E & EP OUTCOME 4 Yr 1

Understand single-phase alternating current(ac) theory

Main parameters: Waveform, amplitude, period, frequency,

instantaneous, rms, average and peak values ofalternating waveforms (sinusoidal andnon-sinusoidal), form-factor, phase angles.

Single phase ac circuit theory: Phasor and algebraic representation of

alternating quantities, graphical and phasoraddition of two sinusoidal voltages. Introduction ofreactance and impedance from circuits containingpure R, L and C and including series combinationsof these elements. Power in an AC circuit,resonance in R, L and C circuits, Q-factor, powerfactor and power triangle.

Circuits: Use of an oscilloscope and multimeter to verify

the relationships between peak, rms and averagevalues in circuits such as half and full waverectification. Observation of the effect of asmoothing capacitor. Derive phasor diagrams fromvoltage measurements taken for all combinationsof series R, L and C circuits.

E & EP Unit 5 Outcome 4

This outcome deals with some of the concepts wecome across in the study and use of AC. We havemostly dealt with DC power until now, as this is farmore simple then AC power. However, AC is morecommon today and is used in sound productionand radio transmission, as well as the mainssupply from power stations.

So why do we use AC power?

AC electricity is easily produced using alternators,which are mechanical devices driven by enginessuch as; petrol/ diesel, heavy diesels, gas/steamturbines, hydro systems etc.AC electricity also has the ability to be easilytransformed, that is to have its voltage/currentratios changed.

If we plot an AC wave form then we find:-,


50Hz

Time(sec)

230312

0

V

RMSPeak

Investigation 9.

Frequency Measurement.

Objective.To investigate the use of a Cathode Ray

Oscilloscope to measure the frequency of aSinusoidal Wave produced by a functiongenerator.

Equipment.C.R.OFrequency Generator.Connecting wires as required.

Procedure.Connect up the CRO and the Fun. Gen. As

shownbelow:-,


CROF. Gen

1) Use the focus and brightness controls on the CRO toproduce a simple, sharp line across the MIDDLE of thescreen.

2) Set the F.Gen output to 1000Hz.

3) Using the y-amp and time base controls on the CROproduce a stable easy to read trace.

4) Record the value of the output from the F.Gen andthe wavelength and time base for the CRO, in the tableprovided.

5) Repeat the procedure for a number of different F.Gensettings including above and below the 1000Hz startvalue.

Readings/Results.

Frequency (Hz) 1

No of Div Time Base

Time baseSetting

Noof divisions for 1 cycle

F (Hz)MULT.RANGE

%Error

C.R.OFunctionGenerator


Calculations.

Using the above readings of F.Gen and CROsettings calculate the Frequency (Hz) and the %Error to complete the previous table.

Note that due to the F.Gen being out of calibrationthe errors produced may be significant, however ifvery large errors occur in excess of 100%.

Then you should consider the units of the timebase i.e. milliseconds (ms) or microseconds (ms)as these errors show that a miss-calculation hasbeen made.

Function Gen.

f Hz MultiRange

C.R.O

f Hz 1wave len cm time base sec

% error fF.GenfCRO

fF.Gen100


Questions 19.

1) With the aid of simple diagrams describe howthe following points of interest in the operation of aCRO are carried out:-,

a) The production and focus of the electron beam.

b) The deflection of the beam for measurementpurposes.


Waves.

A number of different wave forms (saw-tooth,square & Sine ) can be easily produced. Waveshave a number of values of interest to us.

Instantaneous Values - the value of propertyat a point in time.

Peak or Maximum Value - top of the wave form.

Average Value - normally measured over ahalf cycle,

Ave Value Area under wavelength of base line

Effective Value R.M.S - This is the value mostcommonly used for any alternating voltage or current. It is used for the GB Domestic supply which is quoted as being 230 volts RMS., other

common places to see it quoted is;speaker power (e.g. 8 watts RMS./ 12watts peak).It can be calculated using numerical integrationtechniques, and the expression:-,

VRMS V1

2V22Vn

2

n

where, n No of intervals


R C L Circuits.

