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diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb deleted file mode 100755 index 4bad2773..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb +++ /dev/null @@ -1,791 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 22: Three-phase induction motors</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 389</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the synchronous speed of the motor in rev/min.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 50;# in Hz\n",
- "p = 2/2;# number of pairs of poles\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, \n",
- " #f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- "nsrpm = ns*60\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nsynchronous speed of the motor is \",nsrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "synchronous speed of the motor is 3000.0 rev/min"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 389</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the number of poles.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 60;# in Hz\n",
- "ns = 900/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "p = f/ns\n",
- "np = p*2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nnumber of poles is \", round(np,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "number of poles is 8.0"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 390</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the frequency of the supply voltage.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 2/2;# number of pairs of poles\n",
- "ns = 6000/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"frequency is \",f,\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "frequency is 100.0 Hz"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 391</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the synchronous speed and (b) the slip at full load.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 4/2;# number of pairs of poles\n",
- "f = 50;# in Hz\n",
- "nr = 1455/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n",
- "print \"\\n(b) slip is \",s,\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) synchronous speed is 25.0 rev/sec\n",
- "\n",
- "(b) slip is 3.0 percent"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 392</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the synchronous speed,\n",
- "#(b) the speed of the rotor and (c) the frequency of the induced e.m.f.\u2019s in the rotor\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 2/2;# number of pairs of poles\n",
- "f = 60;# in Hz\n",
- "s = 0.02;# slip\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The the rotor runs at\n",
- "nr = ns*(1 - s)\n",
- " #frequency of the e.m.f. induced in the rotor bars is\n",
- "fr = ns - nr\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n",
- "print \"\\n(b) rotor speed is \",nr,\" rev/sec\"\n",
- "print \"\\n(c) frequency of the e.m.f. induced in the rotor bars is is \",fr,\" Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) synchronous speed is 60.0 rev/sec\n",
- "\n",
- "(b) rotor speed is 58.8 rev/sec\n",
- "\n",
- "(c) frequency of the e.m.f. induced in the rotor bars is is 1.2 Hz"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 392</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the synchronous speed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "f = 50;# in Hz\n",
- "nr = 1200/60;# in rev/min\n",
- "s = 0.04;# slip\n",
- "\n",
- "#calculation:\n",
- " #the synchronous speed.\n",
- "ns = nr/(1 - s)\n",
- "nsrpm = ns*60\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n synchronous speed is \",nsrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " synchronous speed is 1250.0 rev/min"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 394</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the slip, and (b) the rotor speed.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 8/2;# number of pairs of poles\n",
- "f = 50;# in Hz\n",
- "fr = 3;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #fr = s*f\n",
- "s = (fr/f)\n",
- " #the rotor speed.\n",
- "nr = ns*(1 - s)\n",
