Why Power Transmission is done at Very High Voltages?

Why Electricity is Transmitted at Higher Voltage?

Once electricity is generated in power plants, it cannot be delivered directly to consumers at the same voltage level. Instead, it is stepped up to higher voltages for transmission and then stepped down again at the user end. The primary reasons for transmitting electricity at higher voltages are to reduce transmission losses and to minimize conductor size and weight.




Reason 1: Reduction in Transmission Losses

During power transmission, energy is lost in the form of heat due to resistance in transmission lines. This loss is given by:

Heat loss=I2R\text{Heat loss} = I^2 R

where:

  • II = Current in the conductor

  • RR = Resistance of the transmission line

The transmission power equation is:

P=V×I×Power FactorP = V \times I \times \text{Power Factor}

Rearranging for current:

I=PV×Power FactorI = \frac{P}{V \times \text{Power Factor}}

From this, you can see that higher transmission voltage (V) results in lower current (I). Since heat loss is proportional to I2I^2, transmitting at a higher voltage drastically reduces losses.


Example: 1 MW Power Transmission at 11 kV vs 33 kV

  • Case 1 (11 kV):

I=1,000,00011,000×PFI = \frac{1,000,000}{11,000 \times \text{PF}}
  • Case 2 (33 kV):

I=1,000,00033,000×PFI = \frac{1,000,000}{33,000 \times \text{PF}}

Now, compare heat losses:

Heat loss (11 kV)=(1,000,00011,000×PF)2R\text{Heat loss (11 kV)} = \left(\frac{1,000,000}{11,000 \times \text{PF}}\right)^2 R Heat loss (33 kV)=(1,000,00033,000×PF)2R\text{Heat loss (33 kV)} = \left(\frac{1,000,000}{33,000 \times \text{PF}}\right)^2 R

Dividing:

Heat loss (11 kV)Heat loss (33 kV)=(3311)2=9\frac{\text{Heat loss (11 kV)}}{\text{Heat loss (33 kV)}} = \left(\frac{33}{11}\right)^2 = 9

Heat loss at 11 kV is 9 times greater than at 33 kV.

This clearly shows why transmission at higher voltage is preferred.


Reason 2: Reduction in Conductor Size and Weight

To reduce transmission losses at lower voltages, thicker conductors (larger diameter) must be used because:

R=ρ×LAR = \frac{\rho \times L}{A}

where:

  • ρ\rho = Resistivity of material

  • LL = Length of conductor

  • AA = Cross-sectional area

Since area (A) is proportional to the square of conductor diameter, larger conductors significantly increase weight and cost.

At higher voltages, current is much lower, so smaller diameter conductors are sufficient. This results in:

  • Lighter conductors

  • Reduced sag in transmission lines

  • Fewer and shorter transmission towers

  • Lower installation and maintenance costs


Example:

From the earlier calculation, heat loss at 11 kV is 9 times greater than at 33 kV. To keep heat loss the same at 11 kV, the conductor’s cross-sectional area must be increased three times compared to 33 kV transmission.

This means heavier conductors, more tower load, and higher costs at lower voltages.


Conclusion

Transmitting electricity at higher voltage offers two major benefits:

  1. Lower transmission losses (since current decreases, reducing I2RI^2R losses).

  2. Reduced conductor size and weight, leading to cost savings in material, installation, and maintenance.

That’s why electricity is transmitted at high voltages (110 kV, 220 kV, 400 kV, and beyond) before being stepped down near consumer locations.


Key Takeaway:
Higher transmission voltage = lower current = lower losses + lighter conductors = efficient, cost-effective power delivery.



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