Alternating Current

24. Alternating Currents

Content

24.1 Characteristics of alternating currents

24.2 The transformer

24.3 Transmission of electrical energy

24.4 Rectification

Learning Outcomes

Candidates should be able to:

(a) show an understanding of and use the terms period, frequency, peak value and root-mean-square value as applied to an alternating current or voltage.

* (b) deduce that the mean power in a resistive load is half the maximum power for a sinusoidal alternating current.

* (c) represent a sinusoidally alternating current or voltage by an equation of the form x = xosinωt.

(d) distinguish between r.m.s. and peak values and recall and solve problems using the

relationship Irms = Io/√2 for the sinusoidal case.

(e) show an understanding of the principle of operation of a simple iron-cored

transformer and recall and solve problems using Ns/Np = Vs/Vp = Ip /Is for an ideal

transformer.

(f) show an appreciation of the scientific and economic advantages of alternating

current and of high voltages for the transmission of electrical energy.

* (g) distinguish graphically between half-wave and full-wave rectification.

(h) explain the use of a single diode for the half-wave rectification of an alternating

current.

(i) explain the use of four diodes (bridge rectifier) for the full-wave rectification of an

alternating current.

* (j) analyse the effect of a single capacitor in smoothing, including the effect of the

value of capacitance in relation to the load resistance.

Direct current

• When a battery is connected to a circuit, the current flows steadily in one direction

• This sort of current is known as direct current or in short d.c.• However the domestic and industrial electricity supply produced

by generators at a power station is one which does not use direct current but alternating current

Direct current Alternating current

0 Time

Alternating currents

• An alternating current (a.c.) or voltage is one which continuously reverses its direction of flow after a certain interval of time, that is, it varies in magnitude and direction with time.

• The amplitude or magnitude of such a current (for a a.c of sinusoidal wave form) at any time frame is as follows:

I = I0sin t I = I0sin 2ft I = I0sin 2(t/T)

where, I = current at any time t in amperes A

I0 = peak current

= 2f, angular frequency, rad s-1

f = /2, frequency in Hz or no. of cycles per second

T = 1/f = period in seconds

• The time for one complete alternation, a cycle, is the period T. The number of cycles in one second is the frequency f. The frequency of an alternating current may range from 50 Hz to 100 Hz. For e.m.f., the same formulae apply

• Sometimes the term peak-to-peak is used which means 2Io or 2Vo i.e. twice the amplitude value

Other forms of alternating currents

The following are some other wave forms of alternating currents produced by specially-designed electronic circuits:

I or V 1 cycle I or V 1 cycle

I0

t t

T/2 T

Saw-toothed wave Square wave

Power in a resistor

• From the picture of a sinosoidal wave although it is clear that the average value of an alternating current is zero, it does not mean that when an a.c source is connected to a resistor, no power is generated in the resistor

• Using I = I0sin t and the power P = I2R generated in a resistance R

P = Io2R sin2 t

• Because Io2 and sin2 t are always positive, the power P is always

positive

• The above expression gives the power at any instant of time, but what is much more useful is the average or mean power which is the power generated in the resistor

Root-Mean-Square (r.m.s.) value of an A.C.

• The root-mean-square value of an alternating current is defined as the equivalent value of the steady d.c. which would dissipate energy at the same average rate in a given resistance. It is also called the effective value of an a.c.

• A direct current with a value of I equal to the r.m.s current Irms of an a.c. circuit will produce exactly the same heating effect in a resistor

• Because the current in an a.c. changes direction many times in a second, theeffective value of the current (Irms or Ir) is thus an average value of currents Ithroughout one whole cycle of the current’s sinusoidal wave.

Pac = Idc2R = Ir

2R

Ir2 = average value of I2

= I02/2

Ir = I0/2 = I0/1.414 = 0.707 I0

• Similarly,Vr = V0/2 V0 = 2 Vr

• In specifying a domestic supply voltage, it is the r.m.s. value that is quoted

• If the voltage of an a.c. mains supply is 240 V (r.m.s.), an electrical appliance must be able to stand up to a peak value of 2 x 240 V (= 339.4 V) in order for it to be able to be used on the mains.

• The r.m.s value of the current or voltage is that value of the direct current or voltage that would produce heat at the same rate in a resistor

R.m.s. the mathematics

The r.m.s value of a function f(x) within the range a x b is given by:

2

1

2

)(1

dxxfab

b

a

If f(x) = sin x and integrating for x between x = 0 and x = 2

gives

2

12

2

0sin

02/

1

dxx

=

2

1

2

0)2cos1(

2

dxx

= 2

1

2

1

=

2

1

Mean power of a.c.

