Power transmission, transformer efficiency and the DC motor
Answers are for parents and teachers.
1
A station sends 80 kW along a line of resistance 1.5 Ω. Find the current and the loss at 4 kV and at 8 kV.
At 4 kV: I = 80 000 ÷ 4000 = 20 A and loss = 20² × 1.5 = 600 W. At 8 kV: I = 10 A and loss = 10² × 1.5 = 150 W. Twice the voltage, a quarter of the loss.
2
A transformer takes 2.5 A at 120 V and delivers 10 A at 24 V. Find its efficiency and the power lost as heat.
Input = 120 × 2.5 = 300 W, output = 24 × 10 = 240 W. η = 240 ÷ 300 × 100 = 80 %. Lost power = 300 − 240 = 60 W.
3
A bell transformer of efficiency 75 % changes 220 V to 6 V. The primary has 1100 turns and carries 0.2 A. Find the number of secondary turns and the secondary current.
Ns = 1100 × 6 ÷ 220 = 30 turns. Input = 220 × 0.2 = 44 W, output = 0.75 × 44 = 33 W, Is = 33 ÷ 6 = 5.5 A.
4
A motor coil has N = 40 turns, sides ℓ = 0.1 m in the field and width 0.05 m. B = 0.2 T and I = 5 A. Find the force on one side and the torque when the plane is parallel to the field. Then find the torque when the normal to the coil makes 30° with the field.
F = 40 × 0.2 × 5 × 0.1 = 4 N. Torque with the plane parallel = 4 × 0.05 = 0.2 N·m. With the normal at 30°: 0.2 × sin 30° = 0.1 N·m.
5
A student says: “A transformer of efficiency 90 % needs 0.9 times the turns that the voltage ratio gives.” Explain in two sentences what is wrong and what the efficiency really changes.
The turns ratio is fixed by the voltages, Ns ÷ Np = Vs ÷ Vp, and does not depend on the efficiency. The efficiency only decides how much of the input power reaches the output, so it changes the currents (Is = η·Vp·Ip ÷ Vs).