3.7.5.6
The National Grid transmits at high voltage because P_loss = I²R
Transformers — AQA A-Level Physics
- The National Grid is the network of cables, transformers, and pylons that distributes electrical energyThe capacity to do work. Measured in joules (J). from powerThe rate of energy transfer. Measured in watts (W). stations to homes and businesses across the country.
- PowerThe rate of energy transfer. Measured in watts (W). stations generate electricity at around 25 kV. Step-up transformers at the powerThe rate of energy transfer. Measured in watts (W). station raise this to 400 kV (or 275 kV, 132 kV on sub-networks) for long-distance transmission. Step-down transformers near homes reduce it to 230 V for domestic use.
Why transmit at high voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). Informal term for potential difference.?
- The transmission cables have resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).. Measured in ohms (Ω).. Any currentThe rate of flow of charge. Measured in amperes (A). flowing through them wastes energyThe capacity to do work. Measured in joules (J). as heat.
$$P_{\text{loss}} = I^2 R$$
- $P_loss$: power dissipated in the cables (W)
- $I$: current in the cables (A)
- $R$: resistanceThe opposition to current flow. The ratio of potential difference to current. Measured in ohms (Ω). of the cables (ohm)
- The power being transmitted is fixed (it is whatever the power station produces). Since $P = IV$, for a given power: $I = P / V$.
- If you increase the transmission voltageThe energy transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference., the current decreases proportionally.
- Power loss is proportional to $I^2$. So if you double the voltageThe energy transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference., you halve the current, and the power loss drops to a quarter.
- This is the entire reason the National Grid exists as it does. Transformers make high-voltage transmission possible, and high-voltage transmission makes long-distance power distribution efficient.
- You cannot reduce R to zero (superconducting cables are not practical at scale), so reducing I is the only viable strategy for cutting transmission losses.
The full chain
- Power station: generates at ~25 kV.
- Step-up transformer: raises to 400 kV for transmission on the supergrid.
- Transmission cables (pylons): carry high-voltage, low-current electricity over long distances.
- Step-down transformer (substation): reduces to 33 kV or 11 kV for local distribution.
- Final step-down transformer: reduces to 230 V for homes.
Worked Example
A power station delivers 500 MW of power. The transmission cables have a total resistanceThe opposition to current flow. The ratio of potential difference to current. Measured in ohms (Ω). of 4.0 ohm. Calculate the power lost in the cables when transmitting at (a) 25 kV and (b) 400 kV.
Show Solution
1
Calculate current at 25 kV
$$I = \frac{P}{V} = \frac{500 \times 10^6}{25 \times 10^3} = 20\,000 \text{ A}$$
2
Calculate power loss at 25 kV
$$P_{\text{loss}} = I^2 R = (20\,000)^2 \times 4.0 = 1.6 \times 10^9 \text{ W} = 1600 \text{ MW}$$
That is more than three times the power being transmitted. Clearly impractical.
3
Calculate current at 400 kV
$$I = \frac{P}{V} = \frac{500 \times 10^6}{400 \times 10^3} = 1250 \text{ A}$$
4
Calculate power loss at 400 kV
$$P_{\text{loss}} = I^2 R = (1250)^2 \times 4.0 = 6.25 \times 10^6 \text{ W} = 6.25 \text{ MW}$$
That is only 1.25% of the transmitted power. Raising the voltage by a factor of 16 reduced the current by 16, and the power loss by $16^2 = 256$.
Answer
(a) $P_{\text{loss}} = 1600$ MW at 25 kV (impractical). (b) $P_{\text{loss}} = 6.25$ MW at 400 kV (1.25% of transmitted power).
Examiner Tips and Tricks
- The key equation chain for National Grid questions is: $P = IV gives$ $I = P/V$, then $P_{\text{loss}} = I^{2} R = P^{2} R/V^{2}$.
- This shows P_loss is proportional to 1/V².
- Doubling V quarters the losses.
- State this explicitly for full marks.