edexcel national certificate/diploma unit 5 - electrical ... · pdf file© d.j.dunn 1...

© D.J.Dunn www.freestudy.co.uk 1

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA

UNIT 5 - ELECTRICAL AND ELECTRONIC PRINCIPLES

NQF LEVEL 3

OUTCOME 4 - ALTERNATING CURRENT

4 Understand single-phase alternating current (ac) theory

Single phase AC circuit theory: waveform characteristics e.g. sinusoidal and non-sinusoidal

waveforms, amplitude, period time, frequency, instantaneous, peak/peak-to-peak, root mean

square (r.m.s), average values, form factor; determination of values using phasor and algebraic

representation of alternating quantities e.g. graphical and phasor addition of two sinusoidal

voltages, reactance and impedance of pure R, L and C components

ac circuit measurements: safe use of an oscilloscope eg setting, handling, health and safety;

measurements (periodic time, frequency, amplitude, peak/peak-to-peak, r.m.s and average

values); circuits e.g. half and full wave rectifiers


1. REVISION OF BASIC A.C. THEORY

1.1 SINUSOIDAL WAVE FORMS

A pure alternating current or voltage varies with time sinusoidally as shown.

INSTANTANEOUS VALUES

For a sinusoidal voltage and current the instantaneous value at any moment in time is given by:-

v = V sin (ωt) and i = I sin (ωt) or v = V sin (2πf t) and i = I sin (2πf t)

Note that this assumes that t = 0 when v or i = 0

AMPLITUDE

The maximum value of volts or current is called the peak volts or current and this is the amplitude

of the wave form (V and I). The peak to peak value is double the amplitude as shown on the

diagram.

FREQUENCY

The voltage or current changes from a maximum (plus) in one direction, through zero to a

maximum (minus) in the other direction. This occurs at f times a second. f is the frequency in Hertz.

1 Hz = 1 cycle/second

PERIODIC TIME

The time it takes to complete 1 cycle is T seconds (the periodic time). It follows that

T = 1/f

ANGULAR FREQUENCY

If we think of the voltage and current being generated by a machine that rotates one revolution per

cycle, the 1 cycle corresponds to 360o or 2π radian.

It follows that f cycles/second = 2πf radian/s and this is the angular frequency.

ω = 2πf = 2π/T rad/s


AVERAGE VALUES

The average value of any true alternating current or voltage is zero since half the cycle is negative

and half is positive. When an average value is stated it refers to the average over one half of the

cycle. This may be determined from the area under the graph as illustrated below.

For a sinusoidal waveform, the blue area is exactly 2 when the angle is in radians and the peak

value is 1. The mean value is value that makes the blue rectangle contain the same area as the blue

area of the half cycle. In other words the green area above the average is equal to the light blue area

below the rectangle.

The blue area must be equal to π x average value hence the average value is 2/π = 0.637

If the peak value is something other than 1 then the average is 0.637 x peak value

FORM FACTOR

This is defined as valueAverage

ValuePeak Factor Form

Hence for a sinusoidal voltage or current the form factor is π/2 = 1.571

PHASE and DISPLACEMENT

The sinusoidal graph is produced because we made θ = 0 when t = 0. We could choose to make θ

any value at t = 0. We would then write the equations as:

x = A sin(θ + ) or v = V sin(θ + ) where is the phase angle.

The plot shown has x = 0 at θ = 30o so it follows that (30 + ) = 0 and so = -30

o.

If we made = 90

o we would have a cosine plot.

Often periodic functions are not based about a mean of zero. For example an alternating voltage

might be added to a constant (d.c.) voltage so that V = Vdc + V sin(θ + )


SELF ASSESSMENT EXERCISE No. 1

1 Mains electricity has a frequency of 50 Hz. What is the periodic time and angular frequency?

(0.02 s and 314 rad/s)

2. An alternating current has a periodic time of 0.00035 s. What is the frequency?

(400 Hz)

3. A alternating voltage has a peak to peak amplitude of 300 V and frequency of 50 Hz. What is

the amplitude and average value?

(150 V and 95.493 V)

What is the voltage at t = 0.02 s?

