diff options
Diffstat (limited to 'Engineering_Circuit_Analysis_by_W_Hayt/13-Magnetically_coupled_circuits.ipynb')
| -rw-r--r-- | Engineering_Circuit_Analysis_by_W_Hayt/13-Magnetically_coupled_circuits.ipynb | 203 |
1 files changed, 203 insertions, 0 deletions
diff --git a/Engineering_Circuit_Analysis_by_W_Hayt/13-Magnetically_coupled_circuits.ipynb b/Engineering_Circuit_Analysis_by_W_Hayt/13-Magnetically_coupled_circuits.ipynb new file mode 100644 index 0000000..0ed6937 --- /dev/null +++ b/Engineering_Circuit_Analysis_by_W_Hayt/13-Magnetically_coupled_circuits.ipynb @@ -0,0 +1,203 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13: Magnetically coupled circuits" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.2: Mutual_Inductance.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clc\n", +"//Example 13.2\n", +"disp('Given')\n", +"disp('Input voltage is 10V')\n", +"Viamp=10\n", +"//From figure 13.7\n", +"//Writing the left mesh equations\n", +"disp('(1+10i)*I1-90i*I2=10')\n", +"//Writing the right mesh equations \n", +"disp('(400+1000i)*I2-90i*I1=0')\n", +"i=%i\n", +"A=[1+10*i -90*i;-90*i 400+1000*i]\n", +"i2mat=[1+10*i 10; -90*i 0] \n", +"//Find i2 \n", +"i2=det(i2mat)/det(A)\n", +"[mag Theta]=polar(i2)\n", +"Theta=(Theta*180)/%pi\n", +"//The value of resistor is 400 ohm\n", +"R=400;\n", +"//Let V=V2/V1\n", +"Vamp=R*mag/Viamp\n", +"printf('Ratio of output voltage to input is %3.2f with angle %3.2f degrees',Vamp,Theta);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.4: Energy_considerations.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clc\n", +"//Example 13.4\n", +"disp('Given')\n", +"disp('L1=0.4H L2=2.5H k=0.6 i1=4i2=20*cos(500t-20)mA')\n", +"L1=0.4;L2=2.5;k=0.6;\n", +"disp('a)')\n", +"t=0;\n", +"i2=5*cos(500*t-(20*%pi)/180)\n", +"printf('i2(0)=%3.2f mA \n',i2)\n", +"disp('b)')\n", +"M=k*sqrt(L1*L2)\n", +"//v1(t)=L1*d/dt(i1)+M*d/dt(i2)\n", +"v1=-L1*20*500*10^-3*sin(500*t-(20*%pi)/180)-M*5*500*10^-3*sin(500*t-(20*%pi)/180)\n", +"printf('v1(0)=%3.2f V \n',v1)\n", +"disp('c')\n", +"//The total energy can be found as\n", +"w=(L1*(4*i2)^2)/2+ (L2*(i2)^2)/2+M*(4*i2)*(i2)\n", +"printf('w=%3.2f uJ \n',w)\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.5: T_and_PI_equivalent_networks.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clc\n", +"//Example 13.5\n", +"printf('Given')\n", +"disp('L1=30 mH L2=60 mH M=40 mH')\n", +"L1=30*10^-3; L2=60*10^-3; M=40*10^-3;\n", +"//The equivalent T network is \n", +"UL=L1-M\n", +"UR=L2-M\n", +"CS=M\n", +"printf('The T network has \n')\n", +"printf('%d mH in the upper left arm\n',UL*10^3)\n", +"printf('%3.0f mH in the upper right arm\n',UR*10^3)\n", +"printf('%d mH in the center stem\n',CS*10^3)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.6: T_and_PI_equivalent_networks.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clc\n", +"//Example 13.6\n", +"printf('Given')\n", +"disp('L1=30 mH L2=60 mH M=40 mH')\n", +"L1=30*10^-3; L2=60*10^-3; M=40*10^-3;\n", +"//Let X=L1*L2-M^2\n", +"X=L1*L2-M^2\n", +"//The equivalent PI network is \n", +"LA=X/(L2-M)\n", +"LB=X/M\n", +"LC=X/(L1-M)\n", +"printf('The PI network has \n')\n", +"printf('LA=%3.0f mH\n',LA*10^3)\n", +"printf('LB=%3.0f mH \n',LB*10^3)\n", +"printf('LC=%3.0f mH\n',LC*10^3)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.8: Equivalent_Circuits.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clc\n", +"//Example 13.8\n", +"disp('Given')\n", +"disp('Vin=50V Zg=100 ohm')\n", +"Vin=50;Zg=100;\n", +"//From figure 13.32\n", +"disp('When the secondary circuit and ideal transformer is replaced by a Thevenin equivalent then the primary circuit sees a 100 ohm impedance')\n", +"//The turns ratio is a\n", +"a=10;\n", +"disp('We place the secondary circuit and ideal transformer by a Thevenin equivalent circuit')\n", +"Vth=-a*Vin\n", +"Zth=(-a)^2*Zg\n", +"printf('The secondary circuit has voltage source %d V rms with %d kohm resistance in series with it along with %d kohm load resistance',Vth,Zth*10^-3,10)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |