Power transmission, transformer efficiency and the DC motor
الأسبوع 11 · الفصل الثالث (٤): المحولات في نقل الطاقة الكهربية وفقد الطاقة وكفاءة المحول وأمثلة محلولة، والمحرك الكهربي (الموتور) وفكرة عمله ودورة كاملة له · 3 دروس
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.
Worked examples
Line voltage
Current I
Loss I²R
Share of 100 kW
2 kV
50 A
5000 W
5 %
20 kV
5 A
50 W
0.05 %
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.
Primary
Secondary
voltage
200 V
12 V
current
1 A
15 A
power
200 W
180 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.
P = I²R = 20² × 4 = 400 × 4 = 1600 W.
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.
At 5 kV: I = 50 000 ÷ 5000 = 10 A and loss = 10² × 3 = 300 W. At 10 kV: I = 5 A and loss = 5² × 3 = 75 W. The voltage doubled, so the loss fell to one quarter (300 ÷ 75 = 4).
3
A transformer takes 2 A at 150 V and delivers 8.5 A at 30 V. Find its efficiency.
Input = 150 × 2 = 300 W. Output = 30 × 8.5 = 255 W. η = 255 ÷ 300 × 100 = 85 %.
4
A transformer of efficiency 80 % changes 240 V to 24 V. Np = 1200 and Ip = 0.5 A. Find Ns and Is.
Ns = 1200 × 24 ÷ 240 = 120 turns (the efficiency does not enter). Input = 240 × 0.5 = 120 W, output = 0.8 × 120 = 96 W, Is = 96 ÷ 24 = 4 A.