© Sir Farook
Figure 3.131 shows a simple transformer, whichconsists of two separate coils of insulated wire wound on a laminated core made from sheets of soft iron separated by an insulating varnish. The sheets are glued together so that they are electrically insulated from each other. The two coils are not linked electrically, the only connection between them being any magnetic field in the core.
p.d. = vP turns= NP current= lp a.c.
l:_;•up-p-ly-4~~~~
current= /5Figure 3.131 Simple transformer
,One of the coils is referred to as the primary coil, aud the other is called the secondary coil. The number of turns on each (NP and N) is usually different.
In Section 3.5 on Electromagnetic induction, it was
shown that a changing current in the primary coil induces an e.m.f. in the secondary coiL An alternating. current in the primary will therefore create an alternating magnetic field in the core,
which links with the turns of the secondary coil. This alternating magnetic field therefore induces au e.m.f .. · in the secondary, alternating with the same frequency as that in the primary. This e.m.f. will cause an alternating current to flow in any external circuit connected to the coil.
Assuming 100% efficiency, it cau be shown that the p.d.s across the primary and secondary coils (VP aud Vs respectively) are related to the number of turns of wire (NP and N5 respectively) as follows:
Equation 5
800 kVtransformers at a large substation in South America
If N5 > NP then Vs > VP - a step-up transformer If N5 < NP then Vs < VP - a step-down transformer
Guided examples (2)
1. The primary coil of a transformer has 500 turns and the secondary coi12000 turns. A 5 V alternating supply is connected to the primary. Calculate the voltage across the secondary.
2. In order to light a 12 V lamp, a transformer with 2400 turns in the secondary coil is used to step-down the voltage of the a. c. mains (240 V). Calculate the number of turns in the primary coil.
Guidelines
Use Equation 5 in each case.
Power and current in transformers
In
a perfectly efficient transformer, there would be no energy losses; i.e. the power delivered to the primary would be equal to the power delivered by the secondary.© Sir Farook
l.e. p =P p s
Equation 6
Combining Equation 5 and Equation 6:
You should note that:
• NP, VP and IP are the number of turns, p.d. and currentrespectively in the primary coil and N 8, V5 and Is are the number of turns, p.d. and current respectively in the secondary coil.
• In practice, the-secondary current in the
transformer depends on the resistance of the load connected to it, not the turns ratio. The primary current drawn from the supply is controlled by the turns ratio.
Energy losses
The above equations assume 100% efficiency, i.e. all the electrical energy supplied to the primary is converted to electrical energy in the secondary.
In practice, a small amount of energy is dissipated to the surroundings:
• Coil resistance causes energy to be dissipated due to the heating effect of current in the coils. This is minimised by using high conductivity copper wire. • Eddy currents are circulating electric currents
within the iron core of the transformer. These are induced in the iron core by the alternating
magnetic field. As iron has considerable resistance, energy losses due to heating of the core would be significant if these currents_ were not minimised. The laminated construction of the core cuts down possible paths for the flow of eddy currents {see Figure 3.132).
thin sheets of soft iron coated with insulating varnish and glued together Figure 3.132 Laminated construction of transformer core
• Magnetic flux leakage. Some of the changing magnetic flux produced in the primary may not actually link with the secondary coil. The energy loss due to this effect is minimised by appropriate design - the primary and secondary coils are often overlapping, for optimum flux linkage.
• Hysteresis is the name given to the reluctance of a material to undergo changes in magnetisation.
During each cycle of a.c., the core reverses the polarity of its magnetisation, which requires energy. This is minimised by careful choice of material (with soft magnetic properties). A well-designed transformer should suffer energy losses of only 1% or so.
Transmission of electrical
energy
Since power
=
p.d. x current, it is possible to transmit electrical energy at, forexample, a rate of 1 million watts using an infinite number of combinations of p.d. and current, some of which are shown in thefollowing table: P.d. Current Power N /A /W 1 X -1 000 000
=
1 000 000 10 X 100000 = 1 000 000 100 X 10 000 = 1 000 000 1000 X 1 000 = 1 000 000 10 000 X 100 = 1 000 000 100 000 X 10 = 1000 000 1 000 000 X 1 = 1 000 000Using transformers, the p.d. of an alternating supply can be stepped-up or stepped-down at will with negligible cost in terms of energy wasted.
The choice of which combination of p.d. and current is best is a compromise based on the following factors.
If the conducting wires have resistance R, and carry a current /, the rate of energy loss due to heating of the wires is equal to
PR.