- "nrrpm = nr*60\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) slip is \",s*100,\" percent\"\n",
- "print \"\\n (b) rotor speed is \",nrrpm,\" rev/min\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) slip is 6.0 percent\n",
- "\n",
- " (b) rotor speed is 705.0 rev/min"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 396</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the rotor copper loss, \n",
- "#(b) the total mechanical power developed by the rotor,\n",
- "#(c) the output power of the motor if friction and windage losses are 750 W, and \n",
- "#(d) the efficiency of the motor, neglecting rotor iron loss.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Psi = 32000;# in Watts\n",
- "Psl = 1200;# in Watts\n",
- "s = 0.05;# slip\n",
- "Pfl = 750;# in Watts\n",
- "\n",
- "#calculation:\n",
- " #Input power to rotor = stator input power -\u0006 stator losses\n",
- "Pi = Psi - Psl\n",
- " #slip = rotor copper loss/rotor input\n",
- "Pl = s*Pi\n",
- " #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n",
- "Pr = Pi - Pl\n",
- " #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n",
- "Po = Pr - Pfl\n",
- " #Efficiency of induction motor = (output power/input power)*100\n",
- "eff = (Po/Psi)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n",
- "print \"\\n(b) Total mechanical power developed by the rotor is \",Pr,\" W\"\n",
- "print \"\\n(c) Output power of motor is \",Po,\" Watt\"\n",
- "print \"\\n(d) efficiency of induction motor is \",round(eff,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) rotor copper loss is 1540.0 Watt\n",
- "\n",
- "(b) Total mechanical power developed by the rotor is 29260.0 W\n",
- "\n",
- "(c) Output power of motor is 28510.0 Watt\n",
- "\n",
- "(d) efficiency of induction motor is 89.09 percent"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 397</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the rotor copper loss, and (b) the efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Psi = 32000;# in Watts\n",
- "Psl = 1200;# in Watts\n",
- "Pfl = 750;# in Watts\n",
- "x = 0.35;\n",
- "\n",
- "#calculation:\n",
- " #The slip, s\n",
- "s = 1-x\n",
- " #Input power to rotor = stator input power - stator losses\n",
- "Pi = Psi - Psl\n",
- " #slip = rotor copper loss/rotor input\n",
- "Pl = s*Pi\n",
- " #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n",
- "Pr = Pi - Pl\n",
- " #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n",
- "Po = Pr - Pfl\n",
- " #Efficiency of induction motor = (output power/input power)*100\n",
- "eff = (Po/Psi)*100# in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n",
- "print \"\\n(b) efficiency of induction motor is \",round(eff,2),\" percent\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a) rotor copper loss is 20020.0 Watt\n",
- "\n",
- "(b) efficiency of induction motor is 31.34 percent"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 398</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the synchronous speed, (b) the slip, \n",
- "#(c) the full load torque, (d) the power output if mechanical losses amount to 770 W, \n",
- "#(e) the maximum torque, (f) the speed at which maximum torque occurs,\n",
- "#and (g) the starting torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "tr = 0.85;# turn ratio N2/N1\n",
- "Pl = 770;# in Watt\n",
- "m = 3;# no. of phases\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- " #Phase voltage, E1 = V/(3**0.5)\n",
- "E1 = V/(3**0.5)\n",
- " #Full load torque\n",
- "T = (m*(tr**2)/(2*math.pi*ns))*((s/100)*E1*E1*R2/(R2*R2 + (X2*(s/100))**2))\n",
- " #Output power, including friction losses\n",
- "Pm = 2*math.pi*nr*T\n",
- " #power output\n",
- "Po = Pm - Pl\n",
- " #Maximum torque occurs when R2 = Xr = 0.35 ohm\n",
- " #Slip \n",
- "sm = R2/X2\n",
- " #maximum torque, Tm \n",
- "Tm = (m*(tr**2)/(2*math.pi*ns))*(sm*E1*E1*R2/(R2*R2 + (X2*sm)**2))\n",
- " #speed at which maximum torque occurs\n",
- "nrm = ns*(1 - sm)\n",
- "nrmrpm = nrm*60\n",
- " #At the start, i.e., at standstill, slip, s=1\n",
- "ss = 1\n",
- " #starting torque\n",
- "Ts = (m*(tr**2)/(2*math.pi*ns))*(ss*E1*E1*R2/(R2*R2 + (X2*ss)**2))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)Synchronous speed is \",round(ns,2),\" rev/sec\"\n",
