Cambridge A2 Physics

1. 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.

2. (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.

3. 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

4. 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 2ft I = I0sin 2(t/T)
where, I = current at any time t in amperes A
I0 = peak current
 = 2f, 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

5. 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

6. 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 = I2
R generated in a resistance R
P = Io
2
R sin2 t
• Because Io
2 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

7. 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, the
effective value of the current (Irms or Ir) is thus an average value of currents I
throughout one whole cycle of the current’s sinusoidal wave.
Pac = Idc
2R = Ir
2R
 Ir
2 = average value of I2
= I0
2/2
 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

8. R.m.s. the mathematics
The r.m.s value ofa function f(x) within the range a  x  b is givenby:
 
2
1
2
)(
1






  dxxf
ab
b
a
If f(x) = sin xand integrating for xbetween x= 0 and x=
2

gives
 
2
1
2
2
0
sin
02/
1








  dxx


=  
2
1
2
0
)2cos1(
2






 dxx


=
2
1
2
1











=
2
1

9. Mean power of a.c.
• Mean power Pac = Ir
2R = (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
I0
2
R P =( I0
2
.sin2
t).R
1/2 I0
2
R
0 t

10. 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

11. 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 a
fluctuating magnetic field which in turn induces a back e.m.f. Ep. The current
Ip in the primary coil is given by:
Vp - Ep = Ip.Rp ( Rp = primary coil resistance )
As p = Np
Ep = dp/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 iron
core linking the coils.

12. cont..
• At the secondary coil where it is connected to a load, the output voltage Vs is
given by:
Vs = Es - IsRs ( Rs = secondary coil resistance )
and Es is the mutually induced e.m.f. in the secondary coil.
 Es = ds/dt = Nsd/dt
• Again, for an ideal transformer, Rs  0 giving Vs  Es.
i.e. Vs = Nsd/dt
 Vs/Vp = Ns/Np

13. cont..
• The voltage Vp applied to the primary, from the source current, is used simply
in 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, in
mechanics, that action and reaction are equal and opposite.)
• For an ideal transformer (i.e. 100% efficient), the power supply in the primary
coil 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

14. Energy Losses and Efficiency in Transformer
• There are 4 main losses:
(a)Heat is lost in coils (primary and secondary) due to resistance of the
windings. For transformers handling very high electrical power, the windings
are made of very thick wires to reduce power lost as heat. The windings are
insulated and immersed in oil for cooling purpose.
(b)The alternating flux  in the primary induces eddy current in the iron core that
causes heat loss.
(c)The magnetisation and demagnetisation of the iron core give rise to the
hysteresis 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 secondary
output as flux leakage occurs. (does not pass through iron core).
• Efficiency = (power in secondary/power in primary) x 100%

15. Electrical power transmission
• When electricity is transmitted from a source, such as a power station to a
distant load, such as a factory or household, power is lost as Joule heating
I2R through the transmission cables where R is the total resistance of the
cables.
• Suppose the electrical power generated Pgen is to be delivered at a p.d. of V
by the supply lines of total resistance R. The current in the supply line will
be:
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 result
in higher insulation cost. The result is a solution taking cable resistance, the
voltage of transfer and insulation cost into consideration.

16. 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 of
overhead lines whose total resistance is 5.0 .
(a) Calculate (i) the current in the wires (ii) the voltage between the terminals
at the far end of the lines.
(b) State, in each case, a reason why the designers of the transmission
system 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 power
cables underground. (ii) State one disadvantage, other than high cost, of
laying 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, possible
discharge in air.
(c) (i) Airfields; wide stretches of water.
(ii) Difficult to dissipate heat; insulation problems; risk of damage by
digger; difficulty of access if faults arise; biological effects of
electric/magnetic fields/radiation from currents near ground level.

17. Electrocution
• Electrocution is actually due to the amount of current that flows through the
body.
• The amount of current depends on the resistance offered by the person
between the wire and the earth.
• A current of 0.1 A is able to cause death due to fibrillation (uncontrolled
contractions of the heart).
• People touching live wires may get their hand stuck to the wire due to
contraction of the muscles. It is therefore current, not voltage, which is
dangerous.

18. 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 to
d.c.) is necessary through use of appropriate rectifiers (diodes)
• Alternating current can be converted to direct current (i.e. rectified) by
making use of devices which conduct appreciable amounts of current in one
direction only. Such devices are called rectifiers and include thermionic
diodes, metal rectifiers and semiconductor diodes.
• A rectifier is an electrical device which converts alternating current to
direct current, a process known as rectification. Rectifiers are used as
components of power supplies and as detectors of radio signals.
• Rectifiers may be made of solid state diodes, vacuum tube diodes, mercury
arc valves, and other technologies.

19. 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 throughrectifier
Low-resistance when
Forward-biased
O PD across rectifier
High-resistance whenreverse-biased

20. 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

21. 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

22. 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

23. cont..

24. 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.