Alternating current

The current in a wall socket is alternating: its size and direction go through 50 complete cycles every second. Such current is easy to generate, to step up or down with a transformer and to send over long distances. In this topic we study rms values, inductive and capacitive reactance, the RLC circuit, resonance, the power factor and the transformer.

§1. Generating alternating current

A coil of N turns and area S rotates with angular velocity ω in a uniform field B. The flux Φ = BS·cos ωt changes, and by the law of electromagnetic induction an EMF appears:

• ω = 2πν — angular frequency (rad/s), ν — frequency (Hz), T = 1/ν — period.
• Mains frequency in Uzbekistan is ν = 50 Hz, so ω ≈ 314 rad/s, T = 0.02 s.

Worked example

A 100-turn coil of area 200 cm² rotates at 50 Hz in a 0.1 T field. Peak EMF?

1. S = 0.02 m², ω = 2π·50 ≈ 314 rad/s.
2. E_max = 100·0.1·0.02·314 ≈ 62.8 V.

§2. Effective (rms) values

The effective value of an alternating current is the direct current that would release the same heat in a resistor:

Ammeters, voltmeters and device labels show rms values. For "220 V" mains the peak is U_max = 220·√2 ≈ 311 V.

Reading an equation: from u = 311·sin 314t we get U_max = 311 V, U = 220 V, ω = 314 rad/s, ν = ω/(2π) = 50 Hz.

§3. Resistance, inductive and capacitive reactance

• Resistance R: current and voltage are in phase; energy turns into heat.
• Inductive reactance: self-induction opposes changes in current, so the current lags the voltage by a quarter period (90°).

• Capacitive reactance: a capacitor blocks direct current but "passes" alternating current by recharging; the current leads the voltage by 90°.

As frequency rises, X_L increases and X_C decreases.

Worked example

Find the reactances of a coil L = 0.1 H and a capacitor C = 100 μF at 50 Hz.

1. X_L = 314·0.1 = 31.4 Ω.
2. X_C = 1/(314·10⁻⁴) ≈ 31.8 Ω.

§4. Series RLC circuit and resonance

The voltages across the elements are out of phase, so the resistances add like vectors, not like ordinary numbers:

Resonance occurs when X_L = X_C: Z = R is smallest and the current is greatest.

This equals the natural frequency of the LC circuit.

Worked example

R = 8 Ω, X_C = 9 Ω, X_L = 3 Ω, U = 140 V. Current and power?

1. Z = √(8² + (3 − 9)²) = √(64 + 36) = 10 Ω.
2. I = 140/10 = 14 A.
3. P = I²R = 196·8 = 1568 W (check: cos φ = 0.8, UI cos φ = 140·14·0.8 = 1568 W).

§5. Power in AC circuits

Energy is used up only in the resistance; the coil and capacitor take energy during part of a cycle and give it back.

cos φ is the power factor. If it is small, a larger current is needed for the same power and line losses grow, so factories try to raise cos φ.

§6. The transformer and power transmission

A transformer is two windings on a common steel core. Alternating current in the primary creates a changing flux in the core, which induces an EMF in the secondary. A transformer does not work on direct current.

• k > 1 — step-down, k < 1 — step-up.
• Power lines use high voltage: for the same power the current is smaller and the I²R losses drop sharply.

Worked example

N₁ = 1000, N₂ = 50, U₁ = 220 V. The secondary current is 2 A. Find U₂ and I₁.

1. U₂ = 220·50/1000 = 11 V.
2. I₁ = U₂I₂/U₁ = 11·2/220 = 0.1 A.

§7. Three-phase current and line losses

Power-station generators have not one but three windings, turned 120° relative to one another. So the three emfs have the same amplitude and frequency but are shifted in phase by 120° (a third of a period). This is the three-phase system. Four wires reach the consumer: three phase (live) wires and one neutral.

Homes are usually supplied with one phase and neutral (220 V), industrial motors with all three phases (380 V). The advantages of three-phase supply: more power is carried with fewer wires, with balanced loads the neutral current is almost zero, and three-phase current produces a rotating magnetic field in a motor — which is why induction motors are simple and reliable.

When power is transmitted through line wires of resistance R_l, the loss in them is ΔP = I²R_l. Since the power transmitted is P = UI (with cos φ = 1):

Raising the voltage n times cuts the loss by n². That is why at the power station a transformer raises the voltage to hundreds of kilovolts, and near the town it is stepped down in stages to 10 kV and then to 380/220 V.

Worked example

What is the line voltage of a supply with a phase voltage of 220 V?

1. U_l = √3·U_ph = 1.732·220.
2. U_l ≈ 381 V, usually called “380 V”.

Worked example

100 kW is transmitted along a line of resistance 0.5 Ω (cos φ = 1). Losses at 10 kV and at 400 V?

1. 10 kV: I = 10⁵/10⁴ = 10 A, ΔP = 10²·0.5 = 50 W — 0.05 %.
2. 400 V: I = 10⁵/400 = 250 A, ΔP = 250²·0.5 = 31 250 W ≈ 31 kW — almost a third!
3. Conclusion: sending power far at low voltage is wasteful.

Check yourself

1. u = 311·sin 314t (V). Rms voltage and frequency?

2. Reactance of a 10 μF capacitor at 50 Hz?

3. L = 0.1 H, C = 10 μF. Resonant frequency?

4. R = 30 Ω, X_L = 40 Ω, no capacitor, U = 100 V. Current, cos φ, power?

5. A step-up transformer turns 220 V into 11 000 V, N₁ = 200. Find N₂.

6. The frequency is doubled. How do X_L and X_C change?

7. 50 kW is transmitted at 5 kV along a line of resistance 2 Ω. Loss?

8. A three-phase motor: U_l = 380 V, I = 10 A, cos φ = 0.8. Power (P = √3·U_l·I·cos φ)?

Common mistakes

• Adding R, X_L, X_C arithmetically: they combine as Z = √(R² + (X_L − X_C)²).
• Taking power as UI (forgetting cos φ).
• Confusing peak and rms values: meters show rms.
• Using ν in place of ω: X_L = 2πνL.
• Thinking a transformer also works on direct current.
• Calculating line loss as U²/R: U is the transmission voltage, not the voltage drop along the wires; the correct formula is ΔP = I²R.