ELEMENTS OF ELECTRICAL ENGINEERING

DC CIRCUITS

SARVAJANIK COLLEGE OF ENGINEERING AND TECHNOLOGY

GUJRAT TECHNOLOGICAL UNIVERSITY

PRESENTED BY : ROLL NO NAME 52 HARDIK LATIYA 53 DHVANI SACHVANI 54 ABHIJEET VAGHELA 57 RUSHABH THHAKAR 58 PRAYAG JARIWALA 59 RAJ BAAROT

GUIDED BY : PROF.SANTOSH RANGANI PROF.CHIRAG VADALIYA

BRANCH : COMPUTER ENGINEERING (SHIFT – II)

CIRCUIT ELEMENTS

1-3

• Resistors• Capacitors• Inductors

2

RESISTORS

1-4

Resist the flow of currentEnergy (voltage) is required to force current thru themthis “voltage energy” is transformed to heat

analogous to a valve like a faucet

Ohms’ Law: v = iR

2

CAPACITORS

1-5

Stores up and releases charge (energy) depending on the change in voltage

i = C dv/dt

i is the “displacement current”cuz the dielectric it does not

conduct current2

INDUCTORS ( )

1-6

Stores up and releases (voltage) energy depending on the change in current

v = L di/dt

2

Circuit Elements:Ideal independent voltage source

An ideal dependent voltage source is characterized as having aconstant voltage across its terminals, regardless of the load connected to the terminals.

The ideal voltage source can supply any amount of current. Furthermore, the ideal independent voltage source can supply anyamount of power.

The standard symbols of the ideal independent voltage source are shown below.

+_v(t) ESometimes used

Most often used

16

Circuit Elements:Ideal independent current sources

An ideal independent current source is characterized as providing a constant value of current, regardless of the load.

If the current source is truly ideal, it can provided any valueof voltage and any amount of power.

The standard symbol used for the ideal independent currentsource is shown below.

i(t)

17

1 m e g 1 a m p

+

-

V

V = ?

Circuit Elements:Dependent voltage sourceA dependent voltage source is characterized by depending ona voltage or current somewhere else in the circuit. The symbolFor the current source is shown below. Note the diamond shape.

A circuit containing a dependent voltage source is shown below.

1 0 2 0

3 0

1 2 +_5 V

Iy

1 0 Iy

A circuit with a currentcontrolled dependentvoltage source.

19

Circuit Elements:Dependent current sourceA dependent current source is characterized by depending ona voltage or current somewhere else in the circuit. The symbolfor a dependent current source is shown as follows:

A circuit containing a dependent current source is shown below.1 0 2 0

3 0

1 2 +5 V+

_v x4 v x_

A circuit with a voltage controlled dependent current source

20

RESISTORS IN SERIES AND PARALLEL

As with capacitors, resistors are often in series and parallel configurations in circuits

Series

Parallel

The question then is what is the equivalent resistance

RESISTORS IN SERIES

Since these resistors are in series, we have the same current in all three resistors

IIII 321

We also have that the sum of the potential differences across the three resistors must be the same as the potential difference between points a and b

ybxyaxab VVVV

RESISTORS IN SERIESThen using 321 ;; RIVRIVRIV ybxyax

We have that 321 RRRIVab

Now the equivalent resistor, R, will also have the same potential difference across it as Vab, and it will also have the same current I

RIVab

Equating these last two results, we then have that

i

iRRRRR 321

The equivalent resistance for a sequence of resistors in series is just the sum of the individual resistances

RESISTORS IN PARALLEL

Here we have that the voltage across each resistor has to be the same (work done in going from a to b is independent of the path, independent of which resistor you go through)

abVVVV 321

RESISTORS IN PARALLELWe now deal with currents through the resistors

At point a the current splits up into three distinct currents

We have that the sum of theses three currents must add to the value coming into this point

321 IIII

We also have that3

32

21

1 ;;RVI

RVI

RVI ababab

The equivalent resistor, R, will have also have the current I going through it

RESISTORS IN PARALLELUsing

RVI ab

and combining with the previous equations, we then have

321 RV

RV

RV

RV abababab

or

i iRRRRR11111

321

The inverse of the effective resistance is given by the sum of the inverses of the individual resistances

SOLVING RESISTOR NETWORKS

• Make a drawing of the resistor network• Determine whether the resistors are in

series or parallel or some combination• Determine what is being asked

• Equivalent resistance• Potential difference across a particular resistance • Current through a particular resistor

SOLVING RESISTOR NETWORKS

• Solve simplest parts of the network first• Then redraw network using the just

calculated effective resistance• Repeat calculating effective resistances

until only one effective resistance is left

SOLVING RESISTOR NETWORKS

Given the following circuit

What is the equivalent resistance and what is the current through each resistor

We see that we have two resistors in parallel with each other and the effective resistance of these two is in series with the remaining resistor

