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Power lines basics

Electrical energy transmitted from power station along power lines (Aluminium). Lines are engineered as excellent conductors but huge length involved means some resistance. A current passing through a resistance will cause a power loss:

$$\begin{align} P_{\text{loss}}=I^{2}R\ Js^{-1},W \\ \end{align} $$

This loss comes in the form of primarily heat loss. To minimise power loss, transmit power with as low a current as possible.

$$\begin{align} P=VI \end{align} $$

Note that the voltage drop between two points is:

$$\begin{align} V_{\text{drop}}=IR \end{align} $$

Farmhouse problem

$$\begin{align} P_{\text{in}}=VI \therefore I_{\text{in}} = \frac{P}{V}=\frac{2400}{240}=10A \\ \therefore P_{\text{loss}}=I^{2}R=10^{2}\times2=200W\text{ lost} \\ \therefore P_{\text{out}}=P_{\text{in}}-P_{\text{loss}}=2200W \\ \\ V_{\text{drop}}=IR=10\times 2 = 20V \\ \therefore V_{\text{out}}=V_{\text{in}}-V_{\text{drop}}=240-20=220V \\ \therefore I_{\text{out}}=\frac{P_{\text{out}}}{V_{\text{out}}}=\frac{2200}{220}=10A \\ \text{Which should make sense, as current should be the same.} \end{align} $$ $$\begin{align} P_{\text{in}}=VI \therefore I_{\text{in}}=\frac{P}{V}=\frac{2400}{2400}=1A \\ \therefore P_{\text{loss}}=I^{2}R=1^{2} \times 2=2W \end{align} $$