Course Plan ICT sem-1 BEC

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

Course plan

Subject : Basic Electric circuits(ICT 102) Teaching plan per week 3 lecture-periods Most lectures in tutorial mode Must bring calculator for every lecture Two weeks devoted to project

presentations


Text Book

INTRODUCTION TO ELECTRIC CIRCUITS 9th Edition Richard C. Dorf and James A. Svoboda Chapters 1 to 10 complete Parts of 11th chapter


Syllabus

Electric circuit variables; Circuit elements : Resistive circuits: Methods of analysis of resistive circuits: Circuit theorems : The operational amplifier : Energy storage elements : Response of RLC circuits : Sinusoidal steady state. Power calculations.


Evaluation

Attendance 5% less than 50 % : 0 : 50-59 ; 1: 60-69 ;2 70-79 ; 3: 80-89;4 : 90-100 : 5 Quizzes 20 % Two Quizzes Mid semester test 30% End semester test 30% Project 15 %


Project

Electrical wiring design (Home, theater,malletc) 8 Students in a team Sequence of work for project. Team building workshop A lecture on the project. Team compositions (1) Theory (2) Practice (3) Communication (4) Social skills


Project

Presentations as and when declared. Preparing the layout-design Report writing Presentation of final design & Submission of report


Teams for lab project

For lab two divisions of around 60 each Team of 4 for lab project Team of 2 for lab experiments


Chapter 1

Units Electric circuit variables Current Voltage Power Energy Problem solving Design and analysis

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Tab_1-3-1

International system of units (SI units)

SI base units


Standards of units of measurements

Measurements done at different places and times should be comparable

Standards are needed Early standards were physical objects that

defined the unit as one of the physical properties

Standard bars for kg and meter and a standard candle

Properties change with time Atomic standards are the best


Derived Units in SI


Examples

1 MHz = ? Hz 106 Hz 1 mF = ? F 10-6 Farad 1 KW = ? W 10 3 W

1 mH = ? H 10-3 H 1 volt = ? mV 106 mV 1 A = ? mA 103 mA


Example 1.3-1 SI units

A mass of 150 grams experiences a force of 100 newtons. Find the energy or work expended if the mass moves 10 cms. Also find the power if the mass completes its move in 1 millisecond


Solution to example 1.3-1

Energy = Force x distance = 100 x 0.1 = 10 joulesNote that the distance is in metersPower = Energy/Time period = 10/10-3

= 104 joules/second = 104 watts = 10 kwatts


Exercise 1.3-1

is largest ?

i1 = 45 mA : i2 = .03x103 = 30 mA i3 = 25x10-4x 106 = 2500 mA

i3 is the largest


Electrical circuit

Electrical elements connected together Used in Generation, transmission and

consumption of electric power and energy

Storage , transmission and processing of information

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Fig_1-2-1



Charge on electron

- 1.602x10-19 coulombs -1 coulomb is the charge on 6.24x1018

electrons Electrons flow from –ve terminal of

battery to +ve terminal Conventionally direction of current is

taken from +ve to –ve terminal


Charge and current calculations

calculatedbecanqgivenisiifSo

0tatchargetheisq(0)Where

qididq

havealsoWe

calculatedbecanigivenischargeIfdt

dqi

t t

)0(

:

0



Nature of current

Direct current ; this is a current of constant magnitude

Battery connected to resistors Time varying current Circuit connected to mains Charging of a capacitor Discharging of a capacitor


A direct current of magnitude I


Examples of time varying current


Exercise 1.2-1

Find the charge that has entered the element by time t when

i = 8t2-4t A Assume q(t) = 0 for t < 0


Solution to exercise 1.2-1


Exercise 1.2-2


Solution to Exercise 1.2-2


Voltage

In a battery electrons move from –ve terminal to +ve terminal.

To maintain the +ve charge on the terminal the electrons reaching it should be removed

This is done by chemical reaction in the battery

To take an electron away from the +ve terminal some work has to be done.


Analogy


Work done/Unit charge

Voltage is defined as the work done to move a +ve charge of 1 coulomb from negative terminal of battery to +ve terminal

Unit of energy (Work) is joule So voltage is joule/coulomb General equation for voltage is

dq

dwv



Power and Energy

A light bulb absorbs energy from source The rate of absorbing the energy is

called power

ondsintimeistand

joulesinenergyiswwattsinpowerispWheredt

dwp

sec

,


Significance of power and energy A light bulb is specified in terms of its

power A 300 watt bulb gives more light than

100 watt bulb Energy absorption will depend on power

and time both as it is obtained by integrating power over the given time period

Your electric bill depends on the energy consumed.


Power in terms of voltage and current

ivdt

dq

dq

dw

dt

dwp ..

Power absorption and Power Supplied

When current enters the + ve terminal, power is absorbed by the element

When the current leaves the + ve terminal the element supplies power

Power absorbed = - power supplied


Exercise 1.5-1

Figure E 1.5-1 shows four circuit elements identified by the letters A, B, C, and D. (a) Which of the devices supply 12 W? (b) Which of the devices absorb 12 W? (c) What is the value of the power received by device B? (d) What is the value of the power delivered by device B? (e) What is the value of the power delivered by device D?


Problem types

Two types Problems with unique solutions Problems with more than one solution


Example

Problem with unique solution A battery across a bulb Voltage is known Wattage of bulb known Find the resistance W = V2/R ; R = V2/W


Problem with non-unique solution A battery across a bulb Wattage rating is specified Find out the battery voltage and the

resistance Now we have a equation W = V2/R W is known but V and R are unknown One equation but two unknowns Assume one and calculate the other Many pairs of V and R will satisfy the

equation


Analysis problems

Sources and circuit is known Find the voltage or current through a

given element Unique solution

A simple problem in analysis

V1 = 10 voltsR1 = 10 kohmsR2 = 10 kohmsFind V2V2 = 5 volts

V2 = V1.R2/(R1+R2)


A more general problem

More complicated circuit Many simultaneous equations Equations solved to get the required

voltage or current


Design

One example Input and output is known Find the circuit

V1 = 1 voltV2 = 0.5 voltMany solutionsR1=R2= 1kohmR1=R2=10kohmR1 = R2 = 100 kohms


A more involved design

Output is often user defined Audio amplifier for PA system P0 = 30 watts Good quality Designer has to chose the Gain Frequency response IC or transistor


The problem solving method


Design cycle

1 Get specifications 2 Design 3 Analyze 4 Are specs satisfied? No yes 5 Implement 6 Measure 7 Are specs satisfied? No 8 Stop


P1.2-1


Solution to P1.2-1


P1.2-2


Solution to P1.2-2


P1.2-4


Solution to P1.2-4

P 1.2-5


Solution to 1.2-5


P 1.2-6


Solution to 1.2-6

P 1.3-4

Solution to P 1.3-4

P1.5-2

Solution to P1.5-2

P 1.5-4

Solution P 1.5-4

Solution to P1.5-4 Continued

P1.7-1

Solution to 1.7-1

Documents

Course Plan ICT sem-1 BEC