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Imagine we want to supply a small town with electricity. The average power needed by the town is 100 MW. The National Grid has to supply this power. If we use a small transmission voltage (of 20 kV), there will need to be a large current in the transmission cables (5000 A). This will make the transmission lines very hot and waste energy. The wasted power increases with the square of current.
If we use a high transmission voltage, then the current that needs to flow to the town will be less (and still deliver the same power). A smaller current causes less heating in the transmission cables. So there is less waste.
The table below shows the calculations.
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| Quantity |
Formula used |
Low V |
High V |
| Power requirement of town (W) |
Pout |
100, 000, 000 |
| Supply voltage at the town (V) |
Vout |
20, 000 |
400,000 |
| Current needed to provide the power (A) |
Pout = Vout x I |
5, 000 |
250 |
| Joule heating in power cables (W) |
Pwaste = I2R |
75,000,000 |
187,500 |
| Power that the power station needs to supply (W) |
Pin = Pwaste + Pout |
175, 000, 000 |
100,187,500 |
| Efficiency of system (%) |
eff = (100 * Pout) ÷ Pin |
57.1 |
99.8 |
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| Table 4. Worked examples for a low transmission voltage (20 kV) and a high voltage (400 kV). Resistance of cable is taken to be 3W. |
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Notice that when the transmission voltage is 20 times bigger, the current needed is reduced by a factor of 20 as well. However, the wasted power is reduced by a factor of 400 (= 202). This is because the joule heating depends on the square of the current.
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