Transformers Physics A Level Notes

1

The transformer

laminated soft iron core

Figure 3.131 shows a simple transformer, which consists 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.

k

l:_;•up-p-ly-4~~~~

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.

= vs

oo

Figure 3.131

p.d.

turns= N5 current= / 5

If N5 > NP then Vs > VP - a step-up transformer

If N 5 < NP then Vs < VP - a step-down transformer

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.

Guided examples (2)

1. The primary coil of a transformer has 500

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turns and the secondary coi12000 turns. A 5 V alternating supply is connected to the primary. Calculate the voltage across the secondary.

2.

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 N 5 respectively) as follows:

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

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Use Equation 5 in each case.

Power and current in transformers

In a perfectly efficient transformer, there would be

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Equation 5

800 kVtransformers at a large substation in South America

no energy losses; i.e. the power delivered to the primary would be equal to the power delivered by the secondary.

2

p p =Ps

• 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.

Combining Equation 5 and Equation 6:

A well-designed transformer should suffer energy losses of only 1% or so.

You should note that:

•

NP, VP and IP are the number of turns, p.d. and currentrespectively in the primary coil and N8 , 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

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 the following table:

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•

• 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).

k

Equation 6

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l.e.

P.d.

• 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.

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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:

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• 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

Current /A

N

Power /W

=

1 000 000

100000

=

1 000 000

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

=

1 000 000

1

X

10

X

100

- 1 000 000

Using 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

3

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.

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:

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power loss = current 2 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.

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time

a.c. supply

CJ

~

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Guided example (3)

supply p.d.

k

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.

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.

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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.

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 forwardbiased, and does not conduct. So, for this half of the

B

A

current

time

no current in 'reverse' direction

Figure 3.133

Half-wave rectification

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 of the 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 through the load from A to B, returning to the supply

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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.

supplyp.d.

time

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.

diode 1

diode 2

I

I

current

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.

A

/ +

.----. rm

-~__, A

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+

Smoothing

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centre-tap transformer

k

a.c. supply

I

B

c

time

R

no current in 'reverse' direction

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Figure 3.136 Figure 3.134

Full-W£ZV!? rectificatimt

a.c. supply

u

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~-~~r~v..~~'~'~'~ -1

Figure 3.135

Full-wave bridge rectifier

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

5

initial charge

charge discharge

smoothed output p.d.

\

k

'

'

oo

A unsmoothed rectified p.d.

Figure 3.137

Smoothing of the output p.d. from a full-wave rectifier

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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.

'·

6

At any time (t) the p.d. (V) across the resistor is given by: V =V0 sin OJt

At any time (t) the p.d. (V) across the capacitor is given by: V =V0 sin OJt

And the current I in the circuit is given

R i.e.

and the charge (Q) at this instant is given by:

I= X= V0 sin OJt

R

Q = CV = CV0 sin OJt where I0 =Vo R

I= I 0 sin OJt

therefore the current (I) at this instant is given by: I= dQ dt

Thus both I and V are sine functions which vary with time as shown in Figure 3.138.

• 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

dt

(CV0 sin OJt)

=cvo~ (sin OJt) dt

= 0JCV0 cos OJ t

..

I

=I 0 cos OJ t

(where I 0 = 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.

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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.

=~

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• In a purely resistive a.c. circuit the p.d. (V) and the current (I) are in phase.

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by:

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R

V.I

V= V0 sin rot

I .

t,

V.= V0 sin rot

(a) Circuit

(b) Variation of p.d. and current with time

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Figure 3.138 Alternating p.d. applied across a pure resistor

c

I

'---rv·--.....J

0

V= V0 sin rot

I =10 cos rot (a) Circuit

Figure 3.139 Alternating p.d. applied across a pure capacitor

(b) Variation of p.d. and current with time

7

_2_

• Xc = Vnns = Vo = _ l = Inns !0 coCV0 coC and since co= 2nf, where fis the a. c. frequency:

E=-L dl dt If V =applied p.d. at any time t then applying Kirchhoff's second law, we have that:

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• The opposition to flow of a.c. produced by a pure capacitance is known as the capacitive reactance (Xc)·

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

dl V-L- =0 dt

_i. (10 sin cot) =0

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• 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.

V- L

dt

V- coL/0 cos cot= 0

Equation 1

V = coL/0 cos cot

You should note that:

V

units:

= 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).

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• 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% 1 = V = .Q s- 1 cV- 1 A

Pure inductor of inductance L

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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

• 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/ _ . ·L • XL = ~= ___Q_;: - - 0- - 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

I

L...-----rv--_. ''l

V = V0 sin OJt

(a) Circuit

Figure 3.140 Alternating p.d. across a pure inductor

(b) Variation of p.d. and current with time

8

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.

E • XL= 2nfL = 2nf di / /dt

units:

s-Iy

v

=

n

c

k

R

oo

Vc = IXc

I

~------~~._------~

v

(a) R-C series circuit

Phasor diagrams

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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.

(b) Phasor diagram

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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°.

Figure 3.142

R-C series circuit

Applying Pythagoras's theorem to the phasor diagram we have: V2

R-C series circuit

= V~ + Vd =12R2 + /2Xd = /2 (R2+ XJ)

Consider an alternating p.d. (V) of frequency (j)

c

R

I

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I

..

I Vc = IXc

...I

I

I

v

v I

v I

goo ("h rad)

Vc = IXc (a) Pure resistance

Figure 3.141

L

Phasor diagrams

(b) Pure capacitance

goo ("h rad)

(c) Pure inductance

I

9

from which we can define the impedance (Z) of the R-C circuit as:

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. Equation 3

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:

VL VR

=

IXL IR

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tan t/J=

k

The current (!) in this circuit lags behind the applied p.d. (V) by a phase angle ( t/J) given by:

You should note that:

Equation 6

R-L-c series circuit

tan t/J = Vc = IXc VR IR

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.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:

Equation 4

y2 = V~ + (VL- Vc)z

R-L series circuit

= /2R2 + 12 (XL- Xc)2

= P[ Rz + (XL- Xc)z]

In 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.

~R2 +(XL- Xc) 2

V =I

from which we can define the impedance (Z) of the · R-L-C circuit as: ·

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From the phasor diagram we have:

from which we cari define the impedance (Z) of the R-L 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.

Equation 5

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R

L

I

v (a) R-L series circuit

Figure 3.143

R-L series circuit

(b) Phasor diagram