Page 469 - Physics Coursebook 2015 (A level)

Chapter 29: Alternating currents
To find the r.m.s. value of I, we now take the square root of <I2>. This introduces a factor of the square root
o f 12 , o r 12 .
Summarising this process: to find the r.m.s. value of the current, we find the root of mean of the square of the current – hence r.m.s.
Whyusea.c.forelectricity supply?
There are several reasons for preferring alternating voltages for a national electricity supply system. The
most important reason is that a.c. can be transformed
to high voltages, so that the current flowing is reduced, and this leads to lower power losses in the transmission lines. Typically, the generators at a power station produce electrical power at a voltage of 25 kV. This is transformed up to a voltage of perhaps 400 kV (and as much as 1 MV in some countries). The power is then transmitted along many kilometres of high-voltage power lines (Figure 29.11) before being transformed down to a lower voltage for supply to the millions of consumers. The transformers used for increasing and decreasing the voltage are discussed in detail in the next section.
This high voltage brings problems: the lines must be suspended high above the ground between pylons, and high-quality insulators are needed to prevent current passing from the cables to the pylons. As current flows through transmission lines (wires), it loses power because of the resistance of the lines. The wires become warm; this is resistive or ohmic heating, as discussed in Chapter 9. The smaller the current, the smaller the losses. This is illustrated in Worked example 2.
Figure 29.11 Power lines carry electricity from power station to consumer.
WORKED EXAMPLE
2
A power station generates electrical power at a
rate of 10 MW. This power is to be transmitted along cables whose total resistance is 10 Ω. Calculate the power losses in the cable if the power is transmitted at50kVandat250kV.
Step1 UsingI= PV,calculatethecurrentflowingin each case:
for50kV: for250kV:
I = 10×106 =200A 5×104
I = 10×106 =40A 25×104
Step2 UsingP=I2R,calculatethepowerlossesin each case:
for50kV: for250kV:
P=2002 ×10=4×105 =400kW P=402 ×10=1.6×104 =16kW
Hint: Take care! Note that we have two quantities
for which we are using the symbol P: the total power being transmitted, and the power lost in the wires. Notice that using a higher voltage does not change the resistance of the cables.
We have shown that, by increasing the voltage by a factor of 5, we have reduced the power losses by a factor of 25.
Economic savings
The resistive heating of power lines is a waste of money, in two ways. Firstly, it costs money to generate power because of the fuel needed. Secondly, more power stations are required, and power stations are expensive. The use
of transformers to transform power to high voltages saves a few per cent of a national bill for electrical power, and means that fewer expensive power stations are needed.
It is claimed that having a few, very large power stations gives economies of scale, but this is debated by many environmentalists who would prefer to see many small, local power stations. It is also the case that new developments in technology are making it easier to transform direct current to high voltages. This is more compatible with sustainable electricity generating systems such as photovoltaics (solar cells), so we may see the development of d.c. grid systems in the near future.
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