☰ Year plan · Physics

Power transmission, transformer efficiency and the DC motor

Week 11 · Chapter 3 (D): transformers in power transmission (I²R loss) and their efficiency, the two transformer examples, the DC motor (construction, Fleming’s left-hand rule, the commutator, one revolution, several coils, motor against galvanometer); the summary and exercises (pp. 86–96) are not used · 3 lessons
1

Transformer efficiency and power transmission

Physics Sec 3 Term 1, Chapter 3: pp. 80–85
GoalExplain why power is sent at a high voltage, find the power lost in a line with I²R, and find the efficiency of a transformer and its current.
New words
line loss · the power turned into heat in the transmission wires, P = I²Refficiency · output power ÷ input power × 100 %laminated core · an iron core made of thin insulated sheets, which cuts the eddy currents
Explanation

Power stations send energy along long wires, and the wires have resistance R. A current I in a line of resistance R turns the power I²R into heat, and that power is wasted. The power sent is P = VI, so for the same power a higher line voltage means a smaller current: I = P ÷ V. Because the loss depends on the square of the current, ten times the voltage gives one tenth of the current and only one hundredth of the loss. This is why a step-up transformer raises the voltage at the power station, and step-down transformers lower it to 220 V near the homes. A real transformer is not perfect. Its efficiency is η = (Vs·Is) ÷ (Vp·Ip) × 100 %, the output power over the input power. Energy is lost as heat in the wires of the two coils, as heat from eddy currents in the iron core (thin insulated sheets reduce them), and in the repeated magnetising of the core, which also causes the hum. The voltage ratio always follows the turns, Vs ÷ Vp = Ns ÷ Np, whatever the efficiency. The losses show up in the currents: Is = η·Vp·Ip ÷ Vs, with η written as a fraction. Never climb or fly a kite near power lines or poles.

01234550.80.20.052 kV5 kV10 kV20 kVloss in the line (kW)
Worked examples
Line voltageCurrent ILoss I²RShare of 100 kW
2 kV50 A5000 W5 %
20 kV5 A50 W0.05 %
A line of resistance R = 2 Ω carries 100 kW. At 2 kV (2000 V) the current is I = P ÷ V = 100 000 ÷ 2000 = 50 A, and the loss is I²R = 50² × 2 = 5000 W = 5 kW. At 20 kV the current is 5 A and the loss is 5² × 2 = 50 W. The current fell to one tenth, and the loss to one hundredth.
PrimarySecondary
voltage200 V12 V
current1 A15 A
power200 W180 W
A transformer takes 1 A at 200 V and delivers 15 A at 12 V. Input power = 200 × 1 = 200 W. Output power = 12 × 15 = 180 W. Efficiency η = 180 ÷ 200 × 100 = 90 %, so 20 W is lost as heat. If the primary has Np = 1000 turns, the voltages give Ns = Np·Vs ÷ Vp = 1000 × 12 ÷ 200 = 60 turns.
★Efficiency and turns
The turns ratio comes from the voltages. The efficiency changes the currents, not the number of turns.
A transformer lowers 220 V to 11 V. Np = 2000, its efficiency is 90 % and Ip = 0.5 A. The turns follow the voltages only: Ns = 2000 × 11 ÷ 220 = 100 turns. Input power = 220 × 0.5 = 110 W. Output power = 0.9 × 110 = 99 W, so Is = 99 ÷ 11 = 9 A.
Class activity

Paper work. Fill this table for a line of resistance 4 Ω that carries 50 kW at 1 kV, 5 kV and 10 kV: the current P ÷ V, then the loss I²R. Find by what factor the loss falls each time the voltage rises five times or two times. Then write one sentence on why a power station raises the voltage before the line, and keep away from real power lines and poles.

Practice
1
A line of resistance 4 Ω carries a current of 20 A. Find the power lost in it.
2
A line of resistance 3 Ω carries 50 kW. Find the loss at 5 kV and at 10 kV, and say by what factor it falls.
3
A transformer takes 2 A at 150 V and delivers 8.5 A at 30 V. Find its efficiency.
4
A transformer of efficiency 80 % changes 240 V to 24 V. Np = 1200 and Ip = 0.5 A. Find Ns and Is.