E & EP OUTCOME 1 Yr.1

During this Unit of your course you will berequired to carryout a number ofinvestigations /experiments and answer anumber of questionnaires.

You must complete all of the work set out inthis document. Grades will be based oncontrolled assignments/investigations/phasetests which will be available after you havecompleted this work.

At all stages you are encouraged to producea good standard of presentation for yourwork. This will help you and us in the accurateassessment of your work and help you later ifany work requires additional input.

Note that this document is your basic notes,examples, work sheets etc. It is also, proofthat you have studied the material required topass this part of your NationalDiploma/Cert/Award.

You must not destroy, or lose thismaterial.

National Dip/Cert/Award E & EP

1

UNITSIn engineering we must consider, record and

use the correct units for all physical quantities.Some of these units are in common use, others willbe unusual and have very specialised uses.

TTeslaMag Flux densityWbWeberMagnetic FluxPaPascalPressure PAAmperesCurrent IVVoltsPotential Diff pdssecondsTime tm/s Metres / secondVelocity vmmetreDistance d NNewtonForce FFFaradCapacitance CohmResistance RSYMBOLUNITQUANTITY

Other unit/quantities can be added to this list as weprogress through the work, use this page forreference later, if you forget which unit or quantityyou should be using.


2

Some of the units described in the table are verysmall (e.g. a Pascal for pressure), or are verylarge in some applications (e.g. ampere formeasuring current in micro-electronics). To allowus to write and use these large values or smallvalues we can use pre-fixes on the units.

0.000000000001 X 10-12ppico0.000000001 X 10-9nnano

0.000001 X 10-6micro0.001 X 10-3mmilli0.01 X 10-2ccenti

1,000 X 103kkilo1,000,000 X 106MMega

1,000,000,000 X 109GGiga

1,000,000,000,000 X 1012TTeraDec. PointPowerSymbolPre-fix

These pre-fixes can be used with most units:

E.g. Milli-metre, Mega bytes, Giga hertz etc..

When using a calculator it should be noted thatusing the EXP button puts x10 into the calculator,you do not need to type x10.

E.g. 17mm is 17 x 10-3 m and is typedinto a calculator as :-

17 EXP or 10x or / 3


3

OUTCOME 1. Be able to use circuit theory todetermine voltage, current and resistance inDirect Current (DC) circuits:

Electrical energy is caused by a potentialdifference due to the surplus or lack of electrons inor on a material. This can result in the movementof these electrons through a material in an attemptto re-establish the equilibrium.

Terms used in electrical and electronic principles:

Electron Smallest component of charge.

Conductor A material that allows electrons to floweasily (e.g. Copper, Silver, Aluminium, almost all metals and Carbon which is a non-metallic material).

Insulator A material that does not allow the easyflow of electrons ( e.g. plastics, wood, rubber, almost all non-metals except for Carbon).

Potential Difference The attractive/repulsive force pushing electrons through a material. (Also, known as Voltage and measured in Volts).

Charge The quantity of electrons available/ stored. (Known as Quantity [Q] and measured in Coulombs [C])


4

Current Charge per second (Q/t) (Amperes [A] or [amps])

ResistanceHow difficult is it for electrons to pass through the conductor, measured Ohms[].

Resistance

All components used in an electrical circuit have aresistance, if they don't, then they belong to agroup of materials known as Superconductors,which are not in common use at the present. Wewill consider a number of components connectedtogether in two different circuits: Series andparallel.

Series Connections.

In series connections all electrons must passthrough all of the components.


5

Comp A Comp B

Comp CComp D

Comp E

Note only path around circuit is through each component

SERIES

Parallel Connections.

In parallel connections the electrons have anumber of paths that they can follow. In practicesome electrons go down each path depending onhow easy it is, i.e. low R- easy path, moreelectrons.

In practice most complex devices are wired withcomponents in series and in parallel. We willconsider these later but first we must considersome simple components (RESISTORS) incompletely series and completely parallel circuits.


6

Comp A

Comp B

Comp C

Electrons can go along any ofthe 3 paths and only pass through 1component

Resistance

One of the most commonly used components usedin electronics is the resistor. This componentcomes in many sizes for values of resistance R,and for the power that it can carry (e.g. ¼, ½, 1watt etc.). To allow us to tell the value of resistanceR we use a colour code painted on to thecomponents.

Colour codes.

