Volts\n",
+ "V2 = 225;# in Volts\n",
+ "Bm = 1.4;# in Tesla\n",
+ "f = 50;# in Hz\n",
+ "\n",
+ "#calculation:\n",
+ " #E.m.f. per turn, V1/N1 = V2/N2 = emfpt\n",
+ " #primary turns,\n",
+ "N1 = V1/emfpt\n",
+ " #secondary turns,\n",
+ "N2 = V2/emfpt\n",
+ " #e.m.f. E1 = 4.44*f*Phim*N1\n",
+ " #maximum flux density,\n",
+ "Phim = V1/(4.44*f*N1)\n",
+ " #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
+ " #cross-sectional area\n",
+ "A = Phim/Bm\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)no. of primary and secondary turns are \", N1,\" turns, and \", N2,\" turns respectively\"\n",
+ "print \"\\n (b)cross-sectional area is \", round(A,4),\"m2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)no. of primary and secondary turns are 300.0 turns, and 15.0 turns respectively\n",
+ "\n",
+ " (b)cross-sectional area is 0.0483 m2"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 12, page no. 324</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine the primary current and power factor when the secondary current is 100 A at a power factor of 0.85 lagging\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "N1 = 2000;# prim turns\n",
+ "N2 = 800;# sec turns\n",
+ "I0 = 5;# in Amperes\n",
+ "pf0 = 0.20;# power factor\n",
+ "I2 = 100;# in Amperes\n",
+ "pf2 = 0.85;# power factor\n",
+ "\n",
+ "#calculation:\n",
+ " #Let I01 be the component of the primary current which provides the restoring mmf. Then I01*N1 = I2*N2\n",
+ "I01 = I2*N2/N1\n",
+ " #If the power factor of the secondary is 0.85\n",
+ "phi2 = math.acos(pf2)\n",
+ " #If the power factor on no-load is 0.20,\n",
+ "phi0 = math.acos(pf0)\n",
+ "I1h = I0*math.cos(phi0) + I01*math.cos(phi2)\n",
+ "I1v = I0*math.sin(phi0) + I01*math.sin(phi2)\n",
+ " #Hence the magnitude of I1\n",
+ "I1 = (I1h*I1h + I1v*I1v)**0.5\n",
+ "pf1 = math.cos(math.atan(I1v/I1h))\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n Primary current is \", round(I1,2),\" A, and Power factor is \",round(pf1,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " Primary current is 43.58 A, and Power factor is 0.8"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 13, page no. 328</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine (a) the equivalent resistance referred to the primary winding, \n",
+ "#(b) the equivalent reactance referred to the primary winding,\n",
+ "#(c) the equivalent impedance referred to the primary winding, and \n",
+ "#(d) the phase angle of the impedance.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "N1 = 600;# prim turns\n",
+ "N2 = 150;# sec turns\n",
+ "R1 = 0.25;# in ohms\n",
+ "R2 = 0.01;# in ohms\n",
+ "X1 = 1.0;# in ohms\n",
+ "X2 = 0.04;# in ohms\n",
+ "\n",
+ "#calculation:\n",
+ "tr = N1/N2# turn ratio\n",
+ "vr = tr# voltage ratio = turn raio, vr = V1/V2\n",
+ " #equivalent resistance Re\n",
+ "Re = R1 + R2*(vr**2)\n",
+ " #equivalent reactance, Xe\n",
+ "Xe = X1 + X2*(vr**2)\n",
+ " #equivalent impedance, Ze\n",
+ "Ze = (Re*Re + Xe*Xe)**0.5\n",
+ " #cos(phie) = Re/Ze\n",
+ "pfe = Re/Ze\n",
+ "phie = math.acos(pfe)\n",
+ "phied = phie*180/math.pi# in \u00b0(deg)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)the equivalent resistance referred to the primary winding is \",round( Re,2),\" ohm\"\n",
+ "print \"\\n (b)the equivalent reactance referred to the primary winding is \",round( Xe,2),\" ohm\"\n",
+ "print \"\\n (c)the equivalent impedance referred to the primary winding is \",round( Ze,2),\" ohm\"\n",
+ "print \"\\n (d)phase angle is \",round( phied,2),\"deg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)the equivalent resistance referred to the primary winding is 0.41 ohm\n",
+ "\n",
+ " (b)the equivalent reactance referred to the primary winding is 1.64 ohm\n",
+ "\n",
+ " (c)the equivalent impedance referred to the primary winding is 1.69 ohm\n",
+ "\n",
+ " (d)phase angle is 75.96 deg"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 14, page no. 329</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the regulation of the transformer.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "V1 = 200;# in Volts\n",
