# Low power loss in electricity transmission lines [duplicate]

To reduce the heat lost during transmission of electricity, we say we increase the voltage of transmission, taking the formula $I^2R$ in consideration. Couldn't I consider $V^2/R$? If I consider the second form, increasing voltage will increase the power dissipated. No?

• You have assumed that $V$ is the potential difference across the transmission line. This is wrong. You aren't increasing that $V$. $V$ across the power line adjusts itself so that $V=IR$. – Yashas Mar 12 '17 at 11:56
• Possible duplicate of Electric power transmission – Yashas Mar 12 '17 at 12:38
• Duplicate physics.stackexchange.com/q/248229 – Farcher Mar 12 '17 at 12:58
• Sorry for duplicate. I tried searching for it at first, but didn't reached the exact post. – Reeshabh Ranjan Mar 12 '17 at 13:02

$P = I^2R$ gives the power consumed by the transmission lines if $I$ is the current through the transmission line and $R$ is the resistance of the transmission line.

$P = \frac{V^2}{R}$ gives the power consumed by the transmission lines if $V$ is the voltage across the transmission line only and $R$ is the resistance of the transmission line.

We aren't increasing the voltage across the transmission line. We are increasing the voltage of the source. The following image should clear your doubts.

The above diagram is an oversimplification of real world transmission lines and load. However, the above diagram is adequate enough to show where the question asker has made a mistake without complicating the answer.

Let the resistance (or impedance) of the transmission line be $R_t$

Let the resistance (or impedance) of the load (the devices used by industries, homes, etc) be $R_l$

Let the R.M.S voltage drop (potential difference) across the transmission line be $V_t$

Let the R.M.S voltage of the source be $V$

Let the R.M.S current through the transmission line be $I$

From Ohm's law, we have:

$$I = \frac{V}{R_t + R_l}$$

The total power consumed by all processes is given by

$$P_{total} = VI$$

The voltage across the transmission line is given by:

$$V_{t} = IR_{t} \tag{1}$$

The power dissipated by the transmission line is given by:

$$P_t = \frac{V_{t}^2}{R_{t}}$$

Since $I$ is reduced, the voltage across the transmission line is reduced (deducible from equation $(1)$). Therefore, $\frac{V_{t}^2}{R_{t}}$ also reduces. Thus, we are saving power.

You need to consider what an electricity transmission line is actually for. It is to transmit power from one place to another and will be designed to have a maximum power rating based upon its specific function in the transmission grid. For example a new 200MW electricity generating station in coastal town A needs to connect to the existing grid which is 20km away. So the designers of the line need it to cope with up to 200MW in the most cost effective manner. They will have a choice of various transmission voltages based upon "standard" values which are related to things like the existing system voltages and transformers typically manufactured. You might have a choice of, say, 66kV, 132kV or 275kV. For transmission of the same amount of power (200MW) choice of a higher voltage means the line current will be lower which means conductors can be smaller hence cheaper however this is offset by the cost of the insulators which will be bigger and more expensive. The towers may also need to be higher owing to greater clearance from the ground required for higher voltages. In the real world a number of financial factors will be taken into account not just the cost of the power loss otherwise we would just make the conductors thicker and the voltage as high as possible which would of course be extremely expensive and totally uneconomic.

No, I believe you are misinterpreting something. For same power to be transmitted in simple terms,

$$VI=constant$$

$V$ = voltage between source and sink.

So higher the votage drop, lesser would be the current.

The tricky part is that when we talk about current, it's passing through the transmission line and being a long wire it has a good amount of resistance so some voltage drop would also be present over it and some votage drop across the town.

Say $R$ is the resistance of the wire and $r$ be of the town.

power loss in wire $= I^2R$

Voltage drop across wire = $IR$

So here you see that some voltage drop across wire is getting wasted.

But by making the voltage of transmission very large, we make the current small and voltage drop across the wire decreases and hence on using $(V')^2/R$, we get a smaller power consumption.

Note that $V'$ is the voltage across the wire.

You would get similar result as the voltage across the wire is in the term and not the voltage across source and sink.