Into -Electric Circuit- Pres

8/14/2019 Into -Electric Circuit- Pres

1/48

Electrotechnology

DC Circuit


2/48

DC CircuitSpecific Learning Objectives

Apply circuit law to solve problem in dc network-series, parallel, combined dc and parallel circuits

Solve problem to find power and efficiency inelectromechanical system and consideration requiredwhen apply the laws in network problems

Solve problem regarding resistance required to extentrange of ammeter and voltmeter and verify the

experiments


3/48

Electric Circuit.


4/48

Electric Circuit


5/48

Electric Circuit


6/48

Electric Circuit


7/48

Electric Circuit


8/48

Electric Circuit


9/48

Electric Circuit


10/48

Electric Circuit


11/48

DC Circuit


12/48

DC Circuit


13/48

DC CircuitOhms law

states that the current I flowing in a circuit is directly proportional to the appliedvoltage V and inversely proportional to the resistance R, provided thetemperature remains constant. Thus,

Electrical power

PowerPin an electrical circuit is given by the product of potential difference Vand current I. The unit of power is the watt, W. Hence

Electrical energy

Electrical energy = power time

If the power is measured in watts and the time in seconds then the unit of energy

is watt-seconds orjoules. If the power is measured in kilowatts and the time in hours then the unit of

energy is kilowatt-hours, often called the unit of electricity. The electricity meter in the home records the number of kilowatt-hours used

and is thus an energy meter.


14/48

DC CircuitMain effects of electric current

The three main effects of an electric current are:

Magnetic effect : bells, relays, motors,

generators, transformers, telephones, car-ignition

and lifting magnets.

Chemical effect : primary and secondary cells and

electroplating

Heating effect : cookers, water heaters, electricfires, irons, furnaces, kettles and soldering irons


15/48

DC Circuit Series circuits

Figure 5.1 shows three resistors R1,R2 and R3 connected end to end, i.e.,in series, with a battery source of Vvolts.

Since the circuit is closed a current I

will flow and the p.d. across eachresistor may be determined from thevoltmeter readings V1, V2 and V3

In a series circuit the current I is thesame in all parts of the circuit and

hence the same reading is found oneach of the two ammeters shown.

the sum of the voltages V1, V2 andV3 is equal to the total appliedvoltage

from Ohms Law :


16/48

DC Circuit

Potential divider

The voltage distribution for thecircuit shown in Figure 2.2(a) isgiven by:Figure 2.2 : dividervoltage - series

Source : Electrical andElectronic 8th Edition (2002)

The circuit shown in Figure 2.2(a)is often referred to as a potentialdivider circuit. Such a circuit canconsist of a number of similar

elements in series connectedacross a voltage source, voltages

being taken from connectionsbetween the elements.

Frequently the dividerconsists of two resistors asshown in Figure 2.2(a), where


17/48

DC CircuitParallel networks

Figure 2.3 shows three resistors, R1, R2and R3 connected across each other, i.e., in

parallel, across a battery source of V volts.

In a parallel circuit:

a) the sum of the currents I1, I2 and I3 isequal to the total circuit current, I, i.e.I=I1 +I2 +I3, and

b) the source p.d., V volts, is the sameacross each of the resistors.

From Ohms law:Figure 2.3 : Parallelcircuit

Source : Electrical and Electronic 8thEdition (2002)


18/48

DC Circuit

Current division

For the circuit shown in Figure

2.4, the total circuit resistance,

RT is given by: Figure 2.4 :Parallel circuit

Source : Electrical and

Electronic 8th Edition (2002)

2.6, E1 is positive and E2 is

negative.)


19/48

Light Circuit


20/48

DC Circuit


21/48

DC Circuit


22/48

DC Circuit


23/48

DC Circuit


24/48

DC Circuit


25/48

DC Circuit Kirchhoffs laws

Kirchhoffs laws state: Current Law. At any junction in

an electric circuit the total currentflowing towards that junction isequal to the total current flowing

away from the junction, Thus,referring to Figure 2.5:

or Voltage Law. In any closed loop

in a network, the algebraic sum ofthe voltage drops (i.e. products of

current and resistance) takenaround the loop is equal to theresultant e.m.f. acting in that loop.Thus, Referring to Figure 2.6:

Note that if current flows away from the positive terminal of a source,that source is considered by convention to be positive. Thus moving

anticlockwise around the loop of Figure


26/48

DC Circuit

QUESTIONS

Define the Ohms Law and KirchoffsLaw.

Three resistors A = 50 , B = 20 & C= 25 . Calculate the total resistance if

the resistors connected in series andparallel.

A three resistor R1 = 2 , R2 = 3 andR3 = 8 are connected in series. Findthe voltage across each of the resistorsand supply voltage if the current supplyis 1.5 A.

For the network shown below I1 = 2.5A, I2 = -1.5 A. Calculate the current I3.


27/48

DC Motor


28/48

RESISTORS IN SERIES

A simple SERIES CIRCUIT is shown in the

diagram below. The current (I) at every point

in a series circuit equals the current leaving

the battery.

I1= I2=I3=ITotal


29/48

RESISTORS IN SERIES

Assuming that the connecting wires offer no resistance tocurrent flow, the potential difference between the terminals

of the battery (V) equals the sum of the potential

differences across the resistors, i.e.,

V=Vl+ V2+ V3The equivalent electrical

resistance (R) for this

combination is equal to the

sum of the individualresistors, i.e.,

R=R1+ R2+ R3


30/48

RESISTORS IN PARALLEL

In a simple PARALLEL CIRCUIT, the current leaving the

battery divides at junction point A in the diagram shown

below and recombines at point B. The battery current (I)

equals the sum of the currents in the branches. In general

I = I1 + I2 + I3


31/48


The potential

difference across each

resistor in thearrangement is the

same, i. e.