In AC circuits we must consider the frequencyof the supply, we know that the GB domesticsupply is very stable at 50 Hz, and that othercountries, e.g. U.S. have stable 60 Hz supplies. Insome situations we can use different supplyfrequencies such as 180 volts at 100 Hz which willgive us a similar RMS. value as 230V at 50Hz.

We must also consider other wider, normallyhigher frequency's that we may have to designcircuits/components for. These include applicationssuch as radio (approx. 1 MHz to 150 MHz), soundsystems (16 Hz to 25 kHz) and microwaveapplications (ovens, phones, radar etc.) (up to 900Ghz).

Due to the nature of AC power, the voltage andthe current can be considered as vectors. That isthey have a magnitude and a direction. Thiscauses us problems in calculations as we can notsimply just add their magnitudes together, butmust consider the complete vector quantity.

We will consider and investigate theResistor-Capacitance-Inductor Series circuit. Thisis a common circuit where we have motors(inductors), motor starters (capacitors), tube lights(caps & starting coils), etc. We have dealt with thecapacitor earlier in this unit, and you should have a


good understanding of its operation in DC circuits.Where it simply charges and is then dischargedthrough transformers, flash tubes, resistors etc.

The full workings of the inductor has beenconsidered in Outcome 3 Magnetic Fields. In briefan inductor is a component which consists of a coilof wire, this coil may be part of a device such asthe windings of a motor, or transformer, or it canbe a discrete component manufactured, andinstalled in a circuit to provide an inductance value.To give a more effective value to a winding of anytype it is normally is given a core of non-permanentmagnetic material, e.g. Soft iron core.

The unit for inductance is the Henry (H), 1 H isa large quantity and it is normal for inductors tohave values in the milli-Henry (mH) or in small lowcurrent circuits micro-Henry (mH). The operation ofan inductor depends on the growth and collapse ofmagnetic fields within the coil. This allows the coilto store energy within it as magnetic flux. Due tothe rise and fall of the field being necessary for itsoperation, the inductor is dependent on thefrequency of the power passing through it.

We must also consider that due to theconstruction of the inductor being that of wire coil,then as well as the magnetic field effect, the wirehas its own resistance.

An inductor in an A.C circuit has a reactance, thevalue of this reactance is a function of the


inductors size (Henrys) and the frequency of thesupply. The value of reactance is given the unitsOHMS as it impedes the progress of electronswithin the circuit, in a similar way to resistance.

However, as well as impeding the flow of electronsdue to the building up of a magnetic field during thecharging part of the supply, an inductor has storedenergy in the form of magnetism which it willreadily give up. This is a similar situation tocapacitors being used in rectifiers to smooth theDC produced by the diodes. The storage howeveris very different physically as a capacitor storespower in the form of electrons, and the inductor asa magnetic field.

This difference results in what is known as phaseshifts, where the current and voltage of the circuitare put out of phase with each other. That is to saythe wave forms produced will not peak at the sametime, we will see this later during an investigation.


The reactance of an inductor and capacitor in anAC circuit is given by:

AND XL 2fL XC 12fC

where, f frequency Hz, L inductance H

C Capacatance F, XL & XC arethe

reactances due to theL & C.

When have calculated the values of the reactanceof each of the inductors and capacitors in a circuitwe can add these to the resistors within the circuit.This will give us a value that impedes the flow ofcurrent within the circuit in OHMS.

This total value is given the nameIMPEDANCE and is normally represented by theletter Z, and has units ohms.

Because reactance is a vector quantity we cannot simply just add up the value in ohms of thereactance's and resistance present in a circuit.They must be added using vectors, or vectoraddition. This is shown graphically later. However,we can use simple pythagoras to calculate thevalue of Z.

