{
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 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 2: The circuit elements"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.1_a: Determine_the_current_flow_and_voltage_drop_across_the_resistor.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"V = 1; // voltage supply \n",
"R = 10; // resistance in ohms \n",
"I = V/R //current flowing through R\n",
"disp('a)')\n",
"disp(V,'voltage across the resistor (in volts)=')\n",
"disp(I,'current flowing through the resistor (in amps) =')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.1_b: Determine_the_current_flow_and_voltage_drop_across_each_resistor.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"V = 1; // voltage supply \n",
"R1 = 10; // first resistance in ohms \n",
"R2 = 5; //resistance of the second resistor \n",
"Vr1 = V * (R1/(R1 + R2)); //voltage across R1\n",
"Vr2 = V - Vr1; //voltage across R2\n",
"Ir = Vr1/R1; //current flowing through R\n",
"\n",
"disp(Vr1,'voltage across the first resistor (in volts)=')\n",
"disp(Vr2,'voltage across the second resistor (in volts)=')\n",
"disp(Ir,'current flowing through the resistor (in amps) =')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.1_c: Repeat_parts_A_and_B_with_the_voltage_source_replaced_by_a_current_source_of_1A.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//c - a\n",
"R1 = 10; // first resistance in ohms\n",
"I = 1; // current source \n",
"V = I*R1; // voltage across R\n",
"disp('c - a)')\n",
"disp(V,'voltage across the resistor (in volts)=')\n",
"disp(I,'current flowing through the resistor (in amps) =')\n",
"\n",
"//c - b\n",
"Vr1 = I*R1; // voltage across R1\n",
"Vr2 = I*R2; //voltage across R2\n",
"disp('c - b)')\n",
"disp(Vr,'voltage across the resistor (in volts)=')\n",
"disp(I,'current flowing through the resistor (in amps) =')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.2: EX2_2.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"R = 100; // resistance in ohms\n",
"I = 0.3; // current in amps \n",
"P = I^2 * R; // power \n",
"//power specification of the resistors available in the stock \n",
"Pa = 5;\n",
"Pb = 7.5;\n",
"Pc = 10;\n",
"\n",
"if Pa > P then\n",
"    disp('we should select resistor a')\n",
"end\n",
"if Pb > P then\n",
"    disp('we should select resistor b')\n",
"end\n",
"if Pc > P then\n",
"    disp('we should select resistor c')\n",
"end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.3: EX2_3.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"L = 1; //length of the copper wire in meters\n",
"A = 1 * 10^-4; // cross sectional area of the wire in meter square \n",
"rho = 1.724 * 10^-8; // resistivity of copper in ohm meter\n",
"R = rho*L / A; // resistance of the wire in ohm \n",
"\n",
"disp(R, 'resistance of the wire (in ohms)=') "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.4: EX2_4.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//1 inches = 0.0254meters\n",
"//1 foot = 0.3048 meters\n",
"d = 0.1*0.0254; // diameter of the wire in meters\n",
"L = 10*0.3048; //length of the wire in meters \n",
"rho = 1.724*10^-8; // resistivity of the wire in ohm-meter\n",
"A = %pi*(d/2)^2; // cross sectional area of the wire \n",
"R = rho*L/A; //resistance of the wire in ohm \n",
"disp(R,'resistance of the wire (in ohm)=')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.5: EX2_5.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"L = 0.1; // inductance of the coil in henry \n",
"t1 = [0:0.001:0.1];\n",
"t2 = [0.101:0.001:0.3];\n",
"t3 = [0.301:0.001:0.6];\n",
"t4 = [0.601:0.001:0.7];\n",
"t5 = [0.701:0.001:0.9]\n",
"//current variation as a function of time \n",
"i1 = 100*t1;\n",
"i2 = (-50*t2) + 15;\n",
"i3 = -100*sin(%pi*(t3-0.3)/0.3);\n",
"i4 = (100*t4) - 60;\n",
"i5 = (-50*t5) + 45;\n",
"\n",
"t = [t1,t2,t3,t4,t5];\n",
"i = [i1,i2,i3,i4,i5];\n",
"plot(t, i)\n",
"\n",
"dt = 0.001;\n",
"di = diff(i);\n",
"V = L*di/dt; //voltage drop appearing across the inductor terminals\n",
"Tv = [0:0.001:0.899];\n",
"plot(Tv, V, 'green')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.6: Find_the_time_variation_of_the_capacitor_voltage.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"C = 0.01; // capacitance of the capacitor in Farads\n",
"t1 = [0:0.001:0.1];\n",
"t2 = [0.101:0.001:0.3];\n",
"t3 = [0.301:0.001:0.6];\n",
"t4 = [0.601:0.001:0.7];\n",
"t5 = [0.701:0.001:0.9]\n",
"//current variation as a function of time \n",
"i1 = 100*t1;\n",
"i2 = (-50*t2) + 15;\n",
"i3 = -100*sin(%pi*(t3-0.3)/0.3);\n",
"i4 = (100*t4) - 60;\n",
"i5 = (-50*t5) + 45;\n",
"\n",
"t = [t1,t2,t3,t4,t5];\n",
"i = [i1,i2,i3,i4,i5];\n",
"plot(t, i)\n",
"\n",
"// voltage across the capacitor as a function of time \n",
"V1 = (1/C)*integrate('100*t','t',0,t1);\n",
"V2 = (1/C)*integrate('(-50*t)+15','t',0.101,t2);\n",
"V3 = (1/C)*integrate('-100*sin(%pi*(t-0.3)/0.3)','t',0.301,t3);\n",
"V4 = (1/C)*integrate('(100*t) - 60','t',0.601,t4);\n",
"V5 = (1/C)*integrate('(-50*t) + 45','t',0.701,t5);\n",
"V = [V1, V2, V3, V4, V5];\n",
"\n",
"plot(t, V, 'green')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.7: EX2_7.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//a\n",
"Ri = 1; \n",
"Rf = 39;\n",
"A = 10^5; //open loop gain of the op-amp\n",
"G = A/(1 + (A*Ri/(Ri+Rf))); //actual voltage gain of the circuit \n",
"disp('a')\n",
"disp(G,'actual voltage of the circuit =')\n",
"\n",
"//b\n",
"G1 = 1 + (Rf/Ri); // voltage gain of the circuit with infinite open loop gain\n",
"disp('b')\n",
"disp(G1,'for ideal case the voltage gain =')\n",
"\n",
"//c\n",
"er = ((G1 - G)/G)*100; //percent error \n",
"disp('c')\n",
"disp(er,'percent error of the ideal value compared to the actual value=')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.8: Design_a_non_inverting_opamp_circuit_of_voltage_gain_4.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"G = 4; // voltage gain of the circuit \n",
"r = G -1; // ratio of the resistances in the non-inverting op-amp circuit\n",
"disp(r,'Rf/Ri =')\n",
"//Result:\n",
"//A suitable choice for R1 is 10K, Hence Rf = 30K"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 2.9: Find_the_input_resistance_of_an_inverting_opamp_circuit_with_voltage_gain_of_4.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"G = 4;\n",
"r = G; // ratio of the resistances in the inverting op-amp circuit\n",
"disp(r,'Rf/Ri')\n",
"//Result;\n",
"//A suitable choice for Rf=30K and R1=7.5K\n",
"//therefore input resistance R1 = 7.5K"
   ]
   }
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