diff options
Diffstat (limited to 'Electronic_Principles_/Chapter_1_New.ipynb')
| -rw-r--r-- | Electronic_Principles_/Chapter_1_New.ipynb | 31 |
1 files changed, 0 insertions, 31 deletions
diff --git a/Electronic_Principles_/Chapter_1_New.ipynb b/Electronic_Principles_/Chapter_1_New.ipynb index e0a4b799..5245d442 100644 --- a/Electronic_Principles_/Chapter_1_New.ipynb +++ b/Electronic_Principles_/Chapter_1_New.ipynb @@ -27,16 +27,11 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#Example 1.1.py\n",
- "#Ac voltage source has Rs 50 ohms. For what load resistance is the source stiff?\n",
"\n",
- "#Variable declaration\n",
"Rs=50 #source resistance(Ohm)\n",
"\n",
- "#Calculation\n",
"RL=100*Rs/1000 #Minimum Load resistance(KOhm)\n",
"\n",
- "#Result\n",
"print 'Minimum Load resistance =',RL,'KOhm'"
],
"language": "python",
@@ -64,17 +59,12 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#Example 1.2.py\n",
- "#Current source of 2 mA has Rs of 10Mohm. Over what range of load resistance is the current source stiff?\n",
"\n",
- "#Variable declaration\n",
"Is=0.002 #Current source(A)\n",
"Rs=10000000 #internal resistance(Ohm)\n",
"\n",
- "#Calculation\n",
"Rl_Max=0.01*Rs/1000 #Maximum load resistance(KOhm)\n",
"\n",
- "#Result\n",
"print 'With 100:1 Rule, Maximum Load resistance =',Rl_Max,'KOhm'\n",
"print 'Stiff range for current source is load resistance from 0 KOhm to',Rl_Max,'KOhm'"
],
@@ -104,17 +94,12 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#Example 1.3.py\n",
- "#If a transistor is pumping 2 mA through RL of 10KOhm then find load voltage.\n",
"\n",
- "#Variable declaration\n",
"Il=0.002 #current source(A)\n",
"Rl=10000 #load resistance(Ohm)\n",
"\n",
- "#Calculation\n",
"Vl=Il*Rl #load voltage(V)\n",
"\n",
- "#Result\n",
"print 'Load Voltage =',Vl,'V'"
],
"language": "python",
@@ -142,24 +127,19 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#Example 1.4.py\n",
- "#What are the Thevenin voltage & resistance in Fig. 1-9a?\n",
"\n",
"print 'As per figure 1-9a, Calculate Thevenin resistor by opening load resistor'\n",
"\n",
- "#Variable declaration\n",
"Vs=72 #source voltage\n",
"R1=6 #Resistance (KOhm)\n",
"R2=3 #Resistance (KOhm)\n",
"R3=4 #Resistance (KOhm)\n",
"\n",
- "#Calculation\n",
"Ro=R1+R2 #Resistance (KOhm)\n",
"I=Vs/Ro\n",
"Vab=R2*Vs/(R1+R2) #Thevenin voltage(V)\n",
"Rth=((R1*R2)/(R1+R2))+R3 #Thevenin Resistance(KOhm)\n",
"\n",
- "#Result\n",
"print 'After removing Rl current flowing through 6KOhm is',I,'mA'\n",
"print 'Thevenin Voltage Vth =',Vab,'V'\n",
"print 'Calculating Thevenin Resistance with considering R1 & R2 in parallel'\n",
@@ -194,10 +174,7 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#Example 1.5.py\n",
- "#In circuit shown in figure 1-11a, measure thevenin voltage and resistance.\n",
"\n",
- "#Variable declaration\n",
"Vs=72 #source voltage\n",
"R1=2 #resistance1 (KOhm)\n",
"R2=2 #resistance1 (KOhm)\n",
@@ -208,12 +185,9 @@ "R7=0.5 #resistance1 (KOhm)\n",
"RL=1 #load resistance (KOhm)\n",
"\n",
- "#Calculation\n",
- "#from multimeter following values will be measured.\n",
"Vth=9 #Thevenin voltage(V)\n",
"Rth=1.5 #Thevenin resistance(KOhm)\n",
"\n",
- "#Result\n",
"print 'Thevenin Voltage Vth =',Vth,'V'\n",
"print 'Thevenin Resistance =',Rth,'KOhm'"
],
@@ -243,18 +217,13 @@ "cell_type": "code",
"collapsed": false,
"input": [
- "#Example 1.6.py\n",
- "#Convert thevenin circuit of figure 1-14a into norton circuit.\n",
"\n",
- "#Variable declaration\n",
"Vs=10 #source voltage\n",
"Rs=2 #series source resistance (KOhm)\n",
"\n",
- "#Calculation\n",
"IN=Vs/Rs #Norton current(mA)\n",
"Rp=Rs #parallel source resistance(KOhm)\n",
"\n",
- "#Result\n",
"print 'Norton current IN =',IN,'mA'\n",
"print 'Norton Resistance =',Rp,'KOhm'"
],
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