+ "V2 = 400;# in Volts\n",
+ "V2L = 387.6;# in Volts\n",
+ "S = 5000;# in VA\n",
+ "\n",
+ "#calculation:\n",
+ " #regulation =(No-load secondary voltage -\u0003 terminal voltage on load)*100/no-load secondary voltage in %\n",
+ "reg = (V2 - V2L)*100/V2\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the regulation of the transformer is \",round(reg,2),\" percent \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the regulation of the transformer is 3.1 percent "
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 15, page no. 329</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the load voltage at which the mechanism operates.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "VnL = 240;# in Volts\n",
+ "reg = 2.5;# in percent\n",
+ "\n",
+ "#calculation:\n",
+ " #regulation =(No-load secondary voltage -\u0003 terminal voltage on load)*100/no-load secondary voltage in %\n",
+ "VL = VnL - reg*VnL/100\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the load voltage at which the mechanism operates is \",round(VL,2),\" V \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the load voltage at which the mechanism operates is 234.0 V "
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 16, page no. 331</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the transformer efficiency at full load and 0.85 power factor.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "S = 200000;# in VA\n",
+ "Pc = 1500;# in Watt\n",
+ "Pi = 1000;# in Watt\n",
+ "pf = 0.85;# power factor\n",
+ "\n",
+ "#calculation:\n",
+ " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
+ " #Efficiency = 1 - losses/input power\n",
+ " #Full-load output power = V*I*pf\n",
+ "Po = S*pf\n",
+ " #Total losses\n",
+ "Pl = Pc + Pi\n",
+ " #Input power = output power + losses\n",
+ "PI = Po + Pl\n",
+ " #efficiency\n",
+ "eff = 1-(Pl/PI)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the transformer efficiency at full load is \",round(eff,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the transformer efficiency at full load is 0.99"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 17, page no. 331</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the efficiency of the transformer\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "S = 200000;# in VA\n",
+ "Pc = 1500;# in Watt\n",
+ "Pi = 1000;# in Watt\n",
+ "pf = 0.85;# power factor\n",
+ "\n",
+ "#calculation:\n",
+ " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
+ " #Efficiency = 1 - losses/input power\n",
+ " #Half full-load power output = V*I*pf/2\n",
+ "Po = S*pf/2\n",
+ " #Copper loss (or I*I*R loss) is proportional to current squared\n",
+ " #Hence the copper loss at half full-load is\n",
+ "Pch = Pc/(2*2)\n",
+ " #Iron loss = 1000 W (constant)\n",
+ " #Total losses\n",
+ "Pl = Pch + Pi\n",
+ " #Input power at half full-load = output power at half full-load + losses\n",
+ "PI = Po + Pl\n",
+ " #efficiency\n",
+ "eff = (1-(Pl/PI))*100\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the transformer efficiency at half full load is \",round(eff,2),\" percent\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the transformer efficiency at half full load is 98.41 percent"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 18, page no. 332</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine the efficiency of the transformer (a) on full load, and (b) on half load.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "S = 400000;# in VA\n",
+ "R1 = 0.5;# in Ohm\n",
+ "R2 = 0.001;# in Ohm\n",
+ "V1 = 5000;# in Volts\n",
+ "V2 = 320;# in Volts\n",
+ "Pi = 2500;# in Watt\n",
+ "pf = 0.85;# power factor\n",
+ "\n",
+ "#calculation:\n",
+ " #Rating = 400 kVA = V1*I1 = V2*I2\n",
+ " #Hence primary current\n",
+ "I1 = S/V1\n",
+ " #secondary current\n",
+ "I2 = S/V2\n",
+ " #Total copper loss = I1*I1*R1 + I2*I2*R2,\n",
+ "Pcf = I1*I1*R1 + I2*I2*R2\n",