Black 0 Violet 7Brown 1 Grey 8Red 2 White 9Orange 3Yellow 4 Gold 5%Green 5 Silver 10%Blue 6


7

1st digit 2nddigit

multipliertol band

Example.

What is the R values of the following combinationsof colour bands;-

1) RED RED RED2 2 00 = 2200W

2) BROWN BLACK BROWN1 0 0 = 100W

3) RED YELLOW GREEN2 4 00000 = 2400000W

PLEASE complete the following colour codes asabove:-

4) RED GREEN BLACK=

5) BLUE BROWN GREEN=

6) WHITE BLACK YELLOW=

7) BLACK RED BLACK=

8) BROWN BLACK BLACK=


8

Resistor addition.

It can be shown that for;

Resistors in Series,

RTotal R1 R2 RnResistors in Parallel,

1RTotal

1R1 1

R2 1

RnNOTE.

This calculation can be done using acalculator, you should keep and use yourcalculators instruction manual. Not all calculatorsare the same but to give you some idea of themethods you can use they will be shown in thistext.

MAKE SURE YOU CAN GET THE CORRECTANSWER USING YOUR CALCULATOR.

(note. [ ] are used to show a button on the calculator)

R1 1x or x1R2 1

x or x1 Rn 1x or x1 THEN 1

x or x1


9

USE OF MULTIMETERS AND RESISTORS IN SERIESAND PARALLEL.

MULTIMETERS

A number of different designs of multimeters exist,however, they all have common features. Thesefeatures allow different uses such as measuringVoltage, current and resistance over a number ofdifferent ranges. It should be noted that theyrequire skill and knowledge in their use, and willnot just do it by themselves.


10

DISPLAY

ON/OFF

SCALECONTROL

CONECTIONS

Features.

Display.This displays the value that is being measured

and care should be taken of the position of thedecimal point.

Connections.

A number of holes are provided for theconnection of the wires.

1) The COM (normal black) is used for the black ornegative connection wire, for all measurements.

2) The V/ohms connection is used for measuringVoltage or resistance (ohms).

3) A or I is used to measure Low currents (amps).

SCALES.

These are set either by a large knob or by aseries of push buttons. The scales may includeA.C & D.C, voltages currents and resistance.These must be selected correctly.

1) Consider A.C or D.C.

2) What is to be measured V, I or R.


11

e.g 10A A COM V/OHM

3) Value to be expected ( i.e. high/low Kilo or millietc.).

If the display shows a 1 on the left then the scalemay require to be moved up a level or two.

When recording values obtained care must betaken to also record the units e.g. mV, KW etc.

Otherwise the values have little or no meaning.

When using the meter; Voltages must bemeasured in parallel with the PSU or resistance.The current must be measured in series with thePSU.


12

PSU

R

V

A+ -Vs

Questions 1.

Complete the following questions using your notesand/or any source such as E & EP books etc.

1) Name 3 materials commonly used asconductors.

2) Name the group of materials commonly usedas insulators.

3) What is the unit of resistance.

4) In a series circuit containing 3 resistors, what isthe formula used to find the total resistance.

5) In a parallel circuit containing 2 resistors what isthe simplified formula that we can use to find thetotal resistance.


13

6) If another resister is added to the circuit in Q5,what formula must be used.

7) Is voltage measured in parallel or series.

8) What is the significance of a 1 on the left handside of the multimeters display.

9) How can a resistors value be checked beforeany measurements are taken.

10) What connections are made on a multimeter ifit is to be used for measuring a low current.( i.e.Where do the black and red wire go to).


14

Investigation 1.Resistors.

Construct the following circuit using the resistorvalues indicated (check all R values by their colourcode and using the meter). Use the multimeter tocheck the total resistance of the circuit and recordin the table provided, calculate the expected valueof total resistance and compare these values with

the measuredvalues.

This procedure must be carried out using 3different sets of resistors for each circuit.

Construct the above circuits and complete thetable on the next page.


15

R1 R2 R3Circuit 1

Circuit 2

R1

R2

R3

READINGS

NOTE

TURN TO NEXT PAGE FOR FORMULA

4701002203

1002202202

1001001001

Parallel

4701002203

1002202202

1001001001% Error

Measured total RCal total RR3R2R1Series


16

CALCULATIONS.

Series.

Rtotal R1 R2 R3PARALLEL.

1RTotal

1R1 1

R2 1

R3

% ERROR

%Error RtotalmeasuredRtotalcalculated

Rtotalmeasured 100

1

Questions.

