To find the power lost in a cable: first find the current with I = P ÷ V, then the loss with P(loss) = I² × R. Raising the transmission voltage lowers the current and therefore cuts the loss sharply.
This lesson uses the ratio from using a transformer ratio and the a.c. idea from alternating current. It belongs to induction and transformers.
Why does high voltage reduce the loss?
The power delivered by a line is P = I × V. To send the same power at a larger voltage, the current can be smaller. The cables have resistance, so they heat up, and the power wasted as heat is I² × R.
Because the current is squared, halving the current cuts the loss to a quarter. Multiplying the voltage by 10 cuts the current to a tenth and the loss to a hundredth.
The three steps
- Current in the line: I = P ÷ V.
- Power lost: I² × R, with R the resistance of the cables.
- Compare: express the loss as a percentage of the power sent if asked.
Worked example
A fictional power station sends 50 kW through cables of total resistance 4.0 Ω. (Invented example data.) Compare sending it at 500 V and at 10 000 V.
At 500 V:
Current = 50 000 ÷ 500 = 100 A.
Loss = 100² × 4.0 = 10 000 × 4.0 = 40 000 W = 40 kW.
Percentage lost = 40 000 ÷ 50 000 × 100 = 80%.
At 10 000 V:
Current = 50 000 ÷ 10 000 = 5.0 A.
Loss = 5.0² × 4.0 = 25 × 4.0 = 100 W.
Percentage lost = 100 ÷ 50 000 × 100 = 0.2%.
Check: the voltage rose by a factor of 20, so the current fell by 20 and the loss by 20² = 400. Indeed 40 000 ÷ 100 = 400. If you want to see how rounding a final answer affects it, the bounds and rounding explainer can help.
The mistake to watch for
Mistaken working: loss = V² ÷ R = 500² ÷ 4.0 = 62 500 W.
The student used the supply voltage as though it were across the cable. The result is larger than the 50 kW being sent, which is impossible.
The correction: use I² × R, and make sure the current comes from P ÷ V first. A quick check is that the loss can never be larger than the power sent.
Check yourself
Try these, then open each answer.
1. A line carries 20 kW at 2000 V through cables of resistance 2.0 Ω. Find the power lost.
Show answer
I = 20 000 ÷ 2000 = 10 A. Loss = 10² × 2.0 = 200 W.
2. The same 20 kW is sent at 400 V through the same cables. Find the power lost and the percentage of the power sent.
Show answer
I = 20 000 ÷ 400 = 50 A. Loss = 50² × 2.0 = 2500 × 2.0 = 5000 W. Percentage = 5000 ÷ 20 000 × 100 = 25%.
3. The transmission voltage is increased by a factor of 4 for the same power and cables. By what factor does the loss fall?
Show answer
The current falls by a factor of 4. The loss depends on I², so it falls by a factor of 4² = 16.
Where this leads next
You now have all five skills in the topic. Test them together in the induction and transformers practice set, then return to the topic overview.
Combining several steps under exam conditions is where guided practice helps most. Our teachers work on this in online one-to-one Physics tuition.