+ " #On full load, total loss = copper loss + iron loss\n",
+ "Plf = Pcf + Pi\n",
+ " # full-load power output = V2*I2*pf\n",
+ "Pof = S*pf\n",
+ " #Input power at full-load = output power at full-load + losses\n",
+ "PIf = Pof + Plf\n",
+ " #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
+ " #Efficiency = 1 - losses/input power\n",
+ "efff = (1-(Plf/PIf))*100\n",
+ "\n",
+ " #Half full-load power output = V*I*pf/2\n",
+ "Poh = S*pf/2\n",
+ " #Copper loss (or I*I*R loss) is proportional to current squared\n",
+ " #Hence the copper loss at half full-load is\n",
+ "Pch = Pcf/(2*2)\n",
+ " #Iron loss = 2500 W (constant)\n",
+ " #Total losses\n",
+ "Plh = Pch + Pi\n",
+ " #Input power at half full-load = output power at half full-load + losses\n",
+ "PIh = Poh + Plh\n",
+ " #efficiency\n",
+ "effh = (1-(Plh/PIh))*100\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)the transformer efficiency at full load is \", round(efff,2),\" percent\"\n",
+ "print \"\\n (b)the transformer efficiency at half full load is \", round(effh,2),\" percent\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)the transformer efficiency at full load is 97.91 percent\n",
+ "\n",
+ " (b)the transformer efficiency at half full load is 97.88 percent"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 19, page no. 333</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine (a) the output kVA at which the efficiency of the transformer is a maximum, and (b) the maximum efficiency,\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "S = 500000;# in VA\n",
+ "Pcf = 4000;# in Watt\n",
+ "Pi = 2500;# in Watt\n",
+ "pf = 0.75;# power factor\n",
+ "\n",
+ "#calculation:\n",
+ " #Let x be the fraction of full load kVA at which the efficiency is a maximum.\n",
+ " #The corresponding total copper loss = (4 kW)*(x**2)\n",
+ " #At maximum efficiency, copper loss = iron loss Hence\n",
+ "x = (Pi/Pcf)**0.5\n",
+ " #Hence the output kVA at maximum efficiency\n",
+ "So = x*S\n",
+ " #Total loss at maximum efficiency\n",
+ "Pl = 2*Pi\n",
+ " #Output power\n",
+ "Po = So*pf\n",
+ " #Input power = output power + losses\n",
+ "PI = Po + Pl\n",
+ " #Efficiency = output power/input power = (input power\u00e2\u20ac\u201dlosses)/input power\n",
+ " #Efficiency = 1 - losses/input power\n",
+ " #Maximum efficiency\n",
+ "effm = (1 - Pl/PI)*100\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the output kVA at maximum efficiency is \",round(So/1000,2),\"kVA\"\n",
+ "print \"\\n max. efficiency is \",round(effm,2),\" pecent\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the output kVA at maximum efficiency is 395.28 kVA\n",
+ "\n",
+ " max. efficiency is 98.34 pecent"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 20, page no. 335</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the equivalent input resistance of the transformer.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "tr = 4;# turn ratio\n",
+ "RL = 100;# in Ohms\n",
+ "\n",
+ "#calculation:\n",
+ " #the equivalent input resistance,\n",
+ "Ri = RL*(tr**2)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the equivalent input resistance is \",round(Ri,2),\" ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the equivalent input resistance is 1600.0 ohm"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 21, page no. 335</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the optimum turns ratio of a transformer which would match a \n",
+ "#load resistance of 7 ohmto the output resistance of the amplifier.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "R1 = 112;# in Ohms\n",
+ "RL = 7;# in Ohms\n",
+ "\n",
+ "#calculation:\n",
+ " #The equivalent input resistance, R1 of the transformer needs to be 112 ohm for maximum power transfer.\n",
+ " #R1 = RL*(tr**2)\n",
+ " # tr = N1/N2 turn ratio\n",
