1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
|
{
"metadata": {
"name": "J.B.Gupta Chapter 6"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": "Chapter 6 Extension of Instrument Range"
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 6.1,Page No 165"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math\n \n#variable declaration\nIm = 50*10**-6; #full scale deflection current in A\nRm = 1000; #instrument resistance in \u03a9\nI = 1; #total current to be measured in A\n \n#calculations\nRs = (Rm/float((I/float(Im))-1)); #resistance of ammeter in \u03a9\n \n \n#result\nprint'resistance of ammeter shunt required Rs = %3.7f'%Rs,'\u03a9';\n \n ",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "resistance of ammeter shunt required Rs = 0.0500025 \u03a9\n"
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 6.2,Page No 165"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math\n \n#variable declaration\nRm = 1; #instrument resistance in \u03a9\nRse = 4999; #series resistance in \u03a9\nV = 250; #full-scale deflection voltage in V\nRs = 0.002004; #Shunt resistance in \u03a9(Rs =1/float(499))\nI1 = 50; #full-scale defelction current in A\n \n#calculations\nIm = V/float(Rm+Rse); #full-scale deflection current in A\nx = 1/float(Rse);\nI = Im*(1+(Rm/float(x))); #current in A\nN = I1/float(Im); \nRsh = Rm/float(N-1); #shunt resistance in \u03a9\n \n#result\nprint'full-scale defelction current in Im = %3.2f'%Im,'A';\nprint'current range of instrument when used as an ammeter with coil connected across shunt is I = %3.2f'%I,'A';\nprint'Shunt resistance for the instrument to give a full-scale deflection of 50A is Rsh = %3.2e'%Rsh,'\u03a9';\n \n ",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "full-scale defelction current in Im = 0.05 A\ncurrent range of instrument when used as an ammeter with coil connected across shunt is I = 250.00 A\nShunt resistance for the instrument to give a full-scale deflection of 50A is Rsh = 1.00e-03 \u03a9\n"
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 6.3,Page No 167"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math\n \n#variable declaration\nRm = 10; #instrument resistance in \u03a9\nIm = 0.05; #instrument current in A\nV = 750; #voltage to be measured in V\nI = 100; #current to be measured in A\n \n#calculations\nR = (V/float(Im))-Rm; #External resistance in \u03a9\nN = I/float(Im); #power of shunt\nRs = Rm/float(N-1); #resistance in \u03a9\n \n \n#result\nprint'external resistance to be connected in sereis to enable the instrument to measure voltage upto 750V is %3.1f'%R,'\u03a9';\nprint'shunt resistance required %3.4f'%Rs,'\u03a9';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "external resistance to be connected in sereis to enable the instrument to measure voltage upto 750V is 14990.0 \u03a9\nshunt resistance required 0.0050 \u03a9\n"
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 6.4,Page No 167"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math\n \n#variable declaration\nRm = 5; #instrument resistance in \u03a9\nIm = 15*10**-3; #full scale defelection current in A\nI =1; #current to be measured in A\nV = 10; #voltage to be measured in V\n \n#calculations\nN = I/float(Im); #power of shunt\nRs = Rm/float(N-1); #resistance in \u03a9\nR = (V/float(Im))-Rm; #series resistance in \u03a9\n \n \n#result\nprint'resistance to be connected in parallel to enable the instrument to measure current upto 1A is %3.4f'%Rs,'\u03a9';\nprint'shunt resistance required for full-scale defelction with 10v is %3.4f'%R,'\u03a9';\n ",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "resistance to be connected in parallel to enable the instrument to measure current upto 1A is 0.0761 \u03a9\nshunt resistance required for full-scale defelction with 10v is 661.6667 \u03a9\n"
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 6.5,Page No 167"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math\n \n#variable declaration\nRm = 2; #instrument coil resistance in \u03a9\nV = 250; #full-scale reading in V\nRs = 5000; #series resistance in \u03a9\nRsh = 2*10**-3; #shunt resistance in \u03a9\n \n \n#calculations\nIm = V/float(Rm+Rs); #current flowing through the instrument for full-scale deflection in A\nIs = (Im*Rm)/float(Rsh); #current through shunt resistance in A \nI = Im+Is; #current range of instrument in A\n \n#result\nprint'current flowing through the instrument for full-scale deflection is %3.4f'%(Im*10**3),'mA';\nprint'current through shunt resistance is %3.2f'%Is,'A';\nprint'current range of instrumentis %3.1f'%I,'A'\n ",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "current flowing through the instrument for full-scale deflection is 49.9800 mA\ncurrent through shunt resistance is 49.98 A\ncurrent range of instrumentis 50.0 A\n"
