{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 17:DC Circuits"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.1:pg-866"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The number of electrons that pass through bulb is=\n",
"9.375e+17\n",
"electrons\n"
]
}
],
"source": [
"#Example 17_1\n",
" \n",
" \n",
"#To find number of electrons flow through bulb\n",
"current=0.15 #Units in C\n",
"q=1.6*10**-19 #Units in C/electron\n",
"noe=current/q #Units in number of Electrons\n",
"print \"The number of electrons that pass through bulb is=\"\n",
"print noe\n",
"print \"electrons\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.2:pg-867"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The resistance in bulb is= 19.4 Ohms\n"
]
}
],
"source": [
"#Example 17_2\n",
" \n",
" \n",
"#To find the resistance in bulb\n",
"v=1.55 #Units in V\n",
"i=0.08 #Units in A\n",
"r=v/i #Units in Ohms\n",
"print \"The resistance in bulb is=\",round(r,1),\" Ohms\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.3:pg-867"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The resistance in wire is= 0.205 Ohms\n"
]
}
],
"source": [
"#Example 17_3\n",
" \n",
" \n",
"#To find the resistance in wire\n",
"row=1.7*10**-8 #Units in Ohm meter\n",
"l=40 #Units in meters\n",
"a=0.0331*10**-4 #Units in meters**2\n",
"r=(row*l)/a #Units in Ohms\n",
"print \"The resistance in wire is=\",round(r,3),\" Ohms\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.4:pg-868"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The appropriate resistance in wire is Ro= 26.6 Ohms\n"
]
}
],
"source": [
"#Example 17_4\n",
" \n",
" \n",
"#To find the appropriate resistance of the wire\n",
"alpha=0.0045 #Units in Centigrade**-1\n",
"t=1780 #Units in Centigrade\n",
"deltaR=240 #Units in Ohms\n",
"ro=deltaR/(1+(alpha*t)) #Units in ohms\n",
"print \"The appropriate resistance in wire is Ro=\",round(ro,1),\" Ohms\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.5:pg-868"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Heat generated in bulb is= 48000.0 J\n"
]
}
],
"source": [
"#Example 17_5\n",
" \n",
" \n",
"#To find out the amount of heat developed in bulb\n",
"t=20.0*60 #Units in sec\n",
"pow=40.0 #Units in W\n",
"heat=t*pow #Units in J\n",
"print \"Heat generated in bulb is=\",round(heat),\" J\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.6:pg-869"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Cost needed to operate is= 0.035 Dollars\n"
]
}
],
"source": [
"#Example 17_6\n",
" \n",
" \n",
"#To calculate the cost needed to operate\n",
"power=0.7 #Units in KW\n",
"time=0.5 #Units in h\n",
"heat=power*time #Units in K Wh\n",
"cost=0.10 #Units in Dollars\n",
"tcost=cost*heat #Units in Dollars\n",
"print \"Cost needed to operate is=\",round(tcost,4),\" Dollars\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.7:pg-869"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The current in circuit is I= -0.82 A\n"
]
}
],
"source": [
"#Example 17_7\n",
" \n",
" \n",
"#To find the current in circuit\n",
"v1=3 #Units in V\n",
"v2=12.0#Units in V\n",
"r1=5.0 #Units in Ohms\n",
"r2=6 #Units in Ohms\n",
"i=(v1-v2)/(r1+r2) #Units in A\n",
