Alternating current and the transformer

In this lesson you will learn

Explain how a generator produces an alternating EMF and read the amplitude, frequency, period and effective value from u(t) and i(t) equations
Calculate inductive and capacitive reactance and use their dependence on frequency to explain changes in lamp brightness
Apply Ohm’s law and the power formula to AC circuits
Calculate a transformer’s turns ratio and efficiency and the power lost in a line

The “220 V” in a socket is a voltage that changes all the time: in every 0.02 s it rises from zero to about +311 V, falls to −311 V and comes back. So why does it say 220? Why does a capacitor block direct current but let alternating current through? Why is electricity sent from power stations to cities along lines at hundreds of thousands of volts? This lesson is the most important application of Faraday’s law from the lesson “Electromagnetic induction”. In 2026 both DİM entrance papers (groups I and IV) had a task on the brightness of lamps in an AC circuit.

The AC generator

In a generator a coil of N turns and area S rotates with angular velocity ω in a uniform field B. The angle between the coil’s normal and B grows steadily, α = ωt, so the flux changes as Φ = BS · cos ωt, and by Faraday’s law a sinusoidal EMF appears in the coil. When the plane of the coil is perpendicular to the field lines, the flux is largest and the EMF is zero; when the plane is parallel to the lines, the flux is zero but changes fastest — the EMF is largest.

e = ℰ_m · sin ωt, ℰ_m = N · B · S · ω

where:

einstantaneous EMF, in V
ℰ_mamplitude (peak value) of the EMF, in V
N, B, Snumber of turns, flux density (T), area of the coil (m²)
ωangular frequency, in rad/s: ω = 2πf = 2π / T

The voltage u = U_m · sin ωt and the current i = I_m · sin(ωt + φ) follow the same law (φ is the phase shift). The mains frequency in Azerbaijan is 50 Hz: T = 0.02 s, ω = 100π ≈ 314 rad/s; the current changes direction 100 times a second.

Interactive

Loading simulation…

Curve 1 is the flux through the coil, Φ ∝ cos ωt, curve 2 the EMF, e ∝ ω · sin ωt (x is time). When the flux is largest the EMF is zero, and when the flux passes through zero the EMF is largest. Increase ω: the faster the rotation, the larger both the amplitude and the frequency of the EMF.

Example 1: a simple generator

A generator coil has 100 turns of area 0.02 m² and makes 50 revolutions per second in a 0.5 T field.
1) Find the amplitude and the effective value of the EMF.
2) How do the amplitude and frequency change if the coil turns twice as fast?
3) If e = 0 at t = 0, what is the EMF at t = 1/600 s?

Show solution

1) ω = 2πf = 2π · 50 ≈ 314 rad/s. ℰ_m = NBSω = 100 · 0.5 · 0.02 · 314 ≈ 314 V; ℰ = ℰ_m / √2 ≈ 222 V — about the mains voltage.
2) ℰ_m ∝ ω and f ∝ ω — both double (≈ 628 V, 100 Hz).
3) ωt = 100π · 1/600 = π/6; e = 314 · sin 30° = 157 V.

Instantaneous, peak and effective values

The heat released in a resistor is proportional to the square of the current: P = i²R. For i = I_m · sin ωt the average of i² over a period is I_m² / 2. So the heating effect of AC is compared with DC. The effective (rms) value is the direct current that releases the same heat in the same resistor in the same time. Ammeters, voltmeters and the ratings on appliances all show effective values.

I = I_m / √2, U = U_m / √2, ℰ = ℰ_m / √2I = I_m / √2, U = U_m / √2, ℰ = ℰ_m / √2

where:

I, U, ℰeffective values of current, voltage and EMF — what meters show
I_m, U_m, ℰ_mamplitudes (peak values)

√2 ≈ 1.41, 1/√2 ≈ 0.71. The mains has U = 220 V, so U_m = 220 · √2 ≈ 311 V.

Example 2: reading the equation

1) A voltage varies as u = 311 · sin 314t (V). Find U_m, U, ω, f and T.
2) A current i = 2 · sin(100πt) (A) flows through a 10 Ω resistor. What does an ammeter show? What average power is released in the resistor?
3) What fraction of the amplitude is the instantaneous voltage u = U_m · sin ωt at t = T/12?

Show solution

1) U_m = 311 V; U = 311 / √2 ≈ 220 V; ω = 314 rad/s (= 100π); f = ω / 2π ≈ 50 Hz; T = 1 / f = 0.02 s.
2) I_m = 2 A → I = 2 / √2 ≈ 1.41 A. P = I² · R = 2 · 10 = 20 W. I_m² · R = 40 W is the peak instantaneous power, not the average.
3) ωt = (2π / T) · (T / 12) = π / 6 → u = U_m · sin 30° = U_m / 2.

Example 3 (multiple choice): what does the voltmeter show?

The voltage across a resistor varies as u = 100√2 · sin(100πt) (V). What does a voltmeter show, and what is the frequency?
A) 141 V; 50 Hz B) 100 V; 100 Hz C) 100 V; 50 Hz D) 141 V; 100 Hz E) 71 V; 50 Hz

Show solution

U_m = 100√2 ≈ 141 V; a voltmeter shows the effective value: U = U_m / √2 = 100 V.
ω = 100π rad/s → f = ω / 2π = 50 Hz.
Answer: C. A takes the amplitude for the meter reading, B takes the 100 in ω for the frequency, and E divides 100 by √2 once more.