(16.4 V)

4. An alternating current is given by the equation I = 5 sin(600t).

Determine the following.

i. the frequency (95.493 Hz)

ii. the periodic time (10.472 ms)

iii. the average value. (3.183 A)

5. Determine the following from the graph shown.

The amplitude.

The offset displacement.

The periodic time.

The frequency.

The angular frequency.

The phase angle.

(Answers 5, 2, 1.57 s, 4 rad/s, 0.637 Hz, 0.2 radian or 11.5o)


1.2 ROOT MEAN SQUARE VALUES (R.M.S.)

The mean value of an alternating voltage and current is zero. Since electric power is normally

calculated with P = V I it would appear that the mean power should be zero. This clearly is not true

because most electric fires use alternating current and they give out power in the form of heat.

When you studied Ohms' Law, you learned that electric power may also be calculated with the

formulae E.P. = I2R or E.P. = V2/R

These formulae work with positive or negative values since a negative number is positive when

squared and power is always positive. In the case of a.c. we must use the average value of V2 or I2

and these are not zero. The diagram shows how a plot of V2 or I

2 is always positive. The mean

value is indicated.

The mean height may be obtained by

placing many vertical ordinates on it as

shown. Taking a graph of current with

many ordinates i12, i2

2 .....in

2.

The mean value of the i2 is:

∑( i12+ i2

2+ i32..... in )/n

If we take the square root of this, we have

a value of current that can be used in the

power formula. This is the ROOT MEAN SQUARE or r.m.s. value.

I(r.m.s.) = ( i12+ i2

2+ i32..... in)/n

It can be shown by the use of calculus that the r.m.s. value of a sinusoidal wave form is Vm/√2. We

use r.m.s. values with a.c. so that we may treat some calculations the same as for d.c. When you use

a voltmeter or ammeter with a.c., the values indicated are r.m.s. values.

Vrms = Vm/√2 = 0.707Vm


1. The periodic time of an ac voltage is 0.002 s. Calculate the frequency.

(500 Hz)

2. The r.m.s. value of mains electricity is 240 V. Determine the peak voltage (amplitude).

(339.5V)

3. An a.c. current varies between plus and minus 5 amps. Calculate the r.m.s. value. (3.535 A)

4. An electric fire produces 2 kW of heat from a 240 V r.m.s. supply. Determine the r.m.s. current

and the peak current.

(8.33 A and 11.79 A)

5. An electric motor is supplied with 110 V r.m.s. at 60 Hz and produces 200W of power.

Determine the periodic time, the r.m.s. current and the peak current.

(16.7 ms, 1.82 A and 2.571 A)


1.3 OTHER WAVE FORMS

Cyclic variations may take many forms such as SQUARE, SAW TOOTH and TRIANGULAR as

shown below.

Square waveforms are really d.c. levels that suddenly change from plus to minus. The r.m.s. value is

the same as the peak value. They are typically used for digital signal transmission.

Saw tooth waves are used for scanning a cathode ray tube. The electron beam moves across the

screen at a constant rate and then flies back to the beginning. Triangular waves change at a constant

rate first in one direction and then the other.


1. Work out the average and form factor figures for a square wave.

2. A triangular voltage has a peak value of 15 V. Work out the average value, the form factor and

the r.m.s. value. Note that shape of the triangle does not make a difference so you can assume a

right angle triangle to make it easier. If you cannot do the maths try plotting and working out

the areas by a graphical method.

(7.5V, 2 and 10.6 V)


2. REACTANCE AND IMPEDANCE

Capacitors and Inductors have a property called Reactance denoted with an X. On their own they

may be used with a form of Ohm’s Law such that V/I = X

Both V and I are r.m.s. values. The value of X depends on the frequency of the a.c. and this is why

they are called REACTIVE. It should be noted that a pure capacitor and inductor does not lose any

energy. A resistor on the other hand, produces resistance by dissipating energy but the value of R

does not change with frequency so a resistor is a PASSIVE component.

When a circuit consists of Resistance, Capacitance and Inductance, the overall impedance is

denoted with a Z. The units of R, X and Z are Ohms.

