Electricity is transferred by the National Grid at a **high potential difference (PD)** (voltage) and **low current** primarily to minimize **energy losses** and ensure efficient transmission over long distances. Here's a detailed explanation of the reasons and the underlying principles:
### 1. **Minimizing Power Losses (Joule Heating Losses)**
The primary reason for transmitting electricity at high voltage and low current is to reduce power losses due to **Joule heating**. These losses occur when electrical energy is converted into heat in the transmission lines, which happens because of the electrical resistance of the wires. The formula for the power loss due to resistance is given by:
\[
P_{loss} = I^2 R
\]
Where:
- \( P_{loss} \) is the power lost as heat,
- \( I \) is the current flowing through the wire,
- \( R \) is the resistance of the wire.
As you can see from the formula, **power loss increases with the square of the current**. If the current is doubled, the losses increase by a factor of four. To minimize these losses, it is essential to keep the current as low as possible.
#### Relationship Between Power, Voltage, and Current:
The power transmitted by the National Grid is related to voltage and current by the formula:
\[
P = VI
\]
Where:
- \( P \) is the power,
- \( V \) is the voltage (potential difference),
- \( I \) is the current.
For a given amount of power \( P \), if we increase the voltage \( V \), the current \( I \) must decrease. Lower current means lower \( I^2R \) losses, as mentioned above.
### 2. **Efficiency of Transmission**
When electricity is transmitted at **high voltage**, the **current** required to transmit the same amount of power decreases. This lower current reduces the heating of the transmission wires, meaning less energy is wasted as heat, and more energy reaches the end-users (homes, industries, etc.). This improves the overall **efficiency of power transmission**.
For example, if you need to transmit 100 MW (megawatts) of power:
- At 1000 V (1 kV), the current would be 100,000 A.
- At 100,000 V (100 kV), the current would be 1000 A.
The second case (with higher voltage) has a much lower current, which leads to significantly reduced resistive losses in the wires.
### 3. **Practical Considerations: Cable Size and Costs**
Low-current transmission has the added benefit of reducing the size and cost of the transmission cables. Higher currents require thicker wires to handle the increased flow of electrons without overheating. Thicker wires mean more material (usually copper or aluminum), which is more expensive and heavier, requiring sturdier support structures.
By reducing the current, the National Grid can use **thinner, lighter wires** and less material, reducing both installation and maintenance costs.
### 4. **High Voltage and Transformers**
While itβs necessary to transmit electricity at high voltage and low current to minimize losses, the voltage used for domestic and industrial use is much lower (e.g., 230 V for homes). To manage this, the National Grid uses **transformers** at both ends of the transmission process:
- **Step-up transformers** increase the voltage for long-distance transmission.
- **Step-down transformers** reduce the voltage for local distribution, making it safe for consumer use.
Transformers work efficiently to change the voltage and current levels while conserving the overall power. This enables electricity to be transmitted at high voltage (for efficiency) and then used at lower voltage (for safety and practicality).
### 5. **Why Not Just Use High Current?**
You might wonder, why not use high current instead of high voltage? As explained, high current would lead to massive energy losses in the form of heat. The transmission lines would need to be made extremely thick to avoid overheating, and this would be highly impractical both in terms of cost and the physical infrastructure required to support such cables.
### Example Calculation:
To understand the significance, let's consider an example. Suppose we want to transmit 10 MW (10 million watts) of power over a distance using a transmission line with a resistance of 1 ohm.
- If the power is transmitted at **10 kV (10,000 V)**:
- The current needed would be \( I = \frac{P}{V} = \frac{10,000,000}{10,000} = 1000 \, \text{A} \).
- The power loss due to resistance would be \( P_{loss} = I^2 R = (1000)^2 \times 1 = 1,000,000 \, \text{W} \) (1 MW lost as heat).
- If the power is transmitted at **100 kV (100,000 V)**:
- The current needed would be \( I = \frac{P}{V} = \frac{10,000,000}{100,000} = 100 \, \text{A} \).
- The power loss due to resistance would be \( P_{loss} = I^2 R = (100)^2 \times 1 = 10,000 \, \text{W} \) (only 10 kW lost as heat).
In this example, the power loss drops drastically from 1 MW to just 10 kW by increasing the transmission voltage from 10 kV to 100 kV.
### Conclusion
The National Grid transfers electricity at a **high voltage (potential difference)** and **low current** to minimize energy losses, increase efficiency, reduce the size and cost of transmission lines, and make long-distance transmission practical. This method ensures that as much of the generated electricity as possible reaches consumers while minimizing the energy wasted as heat.