1) Are all your % errors less then 10%.If not why?

2) Does the value of each R vary much from theexpected (colour code) value.

3) If we can say that the total R in a series circuit isalways greater then the largest R in the circuit.What can we say about the total R in a parallelcircuit.


17

Section 2 Ohm’s Law.

A number of Laws are used in the analysis ofelectrical circuit, the first we are going to considerin known as Ohm’s Law. Ohm’s Law concernsitself with the relationship between the Voltage,current and resistance of a circuit. It can be appliedto single components and complete circuits, it isthe fundamental law used in electrical circuitanalysis.

It is stated as:

V IRwhere,

V voltage VoltsI current ampsR Resistance


18

Investigation 2.Ohms Law.

Objective.To investigate the relationship between

voltage, current and resistance known as OHM'sLAW.

Equipment.Voltmeter, Ammeter, Prototyping board,

PSU, Varies connection leads and varies resistors.

Procedure.Connect up the circuit as shown below:-,

1) Connect up one resistor as shown and set thevoltage to zero.

2) Record the values of voltage and current(amps).

3) Alter the voltage on the PSU, one step or till 1vis flowing.

4) Record the values of voltage and current as inNo 2.


19

R

A

V

PSU

+ -Vs

5) Repeat this until approximately 12 volts hasbeen reached.

Meters

Check that they are set onto DC

Check that they are on the correct settings, i.e Voltage for measuring volts and Current (amps) for measuring the current.

The black wire is plugged into the COM socket.

The red wire is in the correct socket.

PSU

Do not switch the PSU on until all connections are correctly set up.

Do not allow any short circuits of the PSU connecting wires.

Turn the PSU down to minimum setting before switching on.

Resistors.

Record all the values using the colour codes.

Check all R values using the meters.

Do not touch the resistors if they startsmoking. Carefully remove resistors they maybe hot.


20

Take your time and record all the information carefully as you will have only one go at it.

Readings.

Record you result below take care to recordthem correctly and with unit (prefixes) if required.

109876543210

CURRENTVOLTSRESISTOR SIZE 100ohms

109876543210

CURRENTVOLTSRESISTOR SIZE 470 ohms


21

109876543210

CURRENTRESISTOR SIZE 1000ohms


22

Graphs

Plot the information tabulated to plot the followinggraphs on the following page:-,

Calculate thegradient of eachgraph.NOTE.I may be in milli-amps (mA).

Formula

Using the relationship:-,

V IR

gradientR verticalvhorizontalI

Compare the values calculated to the values of theresistors used.

% errorR value from graphR value Measured


23

V

I(amps)

vert

horizontal


24

V

I02468

10

100ohm resistor

20 60 100

mA


25

Questions 2.Use Ohm's Law to calculate the Voltage, currentand/or resistance in the following circuits:-,


26

R=100ohm

A

20v

1)

V

R=2

00

I=0.1

2)

A

A

4) Calculate the current flowing in a circuit if thevoltage source provides 12v, and two resistors of42 are connected in:-,

(b) parallel(a) series

5) Calculate the voltage required to allow a currentof 120mA to flow in a circuit which has a resistanceof 120 ohms.


27

R12v

I=0.3

3)

AA

6) A circuit is constructed using a number ofdifferent resistors wired in series and parallel (seediagram), calculate the current flowing if thevoltage source is set to: (a) 0v, (b) 6v, (c) 9v & (d)12v.

7) The current measured in a simple circuit was0.5 amps, calculate the voltage required if theresistance is given as 1.2k .


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V source A

RR R

R

Each R = 120

Semi-Conductors.

Semi-conductor material is as the name suggests amaterial that will conduct a bit, the ability for it toconduct can be controlled by processes known asdoping. This process involves the introduction ofimpurities into the base material. The common materialsused as semi-conductors are Silicon (Si) andGermanium (Ge). With the doping materials arsenic,phosphorous, antimony used to make n-type materialand aluminium, gallium, boron used to provide p-type.

The use of doping and so producing p- & n-typematerials allows us to produce components such as theDIODE. The diode is a simple component used in alarge number of devices, it can be considered as a oneway valve for electrons.

Let us consider the operation of what is known as thep-n junction:


29

++

++

++- -

--

--

p-type n-type

Dep

letio

n La

yer

B.S Symbol

Line on componentmatches the position of the line on the symbol

The p-n junction diode shown here can be used tocontrol the direction of flow of electrons within a circuit.This is facilitated by the doping of the semi-conductormaterial as either p or n type, the doping of the materialwill make charge carriers available within the diode.