"print \"The current in circuit is I=\",round(i,2),\" A\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.8:pg-870"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current in wire 1 is I1= 1.75 A\n",
"Current in wire 2 is I2= -0.5 A\n",
"Current in wire 3 is I3= 1.25 A\n",
"\n"
]
}
],
"source": [
"#Example 17_8\n",
" \n",
" \n",
"#To find the current in all wires\n",
"v=9 #Units in V\n",
"r1=18.0 #Units in Ohms\n",
"i2=-v/r1 #Units in A\n",
"v1=6.0 #Units in V\n",
"r2=12.0 #Units in Ohms\n",
"i3=(v+v1)/r2 #Units in A\n",
"i1=i3-i2 #Units in A\n",
"print \"Current in wire 1 is I1=\",round(i1,2),\" A\\nCurrent in wire 2 is I2=\",round(i2,2),\" A\\nCurrent in wire 3 is I3=\",round(i3,2),\" A\\n\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.9:pg-871"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The current I= 2.0 A\n"
]
}
],
"source": [
"#Example 17_9\n",
" \n",
" \n",
"#To find the current I in the battery\n",
"r1=3 #Units in Ohms\n",
"r2=6.0 #Units in Ohms\n",
"rbc=(r1*r2)/(r1+r2)#Units in Ohms\n",
"r3=4#Units in Ohms\n",
"rac=r3+rbc #Units in Ohms\n",
"v=12.0 #Units in V\n",
"i=v/rac #Units in A\n",
"print \"The current I=\",round(i),\" A\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.10:pg-872"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"ename": "SyntaxError",
"evalue": "invalid syntax (<ipython-input-10-e97062f38a8b>, line 17)",
"output_type": "error",
"traceback": [
"\u001b[1;36m File \u001b[1;32m\"<ipython-input-10-e97062f38a8b>\"\u001b[1;36m, line \u001b[1;32m17\u001b[0m\n\u001b[1;33m print \"The current in battery is I=\",round( ,2),\" A\",i)\u001b[0m\n\u001b[1;37m ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n"
]
}
],
"source": [
"#Example 17_10\n",
" \n",
" \n",
"#To find the current in battery\n",
"r1=3 #Units in Ohms\n",
"r2=6 #Units in Ohms\n",
"ra=(r1*r2)/(r1+r2)#Units in Ohms\n",
"r3=2 #Units in Ohms\n",
"r4=4 #Units in Ohms\n",
"rb=r3+r4 #Units in Ohms\n",
"r5=6 #Units in Ohms\n",
"rc=(r5*rb)/(r5+rb) #Units in Ohms\n",
"r6=9 #Units in Ohms\n",
"r=r6+rc #Units in Ohms\n",
"v=6 #Units in V\n",
"i=v/r #Units in Ohms\n",
"print \"The current in battery is I=\",round( ,2),\" A\",i)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.11:pg-872"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current in wire 1 is I1= 1.2 A\n",
"Current in wire 2 is I2= 0.6 A\n",
"Current in wire 3 is I3= 0.6 A\n",
"\n"
]
}
],
"source": [
"#Example 17_11\n",
" \n",
" \n",
"#To find the current in the wires\n",
"v1=12.0 #Units in V\n",
"r3=20.0 #Units in Ohms\n",
"v2=6 #Units in V\n",
"r2=10.0 #Units in Ohms\n",
"r1=5 #Units in Ohms\n",
"i3=((v1*r3)-(v2*r1))/((r2*r3)+(r1*r3)+(r1*r2)) #Units in A\n",
"i2=((r2*i3)+v2)/r3 #Units in A\n",
"i1=i3+i2 #Units in A\n",
"print \"Current in wire 1 is I1=\",round(i1,1),\" A\\nCurrent in wire 2 is I2=\",round(i2,1),\" A\\nCurrent in wire 3 is I3=\",round(i3,1),\" A\\n\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.12:pg-873"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current in wire 1 is I1= 0.667 A\n",
"Current in wire 2 is I2= 0.444 A\n",
"Current in wire 3 is I3= -0.222 A\n",
"\n"
]
}
],
"source": [
"#Example 17_12\n",
" \n",
" \n",
"#To find I1, I2 and I3 in the circuit\n",
"v1=40 #Units in V\n",
"r1=10.0 #Units in Ohms\n",
"r2=30.0 #Units in Ohms\n",
"v2=60.0 #Units in V\n",