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 6.6,Page No 168"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math\n \n#variable declaration\nRsh = 0.02; #shunt resistance in \u03a9\nV = 0.5; #potential difference across the shunt in V\nRm = 1000; #resistance in \u03a9\nI1 = 10; #current in A\nI2 = 75; #current in A\nI = 100;\t\t\t #current in A\nx = 40;\t\t\t #deflection %\n \n#calculations\nIs = V/float(Rsh); #current through shunt in A\nIm = V/float(Rm); #current through ammeter for full-scale defelction in A\nV1 = I1*Rsh; #voltage across shunt for 10A in V\nR1 = V1/float(Im);\t\t\t #resistance for the ammeter for a current of 10 A for full-scale defelction in \u03a9\nV2 = I2*Rsh; #voltage across shunt for 75A in V\nR2 = V2/float(Im);\t\t\t #resistance for the ammeter for a current of 75 A for full-scale defelction in \u03a9\nI3 = I*(100/float(x));\t\t\t #full-scale defelction current when 100 A gives 40% defelction\nV3 = I3*Rsh; #voltage across shunt for 250 A in V\nR3 = V3/float(Im);\t\t\t #resistance for the ammeter for a current of 250 A for full-scale defelction in \u03a9\n\n\n#result\nprint'current through ammeter for full-scale defelction is %3.1f'% (Im*10**3),' A'\nprint'Resistance for the ammeter for a current of 10 A for full-scale defelction is %3.2f'%R1,' \u03a9';\nprint'Resistance for the ammeter for a current of 75 A for full-scale defelction is %3.2f'%R2,' \u03a9';\nprint'Resistance for the ammeter for a current of 250 A for full-scale defelction is %3.2f'%R3,' \u03a9';\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "current through ammeter for full-scale defelction is 0.5 A\nResistance for the ammeter for a current of 10 A for full-scale defelction is 400.00 \u03a9\nResistance for the ammeter for a current of 75 A for full-scale defelction is 3000.00 \u03a9\nResistance for the ammeter for a current of 250 A for full-scale defelction is 10000.00 \u03a9\n"
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 6.7,Page No 168"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math\n\n#variable declaration\nB = 0.5;\t\t\t #flux density of the magnetic field in Wb/m**2\nN = 100;\t\t\t #number of turns in coil\nl = 0.04;\t\t #length in m\nr =0.03;\t\t\t #width in m\nTc = 120*10**-6;\t #controlling torque in N-m\nv = 1; \t\t\t #volts per division in V\nn = 100;\t\t\t #number of division on full-scale\nRm = 0;\n\n#calculations\nI = Tc/float(B*N*l*r);\t\t #current for full-scale deflection in A\nV = n*v;\t\t\t #full-scale reading of instrument in V\nR = (V/float(I))-Rm;\t\t\t #External resistance required to be put in series with the coil in \u03a9\n\n#result\nprint'current for full-scale deflection is %3.4f'%I,'A';\nprint'External resistance required to be put in series with the coil is %3.3f'%R,' \u03a9';\n\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "current for full-scale deflection is 0.0020 A\nExternal resistance required to be put in series with the coil is 50000.000 \u03a9\n"
}
],
"prompt_number": 48
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 6.8,Page No 169"
},
{
"cell_type": "code",
"collapsed": false,
"input": "import math\n\n\n#variable decalaration\nRm = 5;\t\t #coil resistance in \u03a9\nRm1 = 0.00075;\t\t #coil resistance in \u03a9\nIm = 0.015;\t\t #full-scale defelction current in A\nI = 100; \t\t #current to be measured in A\nT1 = 0.004;\t\t #temperature coeficient of copper in \u03a9/\u03a9/\u00b0C\nT2 = 0.00015;\t\t #temperature coeficient of manganin in \u03a9/\u03a9/\u00b0C\nT =10; \t\t #rise in temperature in \u00b0C\n#calculations\nN = I/float(Im);\t\t #multiplying power of shunt\nRs = Rm/float(N-1);\t #resistance of manganin shunt in \u03a9\nRc = Rm*(1+(T1*T));\t #coil resitance with 10\u00b0C in temperature in \u03a9\nRsh = Rm1*(1+(T2*T));\t #shunt resitance with 10\u00b0C in temperature in \u03a9\nIn = (Rsh/float(Rc+Rsh))*100;\t#new instrument current in A\nr = (In/float(Im))*100;\t\t#new instrument reading in A\ne = ((r-I)/float(I))*100;\t\t#percentage error in %\n\n\n#result\nprint'percentage error %3.5f'%e,'%';",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "percentage error -3.71583 %\n"
}
],
"prompt_number": 83
},
{
"cell_type": "code",
"collapsed": false,
"input": "",
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