- "print \"\\n(b)Slip is \",round(s,2),\" percent\"\n",
- "print \"\\n(c)Full load torque is \",round(T,2),\" Nm\"\n",
- "print \"\\n(d)power output is \",round(Po,2),\"W\"\n",
- "print \"\\n(e)maximum torque is \",round(Tm,2),\" Nm\"\n",
- "print \"\\n(f)speed at which maximum torque occurs is \",round(nrmrpm,2),\"rev/min\"\n",
- "print \"\\n(g)starting torque is \",round(Ts,2),\" Nm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)Synchronous speed is 25.0 rev/sec\n",
- "\n",
- "(b)Slip is 4.0 percent\n",
- "\n",
- "(c)Full load torque is 78.05 Nm\n",
- "\n",
- "(d)power output is 10998.99 W\n",
- "\n",
- "(e)maximum torque is 113.17 Nm\n",
- "\n",
- "(f)speed at which maximum torque occurs is 1350.0 rev/min\n",
- "\n",
- "(g)starting torque is 22.41 Nm"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 400</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the induction motor in problem 10 at full load, \n",
- "#(a) the rotor current, (b) the rotor copper loss, and (c) the starting current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "tr = 0.85;# turn ratio N2/N1\n",
- "m = 3;# no. of phases\n",
- "\n",
- "#calculation:\n",
- " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n",
- " #p is the number of pairs of poles.\n",
- "ns = f/p\n",
- " #The slip, s\n",
- "s = ((ns - nr)/ns)*100# in percent\n",
- " #Phase voltage, E1 = V/(3**0.5)\n",
- "E1 = V/(3**0.5)\n",
- " #rotor current,\n",
- "Ir = (s/100)*E1*tr/((R2**2 + (X2*(s/100))**2)**0.5)\n",
- " #Rotor copper loss \n",
- "Pcl = m*R2*(Ir**2)\n",
- " #starting current,\n",
- "ss =1\n",
- "I2 = ss*tr*E1/((R2**2 + (X2*ss)**2)**0.5)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(a)rotor current is \",round(Ir,2),\" A\"\n",
- "print \"\\n(b)Total copper loss is \",round(Pcl,2),\" W\"\n",
- "print \"\\n(c)starting current is \",round(I2,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(a)rotor current is 21.61 A\n",
- "\n",
- "(b)Total copper loss is 490.37 W\n",
- "\n",
- "(c)starting current is 57.9 A"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 401</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the power input at full load, \n",
- "#(b) the efficiency of the motor at full load and \n",
- "#(c) the current taken from the supply at full load, if the motor runs at a power factor of 0.87 lagging.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "Psl = 650;# in Watt\n",
- "pf = 0.87;# power factor\n",
- "\n",
- "#calculation:\n",
- "Pm = 11770;# watts from part (d), Problem 22.10\n",
- "Pcl = 490.35;# watts, Rotor copper loss, from part (b), Problem 22.11\n",
- " #Stator input power\n",
- "P1 = Pm + Pcl + Psl\n",
- "Po = 11000# watts, Net power output, from part (d), Problem 22.10\n",
- " #efficiency = (output/input) *100\n",
- "eff = (Po/P1)*100# in percent\n",
- " #Power input, P1 = (3**0.5)*VL*IL*cos(phi)\u000e\n",
- " # pf = cos(phi)\n",
- " #supply current, IL\n",
- "I = P1/((3**0.5)*V*pf)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n(aStator input power is \",round(P1,2),\" W\"\n",
- "print \"\\n(b)efficiency is \",round(eff,2),\" percent\"\n",
- "print \"\\n(c)supply current is \",round(I,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "(aStator input power is 12910.35 W\n",
- "\n",
- "(b)efficiency is 85.2 percent\n",
- "\n",
- "(c)supply current is 20.64 A"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 401</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the resistance of the rotor winding required for maximum starting torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 415;# in Volts\n",
- "f = 50 ;# in Hz\n",
- "nr = 24;# in rev/sec\n",
- "p = 4/2;# no. of pole pairs\n",
- "R2 = 0.35;# in Ohms\n",
- "X2 = 3.5;# in Ohms\n",
- "\n",
- "#calculation:\n",
- " #At the moment of starting, slip, \n",
- "s = 1\n",
- " #Maximum torque occurs when rotor reactance equals rotor resistance\n",
- " #for maximum torque\n",
- "R2 = s*X2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nresistance of the rotor is \",R2,\" Ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "resistance of the rotor is 3.5 Ohm"
- ]
- }
- ],
- "prompt_number": 13
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |