{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 12: Direct Current Circuits"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.1: Current.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"R=[6,6,12]//Resistances from circuit diagram 12.34 on page no.192 in ohms\n",
"V=[5,2]//Voltage in V from circuit diagram 12.20 on page no.192\n",
"\n",
"//Calculations\n",
"Re=((R(2)*R(3))/(R(2)+R(3)))+R(1)//Equivalent resistance in ohms for 5V supply\n",
"I=V(1)/Re//Equivalent current in A for 5V supply\n",
"Ve=((R(2)*R(3))/(R(2)+R(3)))*I//Voltage across 5V supply in V\n",
"I1=(Ve/R(3))//Current in A\n",
"Re2=(1/((1/(R(1)))+(1/(R(2)))))+R(3)//Equivalent resistance in ohms for 2V supply\n",
"I2=V(2)/Re2//Equivalent current in A for 2V supply\n",
"Ix=I1-I2//Current through 12 ohm resistance in A\n",
"Iy=1/Ix//For displaying output in fraction\n",
"\n",
"//Output\n",
"printf('The current through %i ohm resistor is 1/%i A',R(3),Iy)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.2: Equivalent_circuit.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"R=[3,5,6,7]//Resistances from circuit diagram 12.36(a) on page no. 193 in ohms\n",
"V=12//Voltage in V from circuit diagram 12.36(a) on page no. 193\n",
"\n",
"//Calculations\n",
"Vth=(V*R(3))/(R(3)+R(4)+R(2))//Equivalent voltage in V\n",
"Rth=R(1)+(((R(2)+R(4))*R(3))/(R(2)+R(4)+R(3)))//Equivalent resistance in ohms\n",
"\n",
"//Output\n",
"printf('Thevenin equivalent resistance is %i ohms \n Thevenin equivalent voltage is %i V',Rth,Vth)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.3: Norton_equivalent.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Inut data\n",
"R=[2,3,4]//Resistances from circuit diagram 12.37(a) on page no.194 in ohms\n",
"V=5//Voltage in V from circuit diagram 12.37(a) on page no.194\n",
"\n",
"//Calculations\n",
"RN=((R(1)+R(2))*R(3))/(R(1)+R(2)+R(3))//Equivalent resistance in ohms\n",
"IN=V/(R(1)+R(2))//Equivalent current in A\n",
"\n",
"//Output\n",
"printf('Nortons equivalent resistance is %3.2f ohms \n Nortons equivalent current is %i A',RN,IN)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.4: Parameters.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"C=10*10^-6//Capicitance in F\n",
"R=10*10^3//Resistance in ohms\n",
"e=6//Emf of the battery in V\n",
"\n",
"//Calculations\n",
"t=C*R//Time constant in s\n",
"Qm=(C*e)/10^-6//Maximum charge in micro C\n",
"Im=(e/R)*1000//Maximum current in mA\n",
"\n",
"//Output\n",
"printf('Time constant of the circuit is %3.1f s \n Maximum charge on the capacitor is %i micro C \n Maximum current in the circuit is %3.1f mA \n Charge at time t is Q(t) = %i(1-exp(-t/%3.1f)) micro C \n Currrent at time t is I(t) = %3.1f exp(-t/%3.1f) mA',t,Qm,Im,Qm,t,Im,t)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.5: Time_constant.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"L=50//Inductance in mH\n",
"R=5//Resistance in ohms\n",
"V=6//Volatage of the battery in V\n",
"t=5//Time in ms\n",
"\n",
"//Calculations\n",
"t1=(L/R)//Time constant in ms\n",
"I=(V/R)*(1-exp(-t/t1))//Current in A\n",
"\n",
"//Output\n",
"printf('The time constant of the circuit is %i ms \n The current in the circuit is %3.2f A',t1,I)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.6: Parameters.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"L=6//Inductance in mH\n",
"C=12//Capacitance in pF\n",
"V=6//Voltage of the battery in V\n",
"\n",
"//Calculations\n",
"f=(1/(2*3.14*sqrt(L*10^-3*C*10^-12)))/10^5//Frequency of oscillation in Hz*10^5\n",
"Qm=(C*10^-12*V)/10^-12//Maximum charge in C *10^-12\n",
"Im=(2*3.14*f*10^5*Qm*10^-12)/10^-6//Maximum current in micro A\n",
"\n",
"//Output\n",
"printf('Frequency of oscillation is %3.2f *10^5 Hz \n The maximum value of charge on capacitor is %i *10^-12 C \n The current in the circuit is %i micro A',f,Qm,Im)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.e_1: Current.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"V=10//voltage in V from fig.12.7 on page no.175\n",
"R=10//Resistance in ohms from fig.12.7 on page no.175\n",
"\n",
"//Calculations\n",
"I=(V/R)//Current in A\n",
"\n",
"//Output\n",
"printf('Current in the circuit shown in fig.12.7 is %i A',I)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.e_2: Current.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"R=[6,6,3]//Resistances in the circuit from circuit diagram 12.9 on page no. 175 in ohms\n",
"V=[24,16]//Voltages in the circuit from circuit diagram 12.9 on page no. 175 in V\n",
"\n",
"//Calculations\n",
"Re1=1/((1/R(2))+(1/R(3)))//Equivalent resistance for parallel combination in ohms\n",
"Re=R(1)+Re1//Equivalent resistance of the ciriuit in ohms\n",
"I1=(V(1)/Re)//Current across the resistors in A\n",
"pd=(I1*Re1)//Potential difference across A and B from circuit diagram 12.9 on page no. 175 in V\n",
"I2=(pd/R(3))//Current across 3 ohms resistance in A\n",
"I3=(V(2)/(R(1)+R(2)))//Current in A\n",
"I=I2+I3//Total current\n",
"\n",
"//Output\n",
"printf('The current shown in the circiut is %3.1f A',I)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.e_3: Thevenins_equivalent_circuit.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"R=[4,12,2,12]//Resistances from circuit diagram 12.12 on page no. 177 in ohms\n",
"V=12//Voltage in V from circuit diagram 12.12 on page no. 177\n",
"\n",
"//Calculations\n",
"Rth=((R(1)+R(3))*R(2))/(R(1)+R(3)+R(2))//Equivalent resistance in ohms\n",
"Vth=(V*R(2))/(R(1)+R(3)+R(2))//Equivalent voltage in V\n",
"I=(Vth/(Rth+R(4)))//Current in A\n",
"\n",
"//Output\n",
"printf('The current through the resistor is %3.1f A',I)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.e_4: Thevenins_equivalent_circuit.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"R=[2,3,6]//Resistances from circuit diagram 12.15 on page no. 178 in ohms\n",
"I=2//Current in A from circuit diagram 12.15 on page no. 178\n",
"\n",
"//Calculations\n",
"Rth=(R(2)+R(3))//Equivalent resistance in ohms\n",
"Vth=(R(3)*I)//Equivalent voltage in V\n",
"\n",
"//Output\n",
"printf('Thevenin equivalent resistance is %i ohms \n Thevenin equivalent voltage is %i V',Rth,Vth)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.e_5: Thevenins_equivalent_circuit.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"R=[12,8,3,6]//Resistances from circuit diagram 12.17 on page no.179 in ohms\n",
"V=12//Voltage in V from circuit diagram 12.17 on page no.179\n",
"\n",
"//Calculations\n",
"Rth=((R(3)*R(1))/(R(3)+R(1)))+((R(2)*R(4))/(R(2)+R(4)))//Equivalent resistance in ohms\n",
"Vth=2.74//Thevenin voltage taken from the circuit diagram 12.19(a) on page no.179 in V\n",
"\n",
"//Output\n",
"printf('Thevenin equivalent resistance is %3.2f ohms \n Thevenin equivalent voltage is %3.2f V',Rth,Vth)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.e_6: Nortons_equivalent_circuit.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Input data\n",
"R=[4,12,2,12]//Resistances from circuit diagram 12.20 on page no.180 in ohms\n",
"V=12//Voltage in V from circuit diagram 12.20 on page no.180\n",
"\n",
"//Calculations\n",
"RN=((R(1)+R(3))*R(2))/(R(1)+R(3)+R(2))//Equivalent resistance in ohms\n",
"IN=(V/(RN+R(3)))//Equivalent current in A\n",
"\n",
"//Output\n",
"printf('Nortons equivalent resistance is %i ohms \n Nortons equivalent current is %i A',RN,IN)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 12.e_7: Nortons_equivalent_circuit.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Inut data\n",
"R=[4,5,6]//Resistances from circuit diagram 12.22 on page no.181 in ohms\n",
"I=2//Current in A from circuit diagram 12.22 on page no.181\n",
"\n",
"//Calculations\n",
"RN=(R(1)+R(2)+R(3))//Equivalent resistance in ohms\n",
"IN=(R(1)*I)/RN//Equivalent curren in A\n",
"\n",
"//Output\n",
"printf('Nortons equivalent resistance is %i ohms \n Nortons equivalent current is %3.3f A',RN,IN)"
   ]
   }
],
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			 "text": "MetaKernel Magics",
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