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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?

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    $\begingroup$ The 40 kV is between the transmission line wire and ground, not between one end of the wire and the other. $\endgroup$ – The Photon May 6 '18 at 14:19
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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.

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  • $\begingroup$ Thank you very much for your answer. One question: are you saying that the voltage in the transmission lines is not equivalent to the voltage originally given by the source? Also, what exactly do you mean by the voltage across the load? $\endgroup$ – Dan May 6 '18 at 14:59
  • $\begingroup$ @Dan By "voltage across the load", I mean the voltage between a pair of transmission lines at their destination (which is a transformer). (I'll ignore the complexities of three-phase power transmission.) The voltage between a pair of transmission lines is almost as large at their destination as it is at the source (a generator). But the voltage that the heat loss in a transmission line depends on is the voltage between the source end of the transmission line and the destination end of the same transmission line, which is much smaller than the voltage between two transmission lines. $\endgroup$ – Red Act May 6 '18 at 15:49

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