+ "tr = (R1/RL)**0.5\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the optimum turns ratio is \",tr,\": 1.0\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the optimum turns ratio is 4.0 : 1.0"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 22, page no. 335</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the optimum value of load resistance for maximum power transfer\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "tr = 5;# turn ratio\n",
+ "R1 = 150;# in Ohms\n",
+ "\n",
+ "#calculation:\n",
+ " #The equivalent input resistance, R1 of the transformer needs to be 150 ohm for maximum power transfer.\n",
+ " #R1 = RL*(tr**2)\n",
+ "RL = R1/(tr**2)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the optimum value of load resistance is \",round(RL,2),\" ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the optimum value of load resistance is 6.0 ohm"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 23, page no. 335</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine (a) the primary current flowing and (b) the power dissipated in the load resistor.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "V1 = 220;# in Volts\n",
+ "V2 = 1760;# in Volts\n",
+ "V = 220;# in Volts\n",
+ "RL = 1280;# in Ohms\n",
+ "R = 2;# in Ohms\n",
+ "\n",
+ "#calculation:\n",
+ " #Turns ratio, tr = N1/N2 = V1/V2\n",
+ "tr = V1/V2\n",
+ " #Equivalent input resistance of the transformer,\n",
+ " #R1 = RL*(tr**2)\n",
+ "R1 = RL*(tr**2)\n",
+ " #Total input resistance\n",
+ "Rin = R + R1\n",
+ " # Primary current\n",
+ "I1 = V1/Rin\n",
+ " #For an ideal transformer V1/V2 = I2/I1,\n",
+ "I2 = I1*tr\n",
+ " #Power dissipated in load resistor RL\n",
+ "P = I2*I2*RL\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a) primary current flowing is \",round(I1,2),\" A\"\n",
+ "print \"\\n (b) power dissipated in the load resistor is \",round(P,2),\" W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a) primary current flowing is 10.0 A\n",
+ "\n",
+ " (b) power dissipated in the load resistor is 2000.0 W"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 24, page no. 336</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine (a) the value of the load resistance and (b) the power dissipated in the load.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "tr = 25;# teurn ratio\n",
+ "V = 24;# in Volts\n",
+ "R1 = 15000;# in Ohms\n",
+ "Rin = 15000;# in ohms\n",
+ "\n",
+ "#calculation:\n",
+ " #Turns ratio, tr = N1/N2 = V1/V2\n",
+ " #For maximum power transfer R1 needs to be equal to 15 kohm\n",
+ "RL = R1/(tr**2)\n",
+ " #The total input resistance when the source is connected to the matching transformer is\n",
+ "Rt = Rin + R1\n",
+ " #Primary current,\n",
+ "I1 = V/Rt\n",
+ " #N1/N2 = I2/I1\n",
+ "I2 = I1*tr\n",
+ " #Power dissipated in load resistor RL\n",
+ "P = I2*I2*RL\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a) the load resistance is \",round(RL,2),\"ohm\"\n",
+ "print \"\\n (b) power dissipated in the load resistor is \",round(P*1000,2),\"mW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a) the load resistance is 24.0 ohm\n",
+ "\n",
+ " (b) power dissipated in the load resistor is 9.6 mW"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 25, page no. 337</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine the current in each section of the winding.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "V1 = 320;# in Volts\n",
+ "V2 = 250;# in Volts\n",
+ "S = 20000;# in VA\n",
+ "\n",
+ "#calculation:\n",
+ " #Rating = 20 kVA = V1*I1 = V2*I2\n",
+ " #Hence primary current, I1\n",
+ "I1 = S/V1\n",
+ " #secondary current, I2\n",
+ "I2 = S/V2\n",
+ " #Hence current in common part of the winding\n",
+ "I = I2 - I1\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n current in common part of the winding is \", round(I,2),\" A\"\n",
+ "print \"\\n primary current and secondary current are \",round(I1,2),\" A and \",round(I2,2),\" A respectively\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " current in common part of the winding is 17.5 A\n",
+ "\n",