These charge carriers are known as either holes inP-type material, or as electrons in N-type material.These two charge carrier can be used to carry chargethrough the diode but because the diode has a P-typeend and a N-type end it will only carry electrons in onedirection. This then gives rise to the one way valveanalogy used earlier.

The physics behind this which includes the chemistry ofthe materials, as the properties relies on the valence ofthe materials involved, can be found in any goodElectronics book such as Morris Yr. 1 etc.

We normally use the term BIAS to identify which wayaround a diode is within a circuit. We say that a diode isin forward bias when it allows current to flow, and inreverse bias when it is being used to stop the flow ofelectrons. This can be demonstrated by the circuitsbelow:

Note how the bulb only comes on when the diode is inforward bias.


30

DiodeBulb

Batt.+-+-O

FF

ON

REVERSE BIAS FORWARD BIAS

AAK K

Investigation 3.

P-N Junction Diode Characteristics.

Objective.

Measure the forward and reverse bias characteristics ofa number of diodes and plot the V & I on suitable axes.

Procedure.

Construct the following circuit using:

Suitable power supply (0 to 1.0 Volts DC).

Prototyping board (to hold/mount components).

2 X Multimeters (Voltmeter & Ammeter).

Resistor (approx. 50 ohms ).

Diodes (Silicon & Germanium).

Varies connecting wires.

Construct as:

Note. you can reverse the diode bias by just turning itaround in the circuit.


31

DIODE

A

V

Vsupply

-+

-+

Diode in Forward Bias

symbol

R

With the circuit constructed, (please note do not crocclip onto croc clips and/or onto plugs, you will findcorrect wires and connectors if you look properly)complete the READINGS table below by:

1) Set the PSU to zero volts.

2) Set the Voltmeter to read 0 to 2V DC.

3) Set the Ammeter to read 0 to 20 ma.

4) Record both the voltage and the Current in the tableprovided.

5) Slowly increase the voltage until approximately 0.1volts appears on the voltmeter (Note do not use themeters on the PSU).

6) Record the voltage shown (approx. 0.1V) and theAmmeters reading (note this may be zero) also the unitsused i.e. Milli/micro etc.

7) Increase the voltage on the meter in steps of 0.1Vrecording the voltage and current in the table provideduntil approximately 1.0 volts has been reached

(Note you may find that the PSU will not reach this leveland you may only get to 0.9V, just note this in yourtable)

The resistor may get hot take care whentouching it, if it catches fire do not touch it justswitch off your circuit and bring it to yourlecturer attention.


32

READINGS

I would suggest filling these in with a pencil so you can rub it out and startit again if you make a mistake.Silicon Diode

0.90.40.90.4

0.80.30.80.3

0.70.20.70.2

0.60.10.60.1

0.500.50IVIVIVIV

Reverse BiasForward Bias

Germanium Diode

0.90.40.90.4

0.80.30.80.3

0.70.20.70.2

0.60.10.60.1

0.500.50IVIVIVIV

Reverse BiasForward Bias

GRAPHS.

Using the values of voltage and current that youhave got, produce a graph of Voltage vs. Currentusing the axes given. Also use a good qualitypencil.


33

Mark your own scale on the Current (Vertical) axis.


34

FORWARDREVERSE

0.2

0.4

0.6

0.8

1.00

0.2

0.4

0.6

0.8

1.0

Discussion.

1) You should be able to see from your results/graphthat the diode will not conduct electricity when inreverse bias.

2) When in forward bias the diodes will conduct,however, their must be a forward bias voltage present toallow this to occur.

3) This forward bias voltage is approximately 0.6v for asilicon diode and is lower for a germanium diode.

4) The operation of a diode can be shown using theanalogy of a simple one way valve used for a hose pipe.Where the flow is kept in one direction by a ball valve,and a small pressure is required to open the valveagainst a spring to allow water to flow in the desireddirection.

E.g.

5) The resistance of the diode in forward bias isconstantly changing as the line curves up this change ismost noticeable between the, TURN-ON voltage, andapproximately 1 volt in forward bias.

It should also be noted that if the voltage in reverse biasis increased sufficiently then the diode will fail (normallydestroyed) and allow the current to flow.


35

NO Flow ball held in place by water pressure

Flow as water presssure pushes ball back against spring.

Applications of Diodes.

