# Power dissipation in circuits:Why is high voltage used in power lines?

I know this question has been asked before, but there is one doubt I still cannot clear.
Power dissipation is proportional to $I^2R$.
Does this not mean that it is also proportional to $V^2/R$?

If yes, how does increasing the voltage decrease power loss?

## 1 Answer

If you express power loss in a power line as $V^2/R$, the $V$ in that expression is the voltage difference between the two ends of the power line, not the voltage difference between the power line and ground.

To supply a fixed amount of power $P_L$ to a load, if the voltage at the load $V_L$ is larger, the current $I=P_L/V_L$ can be smaller. If the power line has a resistance $R$, the voltage across the line is $V_w=IR$, which is smaller for a smaller current. Thus, to supply a fixed amount of power to the load, if the voltage at the load $V_L$ is higher, the voltage drop across the power line $V_w$ is smaller, so the power loss in the power line, $V_w^2/R$, is smaller.

Putting all those pieces together, the power loss in the power line is

$$P_w=\frac{V_w^2}{R}=\frac{(IR)^2}{R}=I^{2}R=\left(\frac{P_L}{V_L}\right)^2 R\ .$$

• It's also worth noting that distribution transformers typically have greater than 96% efficiency over most of their operating range, so voltage conversion is not very costly in terms of power losses for AC distribution lines. – Edward Oct 2 '14 at 15:14
• Can you expand on the role of DC vs. AC and AC frequency in the R component, and the inductance issues? I've always heard that the U.S. power system would be more efficient if we were running higher than 60Hz but can't (cost) efficiently afford the conversion to higher frequency distribution. – IDisposable Oct 2 '14 at 18:04
• @IDisposable How power line efficiency varies with frequency seems like a different topic to me than how power line efficiency varies with voltage. Could you perhaps turn your comment into a new, separate question? – Red Act Oct 2 '14 at 18:25