Impedance Z R2 XL XC2

Note. If only a Cap or Inductor exists then we canuse:

Impedance Z R2 X?2


Questions 20.1) Calculate the inductive reactance of a 100mHinductor if the frequency of the supply is 200 Hz.

2) Calculate the capacitive reactance of a 150Fcapacitor if the frequency of the supply is 1000 Hz.

3) Calculate the Impedance of a circuit, containinga 8 ohm resistor in series with a capacitor ofcapacitance 200F, if the supply is at 100 Hz.

4) Calculate the impedance of a circuit containinga 120 ohm resistor in series with an inductor of40mH, if the supply is at a frequency of 400 Hz.


5) Calculate the total impedance of a circuitcontaining a resistor of 10 ohms, in series with acapacitor of capacitance 500F and in series withan inductor 12mH, if the supply is 200 volts at 100Hz.

6) Calculate the total impedance, the currentflowing and the voltage across each component(i.e. VR ,VC & VL) of the following circuit:-,


230volts 50Hz

I

C=416uFL=10mH

We have calculated the impedance of the RCLcircuits using the formula:

Z R2 XL XC2

We can also use scale drawing to represent themagnitudes and directions of the resistors, andreactance's of a circuit.E.g.Find the impedance of a circuit constructed with aresistance of 12 ohms, in series with a capacitor ofreactance 4 ohms and an inductor of reactance of9 ohms, using a scale drawing.

The above diagram was developed by using astandard layout (axes). Resistance (R) is shownleft to right (x-axis), and the reactance up-down(y-axis). Because the reactances XL & XC can beconsidered as being in opposition to each other,


R

XL

XC

012

4

5 Z=13ohms

we can plot them in opposite directions (negative &positive). The hypotenuse of the triangle which isdeveloped is the impedance of the circuit. Thedirection of the impedance above or below theR-axis is known as the power factor (p.f). Thepower factor is a value of efficiency and istherefore normally between 0.5 (50%) and 0.9+(90%+).

Example.Find by means of an impedance triangle, and a

voltage triangle, the impedance, voltages acrossthe components, and the power factor. For a circuitcontaining a resistor of 10 ohms, in series with acapacitor giving a reactance of 8 ohms and aninductors with a reactance of 2 ohms. If the circuitis supplied with voltage of 120 volts, 50 Hz, A.c.1)

The above diagram shows the axis and scales.


R

XL

CX

0

2)

In the above diagram we have drawn in theresistance value of 10 ohms, also we havemeasured up the reactance axis, XL 2 ohms. Thisis now our zero point for measuring the capacitorreactance.

3)

The above diagram shows the reactance XC,measured down from the XL point, and thereactance value marked across the graph.


R

XL

CX

02

R

XL

CX

02

4)

The above diagram shows the impedance (Z)value drawn in, and the power factor (p.f.) Alsomarked.We can at this point measure the length of the lineZ, and using our scale find a value for theimpedance. Using a protractor we can alsomeasure the size of the power factors angle. Thisangle can then be used to construct the voltagetriangle, as the power factor will not change andwill therefore give us one of the angles of thevoltage triangle.From the above we find:

Z = 11.8 ohms, and that the power factor is theCosine 30O = 0.85.

With this information we can calculate the currentflowing in the circuit.

I VZ

12011.8 10.2amps.


R

XL

CX

02

p.fZ

5) The voltage triangle will show the voltage acrossthe resistor and the total reactance, as well as thesupply voltage and power factor.


VR =10 X 10.2=102 Volts

VX

=6 X

10.

2 =

61.2

Vol

ts

VS = 120 volts

power factor = 30 o

Questions 21.1) Find using scale drawings the total impedanceof a circuit containing a resistor of 10 ohms, inseries with a capacitor of capacitance 500F andin series with an inductor 12mH. If the supply is200 volts at 100 Hz, calculate the current and drawa voltage triangle.