"r3=15.0 #Units in Ohms\n",
"v3=50 #Units in V\n",
"i1=((-v1*r2)+(-r3*v1)+(60*r3)+(v3*r2))/((r1*r2)+(r2*r3)+(r3*r1)) #Units in A\n",
"i=2 #Units in A\n",
"i2=(i-i1)/3 #Units in A\n",
"i3=i2-i1 #Units in A\n",
"print \"Current in wire 1 is I1=\",round(i1,3),\" A\\nCurrent in wire 2 is I2=\",round(i2,3),\" A\\nCurrent in wire 3 is I3=\",round(i3,3),\" A\\n\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.13:pg-874"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" The current I= 2.5 A\n",
" Resistance is R= 20.0 Ohms\n",
" The value E is= 41.0 V\n"
]
}
],
"source": [
"#Example 17_13\n",
" \n",
" \n",
"#To find the values of e, R and I\n",
"i1=2 #Units in A\n",
"i2=0.5 #Units in A\n",
"i=i1+i2 #Units in A\n",
"v1=6 #Units in V\n",
"v2=16.0 #Units in V\n",
"r=-(v1-v2)/0.5 #Units in Ohms\n",
"v3=25.0 #Units in V\n",
"e=v2+v3 #Units in V\n",
"print \" The current I=\",round(i,1),\" A\\n Resistance is R=\",round(r),\" Ohms\\n The value E is=\",round(e),\" V\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.14:pg-874"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current in wire 1 is I1= 2.0 A\n",
"Current in wire 2 is I2= 4.0 A\n",
"Current in wire 3 is I3= 2.0 A\n",
"\n",
"The charge on the capacitor is q= 1e-05 C\n"
]
}
],
"source": [
"#Example 17_14\n",
" \n",
" \n",
"#To find the I1,I2,I3 values and charge on the capacitor\n",
"v1=12.0 #Units in V\n",
"r1=6.0 #Units in Ohms\n",
"i1=v1/r1 #Units in A\n",
"v2=4.0 #Units in V\n",
"r2=8.0 #Units in Ohms\n",
"i3=(v1+v2)/r2 #Units in A\n",
"i2=i1+i3 #Units in A\n",
"print \"Current in wire 1 is I1=\",round(i1),\" A\\nCurrent in wire 2 is I2=\",round(i2),\" A\\nCurrent in wire 3 is I3=\",round(i3),\" A\\n\"\n",
"v3=10.0 #Units in V\n",
"vfg=-v3+(r1*i1) #Units in V\n",
"c=5*10**-6 #Units in F\n",
"q=c*vfg #Units in C\n",
"print \"The charge on the capacitor is q=\",round(q,5),\" C\" \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.15:pg-874"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The Potential difference between d to c is= 23.8 V\n",
"\n",
"The potential difference between b to a is= 7.8 V\n"
]
}
],
"source": [
"#Example 17_15\n",
" \n",
" \n",
"#To find the terminal potential of each battery\n",
"v=18 #Units in V\n",
"r=9 #Units in Ohms\n",
"i=v/r #Units in A\n",
"r1=0.1 #Units in Ohms\n",
"v1=-i*r1 #Units in V\n",
"v2=24 #Units in V\n",
"v11=v1+v2 #Units in V\n",
"r2=0.9 #Units in Ohms\n",
"v3=i*r2 #Units in V\n",
"v4=6 #Units in V\n",
"v22=v3+v4 #Units in V \n",
"print \"The Potential difference between d to c is=\",round(v11,1),\" V\"\n",
"print \"\\nThe potential difference between b to a is=\",round(v22,1),\" V\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex17.16:pg-875"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The resistance of the recording device is= 1000000.0 Ohms\n"
]
}
],
"source": [
"#Example 17_16\n",
" \n",
" \n",
"#To findout how large a a resistance must the recording device must have\n",
"r1=10000.0 #Units in Ohms\n",
"percent=1.0 #Units in Percentage \n",
"vo=1/(r1*(percent*100)) #Units In terms of Ro\n",
"Ro=1/vo #Units in Ohms\n",
"print \"The resistance of the recording device is=\",round(Ro),\" Ohms\"\n"
]
}
],
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