In other words, the rate of loss of energy is proportional to the square of the current (since R is constant). So reducing the current by a half brings a fourfold saving in energy.However, reducing the current means a correspondin-g increase in p.d. (to transmit the same power). This increases the cost and complexity of the switch gear
© Sir Farook
used to control the distribution of the supply around the country. It also increases the danger to the public
since it is much more difficult to insulate 1 000 000 V
than 100 V.
In the UK, voltages up to 400 kV are used in the 'supergrid' while most electricityis transmitted at 132
kV nationally. This is stepped-down progressively to
33 kV and 11 kV for distribution to industry and
domestic substations, where it is finally stepped-down
to 240 V for homes.
Thus electrical power is transmitted at very high voltage (and hencelow current) to minimise power
losses due to heating of the transmission cables.
Electricity generating stations therefore produce
alternating current due to the ease with which its
voltage can be stepped-up and stepped-down using
transformers.
Guided example (3)
1. Calculate the power loss in transmitting
2 MW of electrical power through cables of
total resistance 2 Q, if the voltage is
(a) 4000 V
(b) 400 kV.
Guidelines
Calculate the current:
current = power/voltage Then calculate the power loss: power loss = current2 x resistance Compare the two answers.
Rectification
The major advantage of a.c, is the ease with which its
voltage can be chaf!ged using transformers. However,
many components require a current in one direction
only - a direct current. Rectification is the process by
which a direct current is produced from an alternating
current.
Half·wave ·rectification
Figure 3.133 shows a simple arrangement for
rectifying an alternating current. A diode· only passes current when it is :forward-biased~. With a single diode in series with the load as shown, current can only pass in the direction A to B. This occurs during the half of the cycle. when
A
'is more positive than B.When the supply p.d. reverses, the secondary p.d. also
reverses. However the diode is no longer
forward-biased, and does not conduct. So, for this half of the
cycle, there is no current. The current supplied is direct current, since its ,direction da.es not alter, but
the magnitude shows con$iderable variation, and for
half the time, there is no current.
supply p.d.
a.c. supply
CJ
~
B Acurrent
no current in 'reverse' direction
Figure 3.133 Half-wave rectification
time
time
A better approximation to a steady direct current is obtained using two diodes.
Full-wave rectification
Two diodes are arranged as shown in Figure 3,134 on
page 300 with a centre-tap transformer. Effectively the load is supplied by two separate half-wave rectifiers. Dtiri{\g one half of the cycle, only rectifier 1 is forward-biased so it is the only one to conduct,
allowing current from A to B through the load. During
the second half of the cycle, only diode 2 conducts, but there is still a current through the load from A to B. Thus the output is a fluctuating direct current
through the load during both harves of
~ach
cycle ofthe a.c. supply.
The full·wave bridge rectifier
Four diode rectifiers 1, 2, 3 and 4 are arranged in a
square as shown in Figure 3.135 on page 300.
During the first half of a cycle, if U is the more positive terminal, there is a current in the direction
U - V - W since diode 1 is forward-biased. Then
© Sir Farook
supplyp.d. diode 1 I current a.c. supply A + I Ino current in 'reverse' direction
Figure 3.134 Full-W£ZV!? rectificatimt
a.c. supply u time centre-tap transformer diode 2 time ~-~~r~v..~~'~'~'~-1
Figure 3.135 Full-wave bridge rectifier
along the path X - Y - Z since diode 2 is also
forward-biased. During this half~cycle, diodes 3 and 4 are reverse-biased, and do not conduct.
During the second half, Z becomes more positive, so diodes 3 and 4 are now forward-biased. So the current path becomes Z - Y - W then through the load from A to B as before, returning along the path X - V - U.
Smoothing
A rectified alternating current still fluctuates between zero and a peak value when the above methods are used. In order to minimise the fluctuations, a smoothing capacitor (sometimes called a reservoir capacitor) can be connected in parallel with the load, as shown in Figure 3.136. unsmoothed rectified d.c.
/
+
.----. rm
-~__, B Ac
RFigure 3.136 Use of capacitor for smoothing
The action of the capacitor can be readily explained by considering Figure 3.137 on page 301.
