# 16.2.1 Power Transmission

Electric power is generated in power plants, but consumed at households and factories. To connect the two, very long transmission lines (easily hundreds of kilometres of copper cables) are required.

Copper may be a fantastic conductor, but over such long distances, the total resistance R is not negligible (think $\displaystyle R=\rho \frac{L}{A}$). As a result, the power lost in transmission cables can be substantial. Since the power lost in the cable is given by $P={{I}^{2}}R$, where I is the current in the cable, it is obvious we should keep I as small as possible to minimise power loss. Since the total power through the transmission line is given by $P=VI$,where V is the voltage the transmission cable is held at, then it is obvious that delivering power using low currents requires high voltages.

For this reason, the outputs of electric generators (at the power station) are first stepped up to extremely hight voltages (up to 500 kV) before they are put on the transmission lines. At the consumer end, the voltage has to be stepped back down to low voltages (rms 230 V for Singapore) for safety and insulation considerations. The device that accomplishes the necessary voltage conversion is called the transformer.

Example

If the total power supplied over a 100 Ω transmission is 2 MW, what is the power lost and voltage drop in the cable, if the transmission line is operated at

a) 500 kV

b) 20 kV

Solution

a)

Current in the transmission cable is $\displaystyle I=\frac{P}{V}=\frac{{2\times {{{10}}^{6}}}}{{500\times {{{10}}^{3}}}}=4.0\text{ A}$

Power dissipated (as heat) in transmission cable is $P={{I}^{2}}R={{4.0}^{2}}(100)=1.6\text{ kW}$

Voltage drop across transmission cable is $\displaystyle V=IR=(4.0)(100)=400\text{ V}$

This is a satisfactory situation. The actual power delivered to the end users is $2\text{ MW}-1.6\text{ kW}=1.9984\text{ MW}$. Only a tiny fraction of the total 2 MW delivered is wasted heating up the cable. The voltage at the end of the transmission cable is $500\text{ kV}-400\text{ V}=499.6\text{ kV}$, so it remains close to the design voltage of 500 kV.

b)

Current in the transmission cable is $\displaystyle I=\frac{P}{V}=\frac{{2\times {{{10}}^{6}}}}{{20\times {{{10}}^{3}}}}=100\text{ A}$

Power dissipated (as heat) in transmission cable is $P={{I}^{2}}R={{100}^{2}}(100)=1.0\text{ MW}$

Voltage drop across transmission cable is $\displaystyle V=IR=(100)(100)=10\text{ kV}$

This is a totally unacceptable situation. 1.0 MW, representing half the total power output, is wasted as heat in the transmission cable. The voltage at the end of the transmission cable is 10 kV, far from the design voltage of 20 kV.

Concept Test

3221

Interesting

War of the Currents: Tesla vs Edison