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Why are high voltage power lines used? Electricity is generated at 11,000V...so is that a potential at the generating station or potential difference across the ends of the power line?

And if it is a potential then I think we cannot apply the formula power = voltage*current.

I don't understand whether it is the potential difference or potential.

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There are essentially, three voltages (potential differences) to consider:

(1) The voltage at the source end of the transmission line

(2) The voltage at the load end of the transmission line

(3) The difference which is the voltage drop along (between the ends of) the transmission line

For example, there may be 11kV at the source and, say, 10.9kV at the load. Assume a current of 10A (I do not know if these numbers are realistic).

For simplicity, assume there is no reactive power. Then, the power delivered by the the source is

$$P_s = 11kV \cdot 10A = 110kW$$

The power delivered to the load is

$$P_l = 10.9kV \cdot 10A = 109kW$$

Thus, the power dissipated by the transmission line is

$$P_s - P_l = 1kW = (11kV - 10.9kV) \cdot 10A$$

In each case, \$P = VI\$ holds. This is an elementary analysis just to give the basic idea of how to apply the power formula.

Here's a simple schematic where the resistance of the transmission line is modelled as the resistor \$R_T\$

enter image description here

Clearly, by Ohm's law, the voltage across the load \$V_L\$ is less than the voltage \$V\$ at the source since

$$V_L = V - I\cdot R_T = 11kV - 10A \cdot 10\Omega = 10.9kV$$

So, as stated earlier, there are three voltage to consider, the source voltage \$V\$, the load voltage \$V_L\$ and the voltage drop across the transmission line (end to end) \$V_T = V - V_L\$.

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  • \$\begingroup\$ why are we taking 11 kv as the potential difference? we should consider the potential difference(voltage at source end-voltage at load end) \$\endgroup\$ Oct 22, 2014 at 2:28
  • \$\begingroup\$ @SouhardyaMondal, you seem to have essentially no idea of the fundamentals. This is truly elementary stuff. There is the source voltage, the load voltage and, since the transmission lines have resistance, the voltage drop in between. Is this really so hard to see? \$\endgroup\$ Oct 22, 2014 at 2:31
  • \$\begingroup\$ i am getting so much confused..:( \$\endgroup\$ Oct 22, 2014 at 2:33
  • \$\begingroup\$ @SouhardyaMondal, are you able to read a schematic diagram? \$\endgroup\$ Oct 22, 2014 at 2:35
  • \$\begingroup\$ Tell me one thing. Why will be the power given by the source will be like that? \$\endgroup\$ Oct 22, 2014 at 2:36
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potential = potential difference relative to something (e.g. ground) = voltage. Voltage is sometimes called 'electromotive force' and it can be considered to be the 'force' that pushes the electrons through the wire.

11,000 volts at the output of a generator at a power station means that at the output of the generator, you can measure 11,000 volts between the wires leading out of the generator.

Electricity is distributed at high voltage because of power = voltage * current and voltage = current * resistance. If you plug one into the other, you get power = resistance * current^2 . Wire has a nonzero resistance, so to minimize power loss you need to minimize the current. Increasing the voltage decreases the current necessary to move the same amount of power, so power loss in transmission can be minimized. Long distance transmission lines can be several hundred kV or even several MV. However, these voltages are not nice to work with, so they are stepped down with transformers at the destination to get a low voltage, high current supply.

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  • \$\begingroup\$ Is the voltage across the power line is 11,00 V? \$\endgroup\$ Oct 22, 2014 at 2:10
  • \$\begingroup\$ if its not a potential difference then we cannot apply p=Voltage *current... \$\endgroup\$ Oct 22, 2014 at 2:11
  • \$\begingroup\$ Maybe. Power lines can sit at various high voltages depending on the purpose. Some standard voltages are: 765 kV, 500 kV, 345 kV, 230 kV, 138 kV, 69 kV, 26 kV, 13 kV, 4 kV. \$\endgroup\$ Oct 22, 2014 at 2:13
  • \$\begingroup\$ how can that be? that means a voltage drop of 11,000 V would occur? so much energy will be lost? \$\endgroup\$ Oct 22, 2014 at 2:13
  • \$\begingroup\$ suppose we are transferring power from the generating station to a substation....what will be the potential difference between the station and substation? \$\endgroup\$ Oct 22, 2014 at 2:15

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