Lessons 20 · 1 lessons · N. Sh. Turdiyev, K. A. Tursunmetov, A. G. Ganiyev, K. T. Suyarov, J. E. Usarov, A. K. Avliyoqulov. Physics Grade 11, 1st edition. Niso Poligraf Publishing House, Tashkent, 2018
20
Work and power of alternating current. The power factor
Textbook: pp. 66–73
GoalExplain the power P = U·I·cosφ and work A = U·I·t·cosφ of alternating current, the meaning of the power factor cosφ = R/Z and the ways of raising it.
New words
power factor · quvvat koeffitsiyentimean power · o‘rtacha quvvatinstantaneous power · oniy quvvatcompensating capacitor · kompensatsiyalovchi kondensator
Explanation
The instantaneous power of alternating current is p = u·i; for u = Uₘcosωt and i = Iₘcos(ωt + φ) it varies in time and is negative during part of the period (energy returns to the source). The mean power over a period, in terms of the effective values of current and voltage, is P = U·I·cosφ (U and I effective), and the work of the current is A = U·I·t·cosφ. The factor cosφ is the power factor; for an R–L–C circuit cosφ = R/Z. In a circuit with only active resistance φ = 0, cosφ = 1 and P = U·I; in a circuit with only a coil or a capacitor φ = ±π/2, cosφ = 0 and P = 0: even though a current flows, the mean power is zero. If cosφ is small, a larger current is needed for the same power (I = P/(U·cosφ)), so the heat losses (I²r) in the wires grow. Electric motors have a large inductive resistance, so their cosφ is low; to raise it, compensating capacitors are connected to the factory's network and motors are not run idle.
Worked examples
A motor draws 5 A from the 220 V mains with cosφ = 0.8. P = U·I·cosφ = 220·5·0.8 = 880 W. If cosφ were raised to 1, the same 880 W would need I = 880/220 = 4 A, and the loss in the wires would be (4/5)² = 0.64 of the previous value, i.e. 36 % less.
Series circuit: R = 30 Ω, X_L – X_C = 40 Ω, U = 100 V (effective). Z = 50 Ω, I = 2 A, cosφ = R/Z = 0.6. P = U·I·cosφ = 100·2·0.6 = 120 W; check: P = I²R = 4·30 = 120 W.
Class activity
Make a table in your notebook: for U = 220 V and P = 1100 W compute the necessary current I = P/(U·cosφ) for cosφ = 1, 0.8 and 0.5 and explain how the heat loss in the wires (proportional to I²) changes. Do not modify any electrical appliances at home.
Practice
1
What is the mean power in a circuit with only a coil and why?
Zero: φ = π/2, cosφ = 0, the energy is returned to the source.
2
U = 220 V, I = 10 A, cosφ = 0.5. What is the power in W?
1100
3
U = 220 V, I = 2 A, cosφ = 0.5, t = 100 s. What is the work in J?
22000
4
R = 12 Ω and Z = 20 Ω. What is cosφ?
0.6
5
Why are compensating capacitors installed in factories?
They partly compensate the inductive (reactive) current of the motors, raising the network's cosφ; the line current and the losses in the wires decrease.