V = VI = V2 = V3

If no other resistance is present, the potential difference

across each resistor equals the potential difference across

the terminals of the battery. The equivalent resistance (R) of a parallel combination

is always less than the smallest of the individual

resistors. The formula for the equivalent resistance is as

follows: 1/R = 1/RI + 1/R2 + 1/R3


32/48


In a simple PARALLEL CIRCUIT, the current leaving the

battery divides at junction point A in the diagram shown

below and recombines at point B. The battery current (I)

equals the sum of the currents in the branches. In general

I = I1 + I2 + I3


33/48

EMF AND TERMINAL

VOLTAGE All sources of emf have what is known as INTERNALRESISTANCE (r) to the flow of electric current. The internal

resistance of a fresh battery is usually small but increases

with use. Thus the voltage across the terminals of a battery is

less than the emf of the battery.

The TERMINALVOLTAGE (V) is given by the equation

V = - Ir, where represents the emf of the source of

potential in volts, I the current leaving the source of emf in

amperes and r the internal resistance in ohms.

The internal resistance of the source of emf is always

considered to be in a series with the external resistance

present in the electric circuit.


34/48

KIRCHHOFF'S RULES

KIRCHHOFF'S RULES are used in conjunction with Ohm's

law in solving problems involving complex circuits:

KIRCHHOFF'S FIRST RULE or JUNCTION RULE: The

sum of all currents entering any junction point equals the sum

of all currents leaving the junction point. This rule is based on

the law of conservation of electric charge.

KIRCHHOFF'S SECOND RULE or LOOP RULE: The

algebraic sum of all the gains and losses of potential around

any closed path must equal zero. This law is based on the lawof conservation of energy.


35/48

SUGGESTIONS FOR USING KIRCHHOFF'S

LAWS

1. Place a (+) sign next the long line of the battery symbol

and a (-) sign next to the short line. Start choosing a direction

for conventional current flow ( flow of positive charge )

If you choose the wrong direction for the flow of current in a

particular branch, your final answer for the current in that

branch will be negative. The negative answer indicates that

the current actually flows in the opposite direction.

I


36/48


LAWS

2. Assign a direction to the circuit in each independent branch

of the circuit. Place a positive sign on the side of each resistor

where the current enters and a negative sign on the side where

the current exits, e.g.; This indicates that a drop in potential

occurs as the current passes through the resistor .


37/48


LAWS

Notice how the

directions of the

currents are labeledin each branch of

the circuit


38/48


LAWS

3. Select a JUNCTION POINT and apply the

junction rule, e.g., at point A in the diagram:

The junction rule may be applied at more than onejunction point. In general, apply the junction rule

to enough junctions so that each branch current

appears in at least one equation.


39/48


LAWS

4. Apply Kirchhoffs loop rule by first taking notewhether there is a gain or loss of potential at each resistor

and source of emf as you trace the closed loop. Remember

that the sum of the gains and losses of potential must addto zero.


40/48


LAWS

For example, for the left loop of

the sample circuit above start

at point B and travel

clockwise around the loop.

Because the direction chosen

for the loop is also the

direction assigned for the

current, there is a gain in

potential across the battery(- to +), but a loss of

potential across each resistor

(+ to -).


41/48


LAWS

Following the path of the

current shown in the

diagram and using the

loop rule, the following

equation can be written:


42/48


LAWS

The direction taken around the

loop is ARBITRARY. Tracing

a counterclockwise path around

the circuit starting at B, as

shown in the above right

diagram, there is gain in

potential across each resistor

to (- to +) and a drop in

potential across the battery (+to -). The loop equation would

then be:


43/48


LAWS

Multiplying both sides of the above equation by - 1 and

algebraically rearranging, it can be shown that the two

equations are equivalent. Be sure to apply the loop rule to

enough closed loops so that each branch current appears in at

least one loop equation. Solve for each branch current using

standard algebraic methods.Solve

simultaneous

equations


44/48

CAPACITORS IN SERIES AND

PARALLEL

A circuit with

CAPACITORS IN

PARALLEL is shown in thediagram below. According to

Kirchhoff s loop rule, the

potential difference (V) of the

source of emf:

V = VI = V2 = V3


45/48

CAPACITORS IN PARALLEL

The total charge stored on the capacitor plates (Q) equals theamount of charge which left the source of:

Q = Ql + Q2 + Q3 ( Charge is additive)

and since Q = CV then CV = CV1 + CV2 + CV3

C= C1+ C2 +C3 (Capacitance is additive)


46/48

CAPACITORS IN SERIES

For CAPACITORS IN SERIES,

the amount of charge (Q) that

leaves the source of emf equals

the amount of charge that forms

on each capacitor:

Q = Ql = Q2 = Q3


47/48

CAPACITORS IN SERIES

From Kirchhoffs looprule, the potential

difference across the

source of emf (V) equals

the sum of the potentialdifferences across the

individual capacitors:


48/48

Circuits containing resistors and

capacitorsAn RC CIRCUIT consists of a resistor and a capacitor

connected in series to a de power source.When switch

1 (S1), shown in the diagram below, is closed, thecurrent will begin to flow from the source of emf and

charge will begin to accumulate on the capacitor.

Using Kirchhoff s loop rule it can be shown that

Into -Electric Circuit- Pres

Documents

Transcript of Into -Electric Circuit- Pres