For example, if 10 kW at a voltage of 100 V is sent to a factory and the transmission wire has a resistance of 100 Ω, we would have the situation shown in Fig. 28.12.
In order to send the 10 kW power at 100 V down this line, the current required could be worked out with the formula P = IV.
Rearranging this, I = P
________ 100
__ V .
Therefore I = 10 000 = 100 A
Putting this current into our formula for power: P = I2R: P = (100)2 × 100 = 1 × 106 W (1 MW)
Basically 100 A would cause the transmission wire to become red hot and lead to the loss of 1 million watts of energy.
Using a transformer to step up your voltage at the generating station to 10 kV, you would have:
I = P
__ V
I = 10 000 = 1 A
_________ 10 000
Using this new current, power loss would be much more efficient: P = (1)2 × 100 = 100 W
A small amount of power is lost when using transformers but overall the benefits outweigh the losses in long-distance electricity transmission.
Appliances that require higher/lower voltages than domestic supply
There are many appliances that require voltage changes in order to work. One of the most common is the large black plug used for charging mobile phones or for computer speakers. These are sealed transformer units.
To demonstrate the operation of a transformer
1. Set up the transformer with an a.c. input source and a voltmeter in parallel to the input and output, in order to measure the voltages (Fig. 28.13).
2. Apply various voltages and note how the ratio of input to output voltages adheres to the ratio of primary and secondary turns.
3. This can be verified by substituting the voltage figures into the following formula:
___ Vo
Vi =
Np ___
Ns soft iron core a.c.
input lower voltage
primary coil FIG. 28.13
alternating magnetic field
induced a.c.
output higher voltage
secondary coil
supply a.c.
Total R = 100
100 V 10 kW
FIG. 28.12 A circuit diagram of a 10 kW, 100 V supply being sent down a 100 Ω wire