2) Use a scale drawing to find the total impedance,the current flowing and the voltage across eachcomponent (i.e. VR ,VC & VL) of the followingcircuit:-,


230volts 50Hz

I

C=416uFL=10mH

Phasors.We have investigated the relationship between

resistors/caps/inductors etc. and calculated valuesof the magnitude of the variables such as voltageand current. We must now consider the directionsof the vectors involved, to allow us to visualise therelationships we use Phasor diagrams. A phasordiagram is a rotating vector diagram. Where due tothe repeating nature of the variables (wave forms),it is easier for us to show them as rotating vectors.

Example.

Phasor diagram showing the voltage and currentfor a circuit with a leading current (thereforecapacitive).


Voltage

current

direction of rotation

angle oflead

In the previous phasor diagram the length ofthe voltage and current vectors represent themagnitudes of the variables. However, to allow usto see both of them, it is common to use twodifferent scales. This is due to the voltagesnormally being: x10 or up to x100, that of thecurrents being considered (e.g. 230 volts, at acurrent of 12 amps, in a kettle/fan heater etc.) . Ina phasor diagram we are mostly interested in theangle between the vectors.

Another big use of phasor diagrams is in thestudy of 3-Phase supplies, as this shows 3 Voltagevectors rotation with a 120O between them.However, this is not covered in this unit and youwill have to wait till the next one Further E & EP.


Questions 22.Sketch the phasor diagrams for the followingsupplies:

1) 220 Volts, with a current of 10 amps lagging by45O.

2) 120 Volts, with a current of 12 amps leading by30O.

3) A current of 6 amps, with a voltage of 110Vleading by 60O.

4) A current of 120 amps, with a voltage of11,000V lagging by an angle of 15O.




We can develop phasors and use them to plot thewave forms of the variables (Voltage and current).This is done in a similar way to the development ofa sine curve using simple harmonic motion. Whenthis has been carried out correctly we see the leador lag of the current/voltage etc. As the waveforms are out of sink with each other. That is theypeak at different angles and cross the time/angleaxis at different points.E.g.


Voltagecurrent

Angle/time

Purely Resistive Circuit No Lead or Lag

Voltagecurrent

Angle/time

current laggingbehind voltageA capacitive circuit

Investigation 10.Construct the following circuit using crocodile clipsusing the component values shown. Using theCRO measure the voltage across the resistor andcapacitor in the circuit. Take care to record thephase difference between the wave forms.


RC

A.C suppy

V V

16n100100k5

0.16u10010k4

1.6u1001k31.6m100100216m100101

ObservedAngle

Lead/Lag Cal

Z()XC ()C (F)R ()f (Hz)TEST

Equations.

XC 12fC Z R2 X2

lead or lag cos1Rz

observed angle CRO DivisionsGAPCRO DivisionsFor 1 cycle

360o


Resonance.

If we consider a simple R-C-L circuit, then we canmake the values of XC & XL the same, this makethe impedance diagram into a flat line.E.g.

When we have a circuit constructed to giveresonance then we find:

andVC VL R Z

power factor 1

XC XL , therefore,1

2fC 2fL

at resonance it can be shown :

fr 12 LC

Hz


XL

RZ

XC

Questions 23.1) Calculate the resonant frequency of a circuitcontaining a 10 ohm resistor, a 1000F capacitorand a 50mH inductor all in series with a 120 voltsupply.

2) Calculate the size of an inductor required toproduce a series circuit with a 10F capacitor,which will resonate at 400 Hz.

3) Calculate the resonate frequency of thefollowing circuit:


20ohm

C=1000pFL=5mH

Questions 24.1) You are required to research what is meant bythe term, “Q-Factor” of a RLC circuit.You should give a simple definition and anyexpressions commonly used in its calculation oruse.


2) The following circuit has been constructed usinga variable frequency power supply, calculate theresonate frequency of the circuit and the voltagemagnification at resonance.


20ohm

C=200nF

L=100mH

120volt

Documents

E & EP OUTCOME 4 Yr 1