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Let us suppose that a transmission line is carrying power to a resistor $R$ The resistor $R$ requires a power of $P=IV$.

Applying the most simple case where a transmission line is supplying a current $I$. The Ohmic loss is $P_c=I^2R_c$ for the cable.

Putting $I=P/V$

$P_c=(P^2/V^2)R_c$ So to decrease Ohmic loss we need to increase $V$ but this is the potential difference across the resistor, right? I have read other questions here, but it still doesn't make sense. Please help.

What my textbook says

Consider a resistor R to which a power P is to be delivered via a transmission cable having res. $R_c$ to be dissipated finally. If $V$ is voltage across $R$ and $I$ the current then

$P=VI$

The connecting wires from the power station to the resistor has res. $R_c$. The Ohmic loss is $P_c=I^2R_c$

$P_c=(P^2/V^2)R_c$

Thus to drive a device with $P$ the power wasted in the connecting wires is inversely proportional to $V^2$.

A few paragraphs later...

The transmission cables from power stations are hundreds of miles long and their resistance $R_c$ is considerable. To reduce $P_c$, these wires carry current at enormous values of $V$

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Daivya Jadeja
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2 Answers2

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to decrease Ohmic loss we need to increase V but this is the potential difference across the resistor, right?

One option is to transmit the power over long distances at high voltage, then use a transformer near the load to reduce the voltage to whatever is required by the load.

For example, power may be transmitted across continents at 100's of kV, then reduced by a sequence of transformers to 110 or 220 V to power your home.

If your power source is DC rather than AC, then you could use a switching converter to do something similar. You'd first want to evaluate whether it's cheaper to include that switching converter in your system, or just use a larger diameter wire to reduce the $R_c$ term.

The Photon
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The V in Ohm's law is the voltage drop across the load.

You are trying to reduce the voltage drop and so the power lost along the line.

Martin Beckett
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