• Mean power Pac = Ir2R = (I0/2)2.R = 1/2 ( I0

2R ) = 1/2 P0

• That is, mean power is half the maximum power or peak power for an a.c. passing through a resistive load.

V = V0.sin t I = I0.sin t

0 t

P

I02R P =( I0

2.sin2t).R

1/2 I02R

0 t

Example

• A 1.5 kW heater is connected to the domestic supply which is quoted as 240 V. Calculate the peak current in the heater, and its resistance.

Solution

From P = Vrms x Irms , Irms = 1500/240 = 6.3 A

Hence peak current I0 is 6.3 x 1.414 = 8.8 A

and the resistance R = Vrms/Irms = 240/6.3 = 38 ohms

Transformer principles

• A transformer is a device used for stepping-up (or down) an a.c. supply voltage using the Mutual Induction Principle. Basically it consists of two coils of wires, one called the primary and the other the secondary, of an appropriate number of turns. These coils normally wind round a laminated soft-iron core for better permeability of the magnetic field or flux linkage of the two coils giving a higher flux.

• When an alternating voltage Vp is applied to the primary coil, it sets up afluctuating magnetic field which in turn induces a back e.m.f. Ep. The currentIp in the primary coil is given by:

Vp - Ep = Ip.Rp ( Rp = primary coil resistance )

As p = Np

Ep = dp/dt = Npd/dt

• For an ideal transformer, Rp 0 giving Vp Ep.

i.e. Vp = Npd/dt

where Np is the number of turns in the primary coil and the flux in the ironcore linking the coils.

cont..

• At the secondary coil where it is connected to a load, the output voltage Vs isgiven by:

Vs = Es - IsRs ( Rs = secondary coil resistance )

and Es is the mutually induced e.m.f. in the secondary coil.

Es = ds/dt = Nsd/dt

• Again, for an ideal transformer, Rs 0 giving Vs Es.

i.e. Vs = Nsd/dt

Vs/Vp = Ns/Np

cont..

• The voltage Vp applied to the primary, from the source current, is used simplyin overcoming the back-e.m.f. Ep., if we neglect the resistance of the wire.Therefore, it is equal in magnitude to Ep. (This is analogous to saying, inmechanics, that action and reaction are equal and opposite.)

• For an ideal transformer (i.e. 100% efficient), the power supply in the primarycoil will be fully transferred to the secondary output.

Hence: VpIp = VsIs or Vs/Vp = Ip/Is

• Thus for an ideal transformer,

• So the transformer steps voltage up or down according to its 'turns-ratio'.The voltage may be stepped up from 25,000 to 400,000 volts for high-tension transmission and stepped down from 240 V to 6 V for ringing bells.

s

p

p

s

p

s

I

I

N

N

V

V

primaryinvoltageApplied

secondaryine.m.f.Induced

Energy Losses and Efficiency in Transformer

• There are 4 main losses:

(a)Heat is lost in coils (primary and secondary) due to resistance of thewindings. For transformers handling very high electrical power, the windingsare made of very thick wires to reduce power lost as heat. The windings areinsulated and immersed in oil for cooling purpose.

(b)The alternating flux in the primary induces eddy current in the iron core thatcauses heat loss.

(c)The magnetisation and demagnetisation of the iron core give rise to thehysteresis loss and hence power loss.

(d)When the flux produced by the primary is not 100% linked to the secondary,then the input electrical power will not be fully transferred to the secondaryoutput as flux leakage occurs. (does not pass through iron core).

• Efficiency = (power in secondary/power in primary) x 100%

Electrical power transmission

• When electricity is transmitted from a source, such as a power station to adistant load, such as a factory or household, power is lost as Joule heatingI2R through the transmission cables where R is the total resistance of thecables.

• Suppose the electrical power generated Pgen is to be delivered at a p.d. of Vby the supply lines of total resistance R. The current in the supply line willbe:

I = Pgen/V

• Hence, the power loss as heat will be given by:

Ploss = I2R = (Pgen/V)2R

• The equation indicates that for lower power loss, V has to be high in value.Hence for economic reasons, transmission must be at high V and low I state.But a low I means a thicker and costlier cable while higher voltage will resultin higher insulation cost. The result is a solution taking cable resistance, thevoltage of transfer and insulation cost into consideration.

Example(1) A power station generates a power of 200 MW at a potential difference of

400 kV. This input power is transmitted to a distant town through a pair ofoverhead lines whose total resistance is 5.0 .

(a) Calculate (i) the current in the wires (ii) the voltage between the terminalsat the far end of the lines.

(b) State, in each case, a reason why the designers of the transmissionsystem did not choose an input voltage of: (i) 240 V (ii) 2.0 MV.