2.1 CAPACITIVE REACTANCE XC

When an alternating voltage is applied to a capacitor, the capacitor charges

and discharges with each cycle. This means that alternating current flows in

the circuit but not across the dielectric. If the frequency of the voltage is

increased the capacitor must charge and discharge more quickly so the

current must increase with the frequency. The r.m.s. current is directly

proportional to the r.m.s. voltage V, the capacitance C and the frequency f.

It follows that Irms = Constant x Vrms x f x C

Vrms/Irms = 1/(constant x f C) The constant is 2 so Vrms/Irms = XC = 1/(2 f C)

Note that when f = 0, XC is infinite and when f is very large XC tends to zero. This means that a

pure capacitor presents a total barrier to d.c. but the impedance to a.c. gets less and less as the

frequency goes up. This makes it an ideal component for separating d.c. from a.c. If we put in a

combined a.c. + d.c. signal as shown, we get out pure a.c. but with a reduced amplitude depending

on the reactance.

WORKED EXAMPLE No. 1

15 V r.m.s. applied across a capacitance of 4.7 μF. Calculate the r.m.s. current when the

frequency is 20 Hz, 200 Hz and 2000 Hz

SOLUTION

20 Hz A 0.008861693

15

X

VIrms Ω 1693

10 x 4.7 x 20 x π2

1

C f π2

1X

C

6-C

200 Hz A 0.08861693

15

X

VIrms Ω 169.3

10 x 4.7 x 200 x π2

1

C f π2

1X

C

6-C

2000 Hz A 0.8861693

15

X

VIrms Ω 16.93

10 x 4.7 x 2000 x π2

1

C f π2

1X

C

6-C


2.2 INDUCTIVE REACTANCE XL

The back e.m.f. produced by a varying current is e = - L x rate of change of current.

In order to overcome the back e.m.f., a forward voltage equal and opposite is required. Hence in

order to produce alternating current, an alternating voltage is needed.

It can be shown that the r.m.s. voltage needed to produce an r.m.s. current is directly proportional to

the current, the inductance and the frequency so that Vrms = Irms (2fL)

Hence Vrms/Irms = XL = 2 f L Ohms

Note that the reactance is zero when f = 0 and approaches infinity when f is very large. This means

that a pure inductor has no impedance to d.c. but the impedance to a.c. increases directly

proportional to frequency. This is the opposite affect to that of a capacitor and an inductor may be

used to reduce the a.c. component of a combined a.c. and d.c. signal as illustrated.


15 V r.m.s. applied across an inductance of 4 μH. Calculate the r.m.s. current when the

frequency is 200 Hz, 200 kHz and 200 MHz

SOLUTION

20 Hz kA 29.8410 x 0.5027

15

X

VIrms mΩ 5027.04x10 x 20 x 2πfL 2πX

3

C

6

C

200 kHz A 2.9845.03

15

X

VIrms Ω03.54x10 x 200 x 2πfL 2πX

C

6

C

200 MHz mA 2.98 5027

15

X

VIrms Ω 50274x10 x 2000 x 2πfL 2πX

C

6

C



1. Calculate the reactance of a capacitor with capacitance 60 F at a frequency of 50 Hz. (53 Ω)

2. The Voltage applied to the capacitor is 110 V r.m.s. Calculate the r.m.s. current. (2.073 A)

3. A capacitor is put in a circuit to limit the r.m.s. current to 2 mA when 10 V r.m.s. at 60 Hz is

placed across it. What should the value of the capacitance be? (530 nF)

4. Calculate the r.m.s. current in an inductor of 60 mH when 110 V r.m.s. is applied at 60 Hz.

(4.863 A)

5. An inductor passes 20 mA rms at 12 V r.m.s. and 1000 Hz. Calculate the inductance.

(95 mH)

6. Calculate the inductance of a coil 25 mm diameter, 100 mm long with 30 turns. The core has a

relative permeability of 2000. (0.0111 H)

Calculate the energy stored when 10 A d.c. flow. (0.555 J)

Calculate the reactance for ac with a frequency of 100 Hz. (6.97 )

Calculate the r.m.s. voltage needed to make 10 A r.m.s. flow. (69.7V rms)


3. PHASOR DIAGRAMS

The way a sinusoidal voltage or current varies with time is given by the following equations.

v = V sin (ωt) or v = V sin (2πf t) i = I sin (ωt) or i = I sin (2πf t)

V and I is the amplitude and v and i are the instantaneous values at time t.

f is the frequency in Hz.

