{
"metadata": {
"name": "",
"signature": "sha256:3404f51fc9045816d89b5507acdfe9797796d47a0805fc064b3ee38cacd900d7"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 9 Chemical effects of current"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.1 Page no 285"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"m=1*10**-3 #kg\n",
"I=2 #A\n",
"z=3.3*10**-7 #kg/C\n",
"\n",
"#Calculation\n",
"t=m/(z*I)\n",
"\n",
"#Result\n",
"print\"Time required is\", round(t,1),\"s\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time required is 1515.2 s\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.2 Page no 285"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"m=0.15*10**-3 #Kg\n",
"z=3.3*10**-7 #Kg/C\n",
"t=900 #S\n",
"I1=0.6 #A\n",
"\n",
"#Calculation\n",
"I=m/(z*t)\n",
"I2=I-I1\n",
"\n",
"#Result\n",
"print\"Correction required for the ammeter reading is\", round(I2,1),\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Correction required for the ammeter reading is -0.1 A\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.3 Page no 285"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"t=0.002 #m\n",
"A=72 #cm**2\n",
"d=8.9 #g/cm**3\n",
"z=33*10**-5 #g/C\n",
"I=5 #A\n",
"\n",
"#Calculation\n",
"V=t*A\n",
"m=V*d\n",
"t1=m/(z*I)\n",
"\n",
"#Result\n",
"print\"Time required is\", round(t1,0),\"S\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time required is 777.0 S\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.4 Page no 286"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"m1=2 #g\n",
"m2=1 #g\n",
"t=1800 #s\n",
"z1=1118*10**-6 \n",
"z2=3294*10**-7 \n",
"a=6\n",
"\n",
"#Calculation\n",
"l1=m1/(z1*t)\n",
"l2=m2/(z2*t)\n",
"l=l1+l2\n",
"p=l*a\n",
"\n",
"#Result\n",
"print\"Power of the current is\",round(p,3),\"W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Power of the current is 16.082 W\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.5 Page no 286"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"m1=0.403*10**-3 #Kg\n",
"z1=1.12*10**-6\n",
"z2=3.3*10**-7\n",
"t=900 #s\n",
"e=12 #V\n",
"\n",
"#Calculation\n",
"m2=(m1*z2)/z1\n",
"d=(e*m1)/z1\n",
"\n",
"#Result\n",
"print\"(a) Mass of the copper deposited is\",round(m2*10**3,3),\"10**-3\"\n",
"print\"(b) The energy supplied by the battery is\",round(d*10**-3,2),\"10**3\",\"J\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Mass of the copper deposited is 0.119 10**-3\n",
"(b) The energy supplied by the battery is 4.32 10**3 J\n"
]
}
],
"prompt_number": 51
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.6 Page no 286"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"r=100 #ohm\n",
"t=600 #s\n",
"z=3.3*10**-7\n",
"m=10**-4\n",
"\n",
"#Calculation\n",
"I=m/(z*t)\n",
"Q=I**2*r*t\n",
"\n",
"#Result\n",
"print\"Heat produced in the resistance coil is\",round(Q,1),\"J\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat produced in the resistance coil is 15304.6 J\n"
]
}
],
"prompt_number": 64
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.7 Page no 286"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"a=1.008\n",
"v=1\n",
"m1=1.05*10**-8\n",
"a1=63.54\n",
"v1=2\n",
"m2=3.29*10**-7\n",
"a2=107.9\n",
"v2=1\n",
"m3=1.12*10**-6\n",
"a3=55.85\n",
"v3=3\n",
"\n",
"#Calculation\n",
"#For water voltameter\n",
"E=a/v\n",
"s=m1/E\n",
"\n",
"#For copper voltameter\n",
"E2=a1/v1\n",
"s1=m2/E2\n",
"\n",
"#For silver voltameter\n",
"E3=a2/v2\n",
"s2=m3/E3\n",
"\n",
"#For iron voltameter\n",
"E4=a3/v3\n",
"m4=(s/a)*E4\n",
"\n",
"#Result\n",
"print\"Mass of hydrogen liberated is\",round(s*10**8,4)*10**-8\n",
"print\"Mass of copper deposited is\",round(s1*10**8,4)*10**-8\n",
"print\"Mass of silver deposited is\",round(s2*10**8,4)*10**-8\n",
"print\"Mass of iron deposited is\", round(m4*10**7,2)*10**-7,\"Kg/C\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Mass of hydrogen liberated is 1.0417e-08\n",
"Mass of copper deposited is 1.0356e-08\n",
"Mass of silver deposited is 1.038e-08\n",
"Mass of iron deposited is 1.92e-07 Kg/C\n"