We can use diodes in a number of simple applications,such as for the protection of circuits from incorrectoperator installation of batteries.

In this case if a power supply such as a battery is put inthe wrong way around then the diode will stop thecurrent from flowing and so protect delicate electroniccomponents.

Another common application is rectification of AC to DCpower supplies.

AC-Alternation Current.

It is common for electricity to be supplied to theconsumer in the form of AC, this is due to the ability ofAC to be easily transformed, that is the ratio betweenthe voltage and the current supplied to be changedusing a simple transformer.

In the IOM the MEA generate electricity at about 600V,this is then transformed to 33,000V for long distancetransmission to areas such as Ramsey, Peel,Castletown etc. and 11,000 for local transmissionaround Douglas.

This high voltage allows the current to be kept relativelylow. The lower the current then the less wire lossesoccur that is heating of the atmosphere around thecables. The high voltage can then be reduced to11,000V for use in towns, and down to 240V for use inpeoples houses ( 380V in factories etc.).

We can reduce this in devices with the home usingpower adapters such as phone chargers, etc.


36

In these devices we may reduce the voltage to 3, 6, 9,12 volts etc. As well as reducing the voltage their areapplications that may require us to increase the voltagefor example a Microwave oven requires 50,000 voltswithin the magnetron. Another device is the CRT (TV orcomputer screen) where 15,000 volts is required to firethe electrons down the tube to the screen.

These changes in voltage are possible by using simpletransformers which are made from simple coils of wire,the big ones are known as SUBSTATIONS and can beseen near housing areas. These are very dangerousand are normally surrounded by high fences due to thelarge voltages causing arcing which an easily reachover a metre and tends to kill people.

DC-Direct Current.

DC power supplies are commonly produced bybatteries, rechargeable or disposable cells can be usedto produce the direct current, the polarity (+ & -)connections do not change, as they do in ac supplies.

Most semi-conductor components such as transistors(amplifier), Integrated Circuits (Silicon Chips), Computerlogic, Processors etc. require a DC supply.

We could supply this from chemical cells however it ismore sensible to use mains power due to cost andavailability. Also we need DC to charge batteries suchas laptop, phones etc.

To change AC to DC we must rectify AC to DC.


37

Power in simple DC circuits.

The power in a DC circuit can be calculated usingthe simple relationship:-

Power watts V Iwhere,

V Voltage.I Current amps

If we consider this relationship in conjunction withOhm’s Law then:-

V IR, then ,P IR IP I2R

This relationship can be at times of more use.

It should be noted that power is measured inelectrical, heat and mechanical devices using thesame UNIT (watt) and can do the same work in thesame time.

Power EnergyTime


38

Let us consider the power in the following circuit:-,

Applying ohm’s law:

V IRI V

RI 12

47 0.25ampsWe can not use the expression we have forpower:-,

P VIP 12 0.25P 3watts.

NOTE. This simple expression is not for use in ACcircuits, as these require the use of vector analysisfor their calculation. (power factor etc.)


39

R=47

12volts

A

Questions 3..1) Calculate the power in the following circuits:-,


40

68ohma)

24volts A

120ohmb)

V AI=0.5amp


41

12ohm 48ohmc)

A

Voltage

I=400mA

Voltage Across R & Current Split.

If we consider a circuit constructed using a numberof resistors in series,

We find that the current (amps) is constant allaround the circuit.

When we measure the voltage across the resistorsas shown, we find:-,

Vs V1 V2

also,V1 IR1 & V2 IR2

This is generalised as,

Vs V1 V2 Vn

andVR IR


42

R1 R2

V2V1

A

Vs=Voltage source

PSU

+ -Vs

If we consider a circuit containing two resistors inparallel supplied with a voltage:-,

We find that the current is split and the voltage isconstant across the resistors, this can bedescribed by the expressions:-,

Atotal A1 A2

VS V1 V2

A1 VS

R1& A2

VS

R2

This can be generalised as:-,

ATotal A1 A2 An

Vs V1 V2 Vn

An VS

Rn


43

R1

R2

A1

A2

V2

V1

A

PSU

+ -Vs

Investigation 4.Series

Let us consider the following circuit:-,

Build up the above circuit using :-,R1 220R2 470R3 680

Note that the voltmeter is moved around the circuitto measure the voltage across each resistor in turnand to measure the total voltage across the 3resistors. The meter can also be used to measurethe current (amps). Record the values in the tableprovided and calculate the unknown values.