SOLVING RESISTOR NETWORKSCurrent through this effective resistor is given by

AmpsRVIeff

3618

The current through the resistors in the intermediate circuit of Step 1 is also 3 Amps with the voltage drop across the individual resistors being given by

VoltsV

VoltsV

623

;1243

2

4

Step 1: Combine the two resistors that are in parallel

yielding

Step 2: Combine the two resistors that are in series

624 effR yielding

SOLVING RESISTOR NETWORKS

2;21

31

611

effeff

RR

To find the current through the resistors of the parallel section of the initial circuit, we use the fact that both resistors have the same voltage drop – 6 Volts

AmpsVoltsI

AmpVoltsI

236

;166

3

6

SOLVING RESISTOR NETWORKS

KIRCHOFF’S RULES• Not all circuits are

reducible

There is no way to reduce the four resistors to one effective resistance or to combine the three voltage sources to one voltage source

KIRCHOFF’S RULESFirst some terminology

A junction, also called a node or branch point, is is a point where three or more conductors meet

A loop is any closed conducting path

KIRCHOFF’S RULESKirchoff’s Rules are basically two statements

1. The algebraic sum of the currents into any junction is zero 0I

A sign convention:

0321 III

A current heading towards a junction, is considered to be positive,

A current heading away from a junction, is considered to be negative

Be aware that all the junction equations for a circuit may not be independent of each other

KIRCHOFF’S RULES2. The algebraic sum of the potential differences in any loop including those associated with emfs and those of resistive elements must equal zero

0VProcedures to apply this rule:

Pick a direction for the current in each branchIf you picked the wrong direction, the current will come out negative

KIRCHOFF’S RULESPick a direction for traversing a loop – this direction must be the same for all loops

Note that there is a third loop along the outside branches

As with the junction equations not all the loop equations will be independent of each other.

KIRCHOFF’S RULESStarting at any point on the loop add the emfs and IR terms

An IR term is negative if we traverse it in the same sense as the current that is going through it, otherwise it is positive

An emf is considered to be positive if we go in the direction - to +, otherwise it is negative

Need to have as many independent equations as there are unknowns

KIRCHOFF’S RULES

For loop I we have 031432111 RIRIRI

For loop II we have 0324332 RIRI

Junction equation at a gives us 0321 III

We now have three equations for the three unknown currents

KIRCHOFF’S RULESAssume that the batteries are: 1 = 19 V; 2 = 6 V; 3 = 2 V

and the resistors are: R1 = 6; R2 = 4; R3 = 4; R4 = 1

you should end up with: I1 = 1.5 A; I2 = -0.5 A; I3 = 2.0 A

The minus sign on I2 indicates that the current is in fact in the opposite direction to that shown on the diagram

Complete details can be found here

STAR-DELTA TRANSFORMATION

EQUIVALENCE

•Equivalence can be found on the basis that the resistance between any pair of terminals in the two circuits have to be the same, when the third terminal is left open.

• First take delta connection: between A and C, there are two parallel paths, one having a resistance of R2 and other having a resistance of ( R1+R3)

Hence resistance between terminal A and C is = R2.(R1+R3)/[R2+( R1+R3)]

• Now take the star connection

The resistance between the same terminal A and C is (RA+RC)Since terminal resistance have to be same so we must have (RA+RC) = R2.(R1+R3)/[R2+( R1+R3)] (1)Similarly for terminals A and B, B and C, we can have the following expression (RA+RB) = R3.(R1+R2)/[R3+( R1+R2)] (2) (RB+RC) = R1.(R2+R3)/[R1+( R2+R3)] (3)

DELTA to STAR

Now subtracting 2 from 1 and adding the result to 3, we will get the following values for R1,R2 and R3.

How to remember?Resistance of each arm of star is given by the product of the resistance of the two delta sides that meet at its ends divided by the sum of the three delta resistance

STAR to DELTAMultiplying 1 and 2, 2 and 3 , 3 and 1 and adding them together and simplifying, we will have the following result.

How to remember: The equivalent delta resistance between any two point is given by the product of resistance taken two at a time divided by the opposite resistance in the star configuration.

PROBLEM• A delta-section of resistors is given in figure.

Convert this into an equivalent star-section.

.5.1;0.1;3:. CBA RRRAns

PROBLEM

The figure shows a network. The number on each branch represents the value of resistance in ohms. Find the resistance between the points E and F.

SOLUTION

Ans. : 5.6 Ω

PROBLEM• Find the current drawn from the 5 volt battery in the network

shown in figure.

SOLUTION :

Ans. : 0.974 A

Note :• During the network reduction or simplification process, some points in the original network are lost.•Care must be taken during this process that no point of ultimate relevance is lost.