ω is the angular frequency in radian/s ω = 2πf.

ωt = θ and this is an angle in radian.

Consider a phasor representing a sinusoidal voltage. It is a vector of length V that can be drawn at

an angle θ = ωt to represent the voltage at any instant in time t. If they were drawn in succession

then they would be rotating anticlockwise (positive) at ω rad/s. Starting from the horizontal position

after a time t it will have rotated an angle θ = ωt.

The vertical component of the phasor is v = V sin (θ)

This corresponds to the value of the sinusoidal graph at that angle.

When θ = π/2 radian (90o) the peak value is V so V is the amplitude or peak value (not the

r.m.s.value).

You will find some animated phasor operations at the following addresses.

http://www.physics.udel.edu/~watson/phys208/phasor-animation.html http://observe.phy.sfasu.edu/SkyLabs/Excel/Phasors.xls Requires Microsoft excel to run http://www.cs.sbcc.cc.ca.us/~physics/flash/optics/phasors4.swf

ADDING AND SUBTRACTING

Phasors can be added or subtracted from each other in the same way as vectors. Consider two

voltage phasors with the same angular frequency. The diagram shows two such phasors V1 and V2

but V2 lags V1 by angle d. This is the PHASE DIFFERENCE BETWEEN THEM. We can express

the vectors as:

v1 = V1 sin(θ) and v2 = V2 sin(θ - d).

http://www.physics.udel.edu/~watson/phys208/phasor-animation.html

http://observe.phy.sfasu.edu/SkyLabs/Excel/Phasors.xls

http://www.cs.sbcc.cc.ca.us/~physics/flash/optics/phasors4.swf


If we add the two vectors we do it as shown.

It is normal to draw one of the phasors horizontal as the phase difference is the same at all angles.

The easiest way to add vectors is to resolve them into horizontal and vertical components.

v2 has a vertical component V2 sin d and a horizontal component V2 cos d

The vertical component of v2 is V2 sin d

The vertical component of v is V2 sin d

The horizontal component of v is V1 + V cos d


If v1 = 10 sin (θ) and v2 = 7 sin (θ - 30o) what is the resultant voltage?

SOLUTION

Let θ = 0o when we add them. (V1 in the horizontal position).

The vertical component of v is v2 sin d = 7 sin (-30o) = -3.5 (down).

The horizontal component of v is V1 + V2 cos d = 10 + 7 sin 30o = 16.06

The resultant voltage is √(-3.52+16.06

2 ) = 16.44

The phase angle is = tan-1

(-3.5/16.06) =-12.3o

We can express the phasor as v = 16.44 sin (θ – 12.3o)


4. CIRCUITS CONTAINING RESISTIVE AND REACTIVE COMPONENTS.

4.1. A.C. AND RESISTANCE

When a.c. is applied to a pure resistance R, Ohm’s Law applies and since it is passive it is the same

at all frequencies at all moments in time. The phasors for voltage and current must rotate together.

They are said to be IN PHASE.

4.2. AC AND INDUCTANCE

This requires a basic knowledge of differentiation. The voltage required to drive a current through

an inductor is v = L x rate of change of current.

L is the inductance in Henries.

Suppose i = I sin t The rate of change of current is obtained by differentiating di/dt = I cos(t)

It follows that v = I L cos(t) and the maximum value is V = I L

If V and I are plotted together we see that that V is ¼ cycle displaced and it is said that the voltage

leads the current by 90o. The voltage phasor is 90

o anticlockwise of the current phasor.


4.3 AC WITH RESISTANCE AND INDUCTANCE

Now consider a.c. applied to a resistor and inductor in series as shown.

The current I flows through both so this is used as the reference.

The voltage over the resistance is VR = I R and on the phasor diagram

this must be in the same direction as the current.

The voltage over the inductor is VL = I XL and this must lead the

current by 90o and also VR by 90

o.

It is not true to say that V = VL + VR because they must be treated as phasors or vectors.

The resultant voltage is VS and this is the hypotenuse of a right-angled triangle so 2

L

2

R VVV

The angle is called the phase angle and is always measured from VR.

It follows that = tan-1

(VL/VR)


4.4. AC AND CAPACITANCE

Thgis requires knowledge of integration. When a.c. is applied

across a capacitor, the voltage is given by the equation

vC = q/C where q is the charge stored and C is the capacitance

in Farads. Since idtq then C

idtvC

i varies sinusoidally so that i = I sin (t) tωcos ω Iidt

Substitute and tωcosC

ω IvC

The maximum value of vC is I/c so this will be the length of the phasor representing VC. If we plot VC and I we find that VC lags the current by ¼ cycle or 90

o. This is opposite to an inductor which

leads by 90o.

4.5. AC WITH RESISTANCE AND CAPACITANCE

Now consider a resistor and capacitor in series as shown.

The voltage over the resistance is I R and on the phasor diagram this

must be in the same direction as the current. It follows that VC lags VR

by 90o. It is not true to say that Vs = VC + VR because they must be

treated as vectors.

The resultant voltage is the hypotenuse of a right-angled

triangle so 2

C

2

Rs VVV

The angle is called the phase angle and is always

measured from VR. It follows that = -tan-1

(Vc/VR)

The only difference between this and the R L circuit is that

VC lags VR and VL leads VR. This means that VL and VC

are 180o out of phase in a series circuit.


4.6. R L C IN SERIES

The 3 voltages VR VL and VC are drawn as 3 phasors

and the vector sum is found. It is convenient to draw VR

horizontally but the vector diagram stays the same for

all angles of rotation.

Examining the small triangle, we see the vertical height is VL - VR and the horizontal length is VR.

It follows that the resultant voltage is given by

2

R

2

cLs VVVV and

R

CL1

V

VVtan

4.7. REACTANCE AND IMPEDANCE REVISITED

We know from previous studies that the relationship between current and voltage for any

component is related as a ratio X = V/I. For a resistor this ratio is resistance R but for an inductor it

is called inductive reactance XL and for a capacitor capacitive reactance XC.

Inductive reactance increases with frequency and is given by XL = 2fL

Capacitive reactance decreases with frequency and is given by XC= 1/ 2fC

When current flows in a RLC circuit, the relationship between it and the resulting voltage is called

the IMPEDANCE Z. Z = V/I where V and I are the resulting r.m.s. volts and current.

Since reactance is V/I it follows that it is also a phasor. The phasor diagram for a series R L C

circuit may be drawn as shown with R drawn horizontally to make it easier.

22

CL RXXZ and

R

XXtan CL1



A resistor of value 470 is connected in series with a capacitor of 22 F and an inductor of

50 mH and a voltage is applied across it. A current of 100 mA (rms) is produced.

Determine the impedance, the phase angle between the voltage and current and the applied

voltage when the frequency is 50 Hz

SOLUTION

XL = 2πfL = 2π x 50 x 50 x 10-3

=15.71 Ω

XC = 1/2πfC = 1/(2π x 50 x 22 x 10-6

) =144.6 Ω

Ω 487.44706.14415.71RXXZ 2222

CL

o1CL1 15.3470

6.14415.71tan

R

XXtan

VS = I Z = 0.1 x 487.4 = 48.7 V rms


1. A resistor of value 4 is connected in series with a capacitor of 47 F and an inductor of 20 μH and a voltage is applied across it. A current of 50 mA (r.m.s.) is produced.

Determine the impedance, the phase angle between the voltage and current and the applied

voltage when the frequency is 100 Hz. (34 Ω, -83.3o and 1.7 V)

2. A resistor of value 0.2 is connected in series with a capacitor of 4.7 F and an inductor of 5

mH and 0.5 V r.m.s. is applied across the ends.

Determine the impedance, the phase angle between the voltage and current and the rms current

when the frequency is 1000 Hz. (2.455 Ω, -85.3o and 204 mA)

edexcel national certificate/diploma unit 5 - electrical ... · pdf file© d.j.dunn 1...

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