If the capacitor is sufficiently large, then the time constant of the capacitor and the load resistor ( CR) will be large. Consider the sequence of events shown in Figure 3.137. The dashed line is the unsmoothed rectified p.d. The line AB shows the rising p.d. at switch on. Theheavy line is the p.d. across the capacitor (and hence the p.d. across the load). Initially the rising supply p.d. sends a current through R in Figure 3.136, and causes the p.d. across the capacitor to increase to the peak value (almost). This is shown by the line AB in Figure 3.137. As the supply p.d. decreases from its peak, the capacitor cannot discharge through the supply (wrong polaricy), only through the load R. The p.d. can only fall slowly (due
to the large time constant, CR) shown by the line BC. Meanwhile the supply p.d. has fallen to zero and started to increase again. At point C, the supply p.d. becomes greater than the p.d. across the capacitor, and the capacitor starts to recharge up to the peak p.d. (line
© Sir Farook
A initial charge charge discharge \ ' smoothed output p.d. unsmoothed rectified p.d.Figure 3.137 Smoothing of the output p.d. from a full-wave rectifier
CD). The process repeats, giving rise to an almost steady p.d., with only a ripple of small amplitude. The capacitor must be matched to the output load R to give a suitable time constant. If C is too small, the time constant will be too short, allowing the capacitor to discharge too much between peaks of p.d. If Cis too large, it could draw too much current on its first charging cycle, possibly damaging the rectifier.
© Sir Farook
At any time (t) the p.d. (V) across the resistor is givenby: V
=
V0 sin OJtAnd the current I in the circuit is given
by: I= X= V0 sin OJt
R
R
i.e. I= I0 sin OJt where I0
=
VoR
Thus both I and V are sine functions which vary with
time as shown in Figure 3.138.
• In a purely resistive a.c. circuit the p.d. (V) and the current (I) are in phase.
• The oppositionto a.c. which this
circuit presents is the resistance,
R-v
- I• The resistance of such a circuit is not affected by the frequency of the supply.
Pure capacitor of capacitance C
This situation is less simple. As you already know, a
capacitor does not conduct electricity. The two plates are separated by an insulator. However, an alternating current does exist in the circuit to which it is
connected, as the charge alternately flows on and off
each plate in tum. The p.d. across the capacitor depends on the amount of charge present on it- more charge, more p.d. But the current is greatest when there is no charge - i.e. zero p.d.! To examine this in more detail, consider an alternating p.d. applied across
a capacitor, as shown in Figure 3.139. R
I .
V .= V0 sin rot
(a) Circuit
At any time (t) the p.d. (V) across the capacitor is given by:
V
=
V0 sin OJtand the charge (Q) at this instant is given by:
Q = CV = CV0 sin OJt
therefore the current (I) at this instant is given by: I= dQ
dt
=
~
(CV0 sin OJt) dt=
cvo~
(sin OJt) dt= 0JCV0 cos OJ t
. . I
=
I0 cos OJ t (where I0 = OJ CV0)This shows that Vis a sine function while I is a
cosine function, and they vary with time as shown in
Figure 3.139(b). So we see that the p.d. is a maximum
when the current is a minimum. This can be
understood if we remember that the charge flowing into a capacitor is only opposed by the p.d. across it. When the capacitor isuncharged, there is no p.d. and therefore no opposition to current, so the current is maximum. When the capacitor is fully charged, the p.d. across it reduces the current to zero.
V.I
V= V0 sin rot
t ,
(b) Variation of p.d. and current with time
Figure 3.138 Alternating p.d. applied across a pure resistor
c
I'---rv
·
--...J
0V= V0 sin rot
I =10 cos rot
(a) Circuit (b) Variation of p.d. and current with time
© Sir Farook
• In a purely capacitive circuit, the current (/) leads the p.d. by 90° (rt/2 rad)- current must flow into the capacitor before a p.d. can be developed across it. • The opposition to flow of a.c. produced by a pure
capacitance is known as the capacitive reactance (Xc)·
• Xc = V nns = Vo =
_ l
=
_2_
Inns !0 coCV0 coC
and since co= 2nf, where fis the a. c. frequency:
Equation 1
You should note that:
• Xc decreases with increasing frequency and capacitance. At high frequencies, there is little time for a large p.d. to build up on the capacitor.
Similarly, larger capacitors can store more charge for a given rise of p.d., so they present less opposition to the current than smaller ones.
1 1
• Xc
=
2nfC=
2nf%units: 1
=
V = .Qs-1cV-1 A
Pure inductor of inductance
L
A pure inductor is one which has zero resistance to direct current - i.e. no p.d. is developed across it due to a direct current. An alternating current, however, causes an alternating magnetic field, which induces a back e.m.f.," which presents opposition to the current. The back e.m.f. will be a maximum when the rate of change of current is a maximum. The rate of change of current is a maximum at the instant that it is changing from positive to negative, i.e. when it is zero. Therefore we have another situation in which current and p.d. are out of phase, as shown in Figure 3.140. L I
L...---rv--_.