+ " primary current and secondary current are 62.5 A and 80.0 A respectively"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 26, page no. 339</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the saving in the volume of copper used in an auto transformer compared with a double-wound transformer\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "V1a = 200;# in Volts\n",
+ "V2a = 150;# in Volts\n",
+ "V1b = 500;# in Volts\n",
+ "V2b = 100;# in Volts\n",
+ "\n",
+ "#calculation:\n",
+ " #For a 200 V:150 V transformer, xa\n",
+ "xa = V2a/V1a\n",
+ " #volume of copper in auto transformer\n",
+ "vca = (1 - xa)*100# of copper in a double-wound transformer\n",
+ " #the saving is\n",
+ "vsa = 100 - vca\n",
+ " #For a 500 V:100 V transformer, xb\n",
+ "xb = V2b/V1b\n",
+ " #volume of copper in auto transformer\n",
+ "vcb = (1 - xb)*100# of copper in a double-wound transformer\n",
+ " #the saving is\n",
+ "vsb = 100 - vcb\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)For a 200 V:150 V transformer, the saving is \", round(vsa,2),\" percent\"\n",
+ "print \"\\n (b)For a 500 V:100 V transformer, the saving is \", round(vsb,2),\" percent\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)For a 200 V:150 V transformer, the saving is 75.0 percent\n",
+ "\n",
+ " (b)For a 500 V:100 V transformer, the saving is 20.0 percent"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 27, page no. 340</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#find the secondary line voltage on no-load when the windings are connected (a) star-delta, (b) delta-star.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "N1 = 500;# prim turns\n",
+ "N2 = 50;# sec turns\n",
+ "VL = 2400;# in Volts\n",
+ "\n",
+ "#calculation:\n",
+ " #For a star-connection, VL = Vp*(3**0.5)\n",
+ "VL1s = VL\n",
+ " #Primary phase voltage\n",
+ "Vp1s = VL1s/(3**0.5)\n",
+ " #For a delta-connection, VL = Vp\n",
+ " #N1/N2 = V1/V2, from which,\n",
+ " #secondary phase voltage, Vp2s\n",
+ "Vp2s = Vp1s*N2/N1\n",
+ "VL2d = Vp2s\n",
+ "\n",
+ " #For a delta-connection, VL = Vp\n",
+ "VL1d = VL\n",
+ " #primary phase voltage Vp1d\n",
+ "Vp1d = VL1d\n",
+ " #Secondary phase voltage, Vp2d\n",
+ "Vp2d = Vp1d*N2/N1\n",
+ " #For a star-connection, VL = Vp*(3**0.5)\n",
+ "VL2s = Vp2d*(3**0.5)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the secondary line voltage for star and delta connection are \",round(Vp2s,1),\" V \"\n",
+ "print \" and \",round(VL2s,0),\" V respectively\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the secondary line voltage for star and delta connection are 138.6 V \n",
+ " and 416.0 V respectively\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 28, page no. 343</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine (a) the reading on the ammeter, \n",
+ "#(b) the potential difference across the ammeter and\n",
+ "#(c) the total load (in VA) on the secondary.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "N1 = 1;# prim turns\n",
+ "N2 = 60;# sec turns\n",
+ "I1 = 300;# in amperes\n",
+ "Ra = 0.15;# in ohms\n",
+ "R2 = 0.25;# in ohms\n",
+ "\n",
+ "#calculation:\n",
+ " #Reading on the ammeter,\n",
+ "I2 = I1*(N1/N2)\n",
+ " #P.d. across the ammeter = I2*RA, where RA is the ammeter resistance\n",
+ "pd = I2*Ra\n",
+ " #Total resistance of secondary circuit\n",
+ "Rt = Ra + R2\n",
+ " #Induced e.m.f. in secondary\n",
+ "V2 = I2*Rt\n",
+ " #Total load on secondary\n",
+ "S = V2*I2\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)the reading on the ammeter is \",round(I2,2),\" A \"\n",
+ "print \"\\n (b)potential difference across the ammeter is \",round(pd,2),\" V \"\n",
+ "print \"\\n (c)total load (in VA) on the secondary is \",round(S,2),\" VA \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)the reading on the ammeter is 5.0 A \n",
+ "\n",
+ " (b)potential difference across the ammeter is 0.75 V \n",
+ "\n",
+ " (c)total load (in VA) on the secondary is 10.0 VA "
+ ]
+ }
+ ],
+ "prompt_number": 28
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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