A current heats the cable it flows through. The heat lost is I²R, so a smaller current wastes much less energy, and stepping the voltage up gives a smaller current for the same power.
This lesson is part of SPM Physics electromagnetism. It uses the ratio from calculating transformer relationships.
What is the chain of reasoning?
Marks go to a chain of reasoning, so write it in order.
- Power delivered is P = VI, so for a fixed power, larger V means smaller I.
- The cable has resistance R, so the power lost as heat is I²R.
- A smaller I gives a much smaller loss, because the loss depends on I².
- So transmit at high voltage, then step down near the user.
Worked example: the same power at two voltages
A power station sends 100 kW through cables with a total resistance of 2 Ω.
At 1000 V. I = 100 000 ÷ 1000 = 100 A. Power loss = 100² × 2 = 20 000 W, which is 20 kW, or 20% of the power sent.
At 10 000 V. I = 100 000 ÷ 10 000 = 10 A. Power loss = 10² × 2 = 200 W, which is 0.2% of the power sent.
The voltage is ten times higher, so the current is ten times smaller, and the loss is one hundred times smaller. That is the I² effect.
The mistake that loses marks
A common slip is to calculate the loss as V²/R using the supply voltage: 1000² ÷ 2 = 500 000 W, which is more than the power sent.
The voltage across the cable is much smaller than the supply voltage. Use I²R, with the current in the cable, for the heat lost in the cable.
Check yourself
50 kW is sent through cables of resistance 4 Ω at 5000 V. Find the current, the power lost and the percentage lost.
Answer
I = P ÷ V = 50 000 ÷ 5000 = 10 A.
Power lost = I²R = 10² × 4 = 400 W.
Percentage lost = (400 ÷ 50 000) × 100 = 0.8%.
What to study next
Test the reasoning in the electromagnetism practice set. If transformer ratios are still slow, return to calculating transformer relationships.
Keep a record of slips in the mistake log. For a teacher to go through the chain of reasoning with you, see online one-to-one Physics tuition or the one-hour trial class (from RM50).