Resistance in the National Grid In the UK national grid, electrical energy is transferred through a circuit with a very high potential difference (400,000V) and a very low current. This is said to be because it reduces the heat loss.
However, resistance is "a measure of the difficulty to pass an electric current through that conductor". Surely the resistance would be proportional to the heat loss because resistance and heat loss would both come as a result of an increase in the number of collisions between electrons in the circuit and the wire.
But resistance isn't proportional to heat loss because, in this situation (with a high potential difference and low current), the resistance is very high (due to $V=IR\rightarrow R=\frac{V}{I}$) and the heat loss is very low.
So my question is: what actually affects the resistance and why isn't it proportional to heat loss?
 A: For a given current, the heat loss in the transmission line is indeed proportional to the transmission line's resistance:
$$P_T=I^2 R_T\ \,$$
where $P_T$ is the power lost in the transmission line as heat, $I$ is the current, and $R_T$ is the resistance of the transmission line.
The transmission line's resistance actually isn't very high; the question arrives at that incorrect conclusion by using the wrong voltage in Ohm's law. What would be valid is
$$R_T=\frac{V_T}{I}\ \ ,$$
where $R_T$ is the resistance of the transmission line, $I$ is the current, and crucially, $V_T$ is the voltage across the transmission line, not the 400 kV source voltage. The 400 kV source voltage is the sum of $V_T$ and the voltage across the load,
$$V_S=V_T + V_L\ \ ,$$
and $V_T \ll V_S$, i.e. $V_L \approx V_S$.
According to Watt's law, the current through the load is
$$I=\frac{P_L}{V_L}\ \ ,$$
where $P_L$ is the power needed by the load. (For simplicity, I'm treating the load as being purely resistive, with no capacitive or inductive reactance.) But the current through the load is the same as the current through the transmission line, so you can plug $I$ into the first equation above to give
$$P_T=\left (\frac{P_L}{V_L}\right )^2 R_T\ \ .$$
I.e., for a given $P_L$ and $R_T$, the way to make $P_T$ be small is to make $V_L$ be large, which means you need to make $V_S$ be large.
