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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 5 : Parallel Circuits"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example No. 5_1 Page No. 149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The Current Resistor R1 = 0.015 Amps\n",
+ "i.e 15 mAmps\n",
+ "The Current Resistor R2 = 0.03 Amps\n",
+ "i.e 25 mAmps\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Solve for branch currents I1 and I2.\n",
+ "\n",
+ "R1 = 1.*10**3# # Resistor 1=1*10**3 Ohms\n",
+ "R2 = 600.# # Resistor 2=600 Ohms\n",
+ "Va = 15.# # Applied Voltage=15 Volts\n",
+ "\n",
+ "I1 = Va/R1#\n",
+ "print 'The Current Resistor R1 = %0.3f Amps'%I1\n",
+ "print 'i.e 15 mAmps'\n",
+ "\n",
+ "I2 = Va/R2#\n",
+ "print 'The Current Resistor R2 = %0.2f Amps'%I2\n",
+ "print 'i.e 25 mAmps'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example No. 5_2 Page No. 150"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The Total Current in the Mainline = 11 Amps\n"
+ ]
+ }
+ ],
+ "source": [
+ "# An R1 of 20 Ohms\u0002, an R2 of 40 Ohms\u0002, and an R3 of 60 Ohms\u0002 are connected in parallel across the 120-V power line. Using Kirchhoff’s current law, determine the total current It.\n",
+ "\n",
+ "# Given data\n",
+ "\n",
+ "R1 = 20.# # Resistor 1=20 Ohms\n",
+ "R2 = 40.# # Resistor 2=40 Ohms\n",
+ "R3 = 60.# # Resistor 3=60 Ohms\n",
+ "Va = 120.# # Applied Voltage=120 Volts\n",
+ "\n",
+ "I1 = Va/R1#\n",
+ "I2 = Va/R2#\n",
+ "I3 = Va/R3#\n",
+ "\n",
+ "It = I1+I2+I3\n",
+ "print 'The Total Current in the Mainline = %0.f Amps'%It"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example No. 5_3 Page No. 151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The Current in R2 branch = 5 Amps\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Two branches R1 and R2 across the 120-V power line draw a total line current It of 15 A. The R1 branch takes 10 A. How much is the current I2 in the R2 branch?\n",
+ "\n",
+ "# Given data\n",
+ "\n",
+ "I1 = 10# # Current in R1 branch=10 Amps\n",
+ "It = 15# # Total Current=15 Amps\n",
+ "\n",
+ "I2 = It-I1#\n",
+ "print 'The Current in R2 branch = %0.f Amps'%I2"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example No. 5_4 Page No. 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The Total Current = 0.6008 Amps\n",
+ "i.e 600.8 mAmps\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Three parallel branch currents are 0.1 A, 500 mA, and 800 \u0002A. Using Kirchhoff’s current law, calculate It.\n",
+ "\n",
+ "\n",
+ "# Given data\n",
+ "\n",
+ "I1 = 0.1# # Branch Current 1=0.1 Amps\n",
+ "I2 = 0.5# # Branch Current 2=500m Amps\n",
+ "I3 = 800*10**-6# # Branch Current 3=800u Amps\n",
+ "\n",
+ "It = I1+I2+I3#\n",
+ "print 'The Total Current = %0.4f Amps'%It\n",
+ "print 'i.e 600.8 mAmps'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example No. 5_5 Page No. 153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The Equivalent Resistance Req = 9 Ohms\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Two branches, each with a 5-A current, are connected across a 90-V source. How much is the equivalent resistance Req?\n",
+ "\n",
+ "# Given data\n",
+ "\n",
+ "I1 = 5# # Branch Current 1=5 Amps\n",
+ "I2 = 5# # Branch Current 2=5 Amps\n",
+ "Va = 90# # Applied Voltage=90 Volts\n",
+ "\n",
+ "It = I1+I2#\n",
+ "Req = Va/It#\n",
+ "print 'The Equivalent Resistance Req = %0.f Ohms'%Req"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example No. 5_6 Page No. 158"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The Value of Rx = 60.00 Ohms\n"
+ ]
+ }
+ ],
+ "source": [
+ "# What Rx in parallel with 40 Ohms\u0002 will provide an Req of 24 Ohms?\n",
+ "\n",
+ "# Given data\n",
+ "\n",
+ "R = 40.0# # Resistance=40 Ohms\n",
+ "Req = 24.0# # Equivqlent Resistance=24 Ohms\n",
+ "\n",
+ "Rx = (R*Req)/(R-Req)#\n",
+ "print 'The Value of Rx = %0.2f Ohms'%Rx"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example No. 5_7 Page No. 158"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The value of R = 50000 Ohms\n",
+ "i.e 50 kohms\n"
+ ]
+ }
+ ],
+ "source": [
+ "# What R in parallel with 50 kOhms will provide an Req of 25 kOhms\u0002\n",
+ "\n",
+ "# Given data\n",
+ "\n",
+ "R1 = 50*10**3# # R1=50k Ohms\n",
+ "Req = 25*10**3# # Req=25k Ohms\n",
+ "\n",
+ "R = (R1*Req)/(R1-Req)#\n",
+ "print 'The value of R = %0.f Ohms'%R\n",
+ "print 'i.e 50 kohms'"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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