44

R1 R2 R3

A

V1 V2 V3

VT

PSU

+ -Vs

Readings for Series

Calculated valuesof V1, V2, V3 & VT

680470220RcalI(Amps)VTV3V2V1RTR3R2R1

Calculations.

RT R1 R2 R3

VT V1 V2 V3

V IRcal

Therefore,Rcal

VTI


45

Parallel.

Let us now consider a parallel circuit:-,

Construct the above circuit.

We can measure the current at different placesaround the circuit by moving the meter around thecircuit.

Use resistors with the following values:

R1 470R2 680


46

R1

R2

A

A

A2

V1&2

1

T

RT

PSU

+ -Vs

Readings for parallel circuit.

Calculated values ofI1, I2 & IT

680470RTITI2I1V2V1R2R1

Calculations.

RT R1R2R1R2

IT I1 I2

V IR


47

Kirchhoff’s Laws.

Current Law:At any junction in an electric circuit the

total current flowing towards that junction is equalto the total current flowing away from the junction.

I 0

When we consider the above system:

I1 I2 I3 I4 I5

I1 I2 I3 I4 I5 0


48

I3

I4

I5

I1

I2

Voltage Law:In any closed loop in a network, the

algebraic sum of the voltage drops (i.e. Products ofcurrent and resistance) taken around the loop isequal to the resultant e.m.f. acting in that loop.

E1 E2 IR1 IR2 IR3

We can also consider these as the voltage dropsacross the resistors.


49

R1

R2R

3

E1

E2

Example 1.Use Kirchhoff’s Laws to determine the currents

flowing in each branch of the following network:

Procedure.1) Using the current law, we can label the currentdirections and magnitude:


50

R2=

1ohm

R1=

2ohm

s R=4

ohm

s

E2=2vE1=4v

R2=

1ohm

R1=

2ohm

s R=4

ohm

s

E2=2vE1=4v

(I + I )1 2

I1 I2

Loo

p 1

Loo

p 2

2) Divide the circuit into two loops and applyKirchhoff’s Voltage Laws to each loop, let usconsider Loop 1:

E1 I1R1 I1 I2R

4 I1 2 4 I1 I2

4 2I1 4I1 4I2

4 6I1 4I2 1Loop 2:

E2 I2R2 I1 I2R

2 1I2 4I1 I2

2 4I1 5I2 23) Solve equations (1) & (2) for I1 & I2:

4 6I1 4I2 4 2 4I1 5I2 6

16 24I1 16I2 12 24I1 30I2

12 24I1 30I2

4 0I1 14I2

I2 414 0.286amp


51

2 4I1 5I2 I2 0.286Amps

2 4I1 5 0.286

2 4I1 1.43

2 1.43 4I1

3.434 I1

I1 0.8575Amps

Therefore current flowing through the third resistorR:

I1 I2 0.8575 0.286

IR 0.5715Amps


52

Investigation 5.

Kirchhoff’s Law.

Construct the following circuit using the voltage(E1 & E2) and resistor values (R1, R2 & R3) listedin table 1. Record the value of current around thecircuit (A1, A2 & A3).

Note: Take care with the polarity of the PSUs, andthe connection of the meters.Construct the circuit using 3 meters and 2 powersupplies, as well as the required resistors. Recordyour values of current carefully.


53

E2E1

R1

R2

R3

A1 A2

A3

+ +

Using Kirchhoff’s Lawcalculate the current ineach of the branches (I1,I2 & I3) and record in thetable.

Calculate the % errorbetween the calculatedand the measured valuesof current using:-

%Error ICalIMeasICal

100

Record errors in table 1.


54

Questions 4.1) Use kirchhoff’s laws to calculate the currentflowing in each branch of the following circuit;-,


55

R=6

R=3

R=1

E=10vE=3

v

2) For the resistor network shown below,determine the currents in each of the ARMS (A1,A2 & A3):-,


56

A1 A2

A3

2010 30

20

10

8v5v

3) Use Kirchhoff’s laws to find the current flowingin the 6 ohm resistor in the following circuit and thepower dissipated in the 4 ohms resistor.


57

5 ohm

4 ohm

6 ohm40v

40v

4) Find the current flowing in the 3 ohm resistor forthe network shown below, also calculate the pd.across the 10 and 2 ohm resistors.

You have now reach an AssessmentPoint.

You should now check your work todate for Assessment 1.

(covering P1, 2, 3)


58

20v


59

Documents

E & EP OUTCOME 1 Yr.1