Resistance, inductive and capacitive reactance

Resistor — resistance R. It turns electrical energy into heat. The current and voltage reach their maxima at the same moment (they are in phase), and R does not depend on frequency.
Coil — inductive reactance X_L. As the current changes, a self-induced EMF appears in the coil and opposes the change. The higher the frequency and the inductance, the stronger the opposition. The current lags behind the voltage by a quarter of a period (T/4). For direct current (f = 0), X_L = 0.
Capacitor — capacitive reactance X_C. Direct current cannot pass through the dielectric. With an alternating voltage, however, the capacitor charges and discharges again and again, and a current flows in the wires. The higher the frequency and the capacitance, the larger this current, so the smaller X_C. The current leads the voltage by T/4.

A coil and a capacitor use no energy on average: for a quarter of a period the energy is stored in their field, and in the next quarter it goes back to the source. That is why X_L and X_C are called reactances.

X_L = ω · L = 2πf · L, X_C = 1 / (ωC) = 1 / (2πf · C)X_L = ω · L = 2πf · L, X_C = 1 / (ωC) = 1 / (2πf · C)

where:

X_Linductive reactance, in Ω
X_Ccapacitive reactance, in Ω
ffrequency, in Hz
L, Cinductance (H) and capacitance (F)

As f grows, X_L grows in proportion and X_C falls in inverse proportion; R does not depend on frequency.

Element	Resistance	As f grows	Current and voltage	Graph X(f)
Resistor	R	unchanged	in phase	horizontal line
Coil	X_L = ωL	grows	current lags by T/4	straight line through the origin
Capacitor	X_C = 1/(ωC)	falls	current leads by T/4	hyperbola

Interactive

Loading simulation…

x is the frequency f (Hz), y the resistance (Ω). Line 1 is R, line 2 is X_L = 2πfL (L in henries), curve 3 is X_C = 1/(2πfC) (C in microfarads). With L = 0.1 H and C = 100 μF at 50 Hz, X_L ≈ 31 Ω and X_C ≈ 32 Ω — almost equal; the frequency where the X_L and X_C lines cross is the resonant frequency. Change L and C and watch the lines move.

Example 4: calculating X_L and X_C

1) Find the inductive reactance of a 0.2 H coil at 50 Hz and at 100 Hz.
2) Find the capacitive reactance of a 10 μF capacitor at 50 Hz. How does it change when the frequency doubles?
3) This capacitor is connected to the 220 V, 50 Hz mains. What current flows (ignore the resistance of the wires)?

Show solution

1) X_L = 2πfL = 2π · 50 · 0.2 ≈ 62.8 Ω; at 100 Hz ≈ 125.6 Ω — proportional to the frequency.
2) X_C = 1 / (2πfC) = 1 / (2π · 50 · 10⁻⁵) ≈ 318 Ω; when the frequency doubles it halves (≈ 159 Ω).
3) I = U / X_C = 220 / 318 ≈ 0.69 A. With a steady voltage the settled current would be zero.

The circuit of Example 5

Example 5 (multiple choice): brightness of the lamps

The frequency of the generator in the diagram is increased (U_m = const). How does the brightness of the lamps change?
A) 1 — decreases, 2 — increases, 3 — unchanged
B) 1 — increases, 2 — decreases, 3 — increases
C) all three increase
D) 1 — increases, 2 — decreases, 3 — unchanged
E) 1 — unchanged, 2 — decreases, 3 — increases

Show solution

Each branch is connected straight to the generator, so I_m = U_m / Z.
Branch 1: X_C = 1 / (2πfC) falls → the current grows, the lamp gets brighter.
Branch 2: X_L = 2πfL grows → the current falls, the lamp gets dimmer.
Branch 3: R does not depend on frequency → the brightness does not change.
Answer: D. Option A comes from mixing up how X_L and X_C depend on frequency.

When a resistor, a coil and a capacitor are in series, the same current flows through them, but the voltages reach their maxima at different moments: U_L and U_C are opposite to each other, and U_R is a quarter of a period away from both. So effective voltages do not simply add up: U² = U_R² + (U_L − U_C)². This gives Ohm’s law for an AC circuit:

I = U / Z, Z = √(R² + (X_L − X_C)²)I = U / Z, Z = √(R² + (X_L − X_C)²)

where:

Zimpedance (total resistance) of the circuit, in Ω
U, Ieffective (or peak) values of voltage and current

With only a resistor Z = R, only a coil Z = X_L, only a capacitor Z = X_C. When X_L = X_C, Z = R is smallest and the current is largest — this is resonance (see “Electromagnetic oscillations and waves”).

P = I² · R = U · I · cos φ, cos φ = R / ZP = I² · R = U · I · cos φ, cos φ = R / Z

where:

Paverage (active) power of the AC circuit, in W
cos φpower factor; φ is the phase shift between current and voltage

Power is released only in the resistance. With only a