(c) (i) Give one example of a situation where it is essential to route powercables underground. (ii) State one disadvantage, other than high cost, oflaying power cables underground.

Solution

(a) (i) I = P/V = 200 x 106/(400 X 103) = 500 A

(ii) Voltage drop = IR = 500 x 5.0 = 2500 = 2.5 kV

Voltage between terminals at far end of lines = 400 – 2.5 = 397.5 kV

(b) (i) Current high, hence requires thick expensive cables.

(ii) Need tall pylons, wide cables’ spacing, costly insulation, possibledischarge in air.

(c) (i) Airfields; wide stretches of water.

(ii) Difficult to dissipate heat; insulation problems; risk of damage bydigger; difficulty of access if faults arise; biological effects ofelectric/magnetic fields/radiation from currents near ground level.

Electrocution

• Electrocution is actually due to the amount of current that flows through thebody.

• The amount of current depends on the resistance offered by the personbetween the wire and the earth.

• A current of 0.1 A is able to cause death due to fibrillation (uncontrolledcontractions of the heart).

• People touching live wires may get their hand stuck to the wire due tocontraction of the muscles. It is therefore current, not voltage, which isdangerous.

Rectification

• A.C. is important and useful in power generation and distribution since a.c.can be stepped up for minimum power-loss transmission.

• For electrical and electronic devices operating on d.c. sources only (e.g.radio, television, computers etc.), rectification of the a.c. (i.e. to change it tod.c.) is necessary through use of appropriate rectifiers (diodes)

• Alternating current can be converted to direct current (i.e. rectified) bymaking use of devices which conduct appreciable amounts of current in onedirection only. Such devices are called rectifiers and include thermionicdiodes, metal rectifiers and semiconductor diodes.

• A rectifier is an electrical device which converts alternating current todirect current, a process known as rectification. Rectifiers are used ascomponents of power supplies and as detectors of radio signals.

• Rectifiers may be made of solid state diodes, vacuum tube diodes, mercuryarc valves, and other technologies.

cont..

• A rectifier is said to be forward-biased when it is connected to a power supply in such a way that it conducts. If connected the other way, the rectifier is reverse-biased. The current-voltage curve of a typical rectifier is shown below:

Current through rectifier

Low-resistance when

Forward-biased O PD across rectifier

High-resistance when reverse-biased

Half-wave rectification by a single diode

• The rectifier conducts only during the half cycle which means that the output across the load will consist of only the positive half-cycles. Although the output is pulsating, it is unidirectional, i.e. direct current.

X

Alternating supply Load

Y

Supply PD

O t

PD across load

O t

Full-wave rectification• It is more satisfactory also to make use of

the negative hal-cycles as well and this can be achieved by using an arrangement of 4 rectifiers (diodes) known as a bridge rectifier.

• When P is positive, diodes across PQ and SR conduct; when R is positive, diodes across RQ and SP conduct. In each case the current through the load is in the same direction – from Q to S. The p.d. across the load has the form shown below.

• Thus, full-wave rectification allows the load to draw current from the supply on each half of each cycle and therefore the power that can be utilized is double that achieved with half-wave rectification.

P.D. across load

O t

Smoothing by a single capacitor

• The pulsating unidirectional rectified current output produced by both half-wave and full-wave rectifiers is still not a good approximation to the steady direct current required for most electronic equipment

• It can be made more steady (smoothed) by inserting a suitable capacitor in parallel with the load or across the output terminals of the bridge circuit

• The effect is to reduce the fluctuations in the unidirectional output• Generally a larger value of the capacitor will give better smoothing although the

more important factor is the resistor-capacitor time-constant

X

Pulsating Smoothing Load Current Rectified p.d. capacitor

cont..

cont..• As the rectifier voltage increases, it charges the capacitor and also supplies current to

the load. At the end of the quarter cycle the capacitor is charged to its peak value Vm of the rectifier voltage. Following this the rectifier voltage starts to decrease as it enters the next quarter cycle. This initiates the discharge of the capacitor through the load.

• At points such as A the p.d. across the load has just reached its maximum value. If the capacitor were not present, the p.d. would start to fall to zero along the broken curve. However, as soon as the p.d. across the load starts to fall, it becomes less than that across the capacitor and the capacitor starts to discharge through the load. Since the charging process causes plate X to be positive, the discharge drives current through the load in the same direction as it flowed during charging.

P.D. across Smoothed p.d.

load A Ripple voltage

O time Unsmoothed half-wave rectified p.d.

P.D. across Load A Smoothed p.d.

O time

Unsmoothed half-wave rectified p.d.