]
}
],
"prompt_number": 104
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.8 Page no 287"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"m=2.5 #g\n",
"I=10 #A\n",
"F=96500 #C/mol\n",
"m1=63.5\n",
"n=2.0\n",
"\n",
"#Calculation\n",
"E=m1/n\n",
"t=m*F/(E*I)\n",
"\n",
"#Result\n",
"print\"Time required is\",round(t,1),\"S\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time required is 759.8 S\n"
]
}
],
"prompt_number": 110
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.9 Page no 287"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"m=16.43 #g\n",
"t=4000 #s\n",
"F=96485 #C/mol\n",
"m1=63.54\n",
"n=2.0\n",
"I1=12.6 #A\n",
"\n",
"#Calculation\n",
"E=m1/n\n",
"I=m*F/(E*t)\n",
"I2=I1-I\n",
"\n",
"#Result\n",
"print\"Error in the ammeter reading is\", round(I2,3),\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Error in the ammeter reading is 0.126 A\n"
]
}
],
"prompt_number": 119
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.10 Page no 287"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"t=600 #S\n",
"m=5.92 #g\n",
"F=96500 #C/mol\n",
"V1=1.62 #V\n",
"V2=1.34\n",
"m1=63.5\n",
"n=2.0\n",
"\n",
"#Calculation\n",
"V=V1-V2\n",
"E=m1/n\n",
"I=m*F/(E*t)\n",
"R=V/I\n",
"\n",
"#Result\n",
"print\"Resistance of the voltmeter is\",round(R*10**3,2),\"m ohm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Resistance of the voltmeter is 9.34 m ohm\n"
]
}
],
"prompt_number": 123
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.11 Page no 287"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"E=8 #V\n",
"r=1 #ohm\n",
"R=15 #ohm\n",
"E1=120\n",
"t=300 #s\n",
"\n",
"#Calculation\n",
"I=(E1-E)/(R+r)\n",
"V=E+(I*r)\n",
"E12=E*I*t\n",
"\n",
"#Result\n",
"print\"(a) Current in the circuit is\", I,\"A\"\n",
"print\"(b) Terminal voltage across the battery is\",V,\"V\"\n",
"print\"(c) Chemical energy stored in the battery is\",E12,\"J\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Current in the circuit is 7 A\n",
"(b) Terminal voltage across the battery is 15 V\n",
"(c) Chemical energy stored in the battery is 16800 J\n"
]
}
],
"prompt_number": 133
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.12 Page no 288"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"I=10 #A\n",
"t=300 #S\n",
"m=2.016\n",
"n=2.0\n",
"n1=1.0\n",
"m1=1.008\n",
"F=96500\n",
"V=22.4\n",
"\n",
"#Calculation\n",
"q=I*t\n",
"M=m/n\n",
"M2=m1/n1\n",
"q1=F*m/m1\n",
"V1=V*q/q1\n",
"\n",
"#Result\n",
"print\"Volume of hydrogen is\",round(V1,4),\"litre\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Volume of hydrogen is 0.3482 litre\n"
]
}
],
"prompt_number": 136
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.13 Page no 288"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"w=107.9 #g/mol\n",
"r=2 #ohm\n",
"E=12 #V\n",
"V=10\n",
"F=96500\n",
"t=1800\n",
"\n",
"#Calculation\n",
"R=r*(V/(E-V))\n",
"I=E/(R+r)\n",
"E=w/I\n",
"z=E/F\n",
"m=z*I*t\n",
"\n",
"#Result\n",
"print\"Silver deposited at the cathode is\", round(m,2),\"g\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Silver deposited at the cathode is 2.01 g\n"
]
}
],
"prompt_number": 143
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.14 Page no 288"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"I=5.0 #A\n",
"a=0.5 #mole\n",
"n=1\n",
"F=96500\n",
"I1=10 #A\n",
"t1=9650*2\n",
"n1=2.0\n",
"m=63.54\n",
"m2=55.85\n",
"n2=3.0\n",
"\n",
"#Calculation\n",
"q=F*a\n",
"t=q/I\n",
"E=m/n1\n",
"m1=(E*I1*t1)/F\n",
"E3=m2/n2\n",
"m3=(E3*I1*t1)/F\n",
"\n",
"#Result\n",
"print\"Molar mass of copper is\", m1,\"equal to its atomic mass i.e 1 mole of copper is liberated.\"\n",
"print\"Molar mass of iron is\",round(m3,3),\"Hence 2/3 mole of iron will be deposited.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Molar mass of copper is 63.54 equal to its atomic mass i.e 1 mole of copper is liberated.\n",
"Molar mass of iron is 37.233 Hence 2/3 mole of iron will be deposited.\n"
]
}
],
"prompt_number": 153
}
],
"metadata": {}
}
]
}