V = V0 sin OJt (a) CircuitFigure 3.140 Alternating p.d. across a pure inductor
The alternating current I= I0 sin cot through the inductor sets up a varying magnetic flux which links up with the coil and induces a back e.m.f. in it whose value is given by
E=-L dl
dt
If V
=
applied p.d. at any time t then applying Kirchhoff's second law, we have that:dl V - L - =0 dt V-L
_i
.
(1 0 sin cot)=
0 dtV- coL/0 cos cot= 0
V
=
coL/0 cos cotV
=
V0 cos cot (where coL/0=
V0) In this case V is a cosine function and I is a sine function. They vary with time as shown in Figure 3.140(b).• In a purely inductive a.c. circuit the p.d. (V) leads the current (/) by 90° (rt/2 rad).
• The opposition to a.c. presented by a pure
inductance is known as the inductive reactance XL. V V coL/0 _ . ·L·
• XL = ~= ___Q_;: - - - - co
Inns Io Io
(and since co= 2nj)
s-1
H
Equation 2
You should note that:
• XL increases with increasing frequency and inductance. Higher frequencies mean a higher rate of change of current, and therefore higher back e.m.f. Higher inductance also causes a higher back e.m.f., which leads to a higher opposition to current - higher reactance;
V,I
' ' l
© Sir Farook
E • XL= 2nfL = 2nf di / /dt s-Iy units:v
=n
Phasor diagrams
In this type of diagram the alternating current in an a. c. circuit and the p.d. across the circuit components are represented by vectors (or phasors) which show. the phase relationship between the two quantities. One ofthe quantities (usually the current) is drawn as the reference vector. The other quantities (the p.d.s) are drawn as vectors at angles representing their phase difference from the reference vector.
Figure 3.141 shows the phasor diagram for a.c. circuits which are: (a) purely resistive, (b) purely
capacitive and (c) purely inductive. In (b) arid (c),
notice that the current and p.d. vectors are drawn at 90°, since the phase difference between p.d. and
current is 90°.
R-C series circuit
Consider an alternating p.d. (V) of frequency (j)
R
c
I
applied acro~s a resistor of resistance (R) and a capacitor o(capacitance (C) connected in series as shown in Figure 3.142(a). The phasor diagram for this circuit is shown in Figure 3.142(b). In any series circuit the current (I) in the circuit is the same through each component and it is therefore used as the
referenc'e phasor. R c Vc = IXc I ~---~~._---~
v
(a) R-C series circuit
(b) Phasor diagram
Figure 3.142 R-C series circuit
Applying Pythagoras's theorem to the phasor diagram we have: V2
=
V~ + Vd =12R2 + /2Xd=
/2 (R2+ XJ) L I..
Vc = IXc..
.
I I I Iv
v
v
I I goo ("h rad) Vc = IXc goo ("h rad) I(a) Pure resistance (b) Pure capacitance (c) Pure inductance
© Sir Farook
from which we can define the impedance (Z) of theR-C circuit as:
Equation 3
You should note that:
1
• Xc=--2rtfC
• The impedance, (Z) is the total opposition to a.c. flow due to resistance (R) and capacitive
reactance (Xc) which the circuit presents - Z is in ohms(Q).
• The current ([) in this circuit leads the applied p.d. (V) by a phase angle (t/J) givenby:
tan t/J
=
V c = IXcVR IR
Equation 4
R-L
series circuitIn this case an alternating p.d. (V) offrequency (j) is applied across a resistor of resistance (R) and· an inductor of inductance (L) connected in series. Figure 3.143 shows the circuit and its phasor diagram. From the phasor diagram we have:
from which we cari define the impedance (Z) of the
R-L circuit as:
Equation 5
R L
I
v
(a) R-L series circuit
Figure 3.143 R-L series circuit
You should note that:
• XL
=
2rtfL• The impedance (Z) in this case is the total opposition to a.c. due to resistance and inductive reactance.
The current (!) in this circuit lags behind the applied p.d. (V) by a phase angle ( t/J) given by:
tan t/J= VL
=
IXLVR IR
Equation 6
R-L-c
series circuit. The alternating p.d. Vis now applied to a resistor, a capacitor and an inductor connected in series as
shown in Figure 3.144 on page 306. From the phasor diagram we have:
y2 = V~ + (VL-Vc)z
=
/2R2 + 12 (XL-Xc)2=
P[
Rz + (XL-Xc)z]V =I
~R
2 +(XL-Xc)2from which we can define the impedance (Z) of the·
R-L-C circuit as: ·
Equation 7
The impedance (Z) in this case is the total opposition to a.c. due to resistance, inductive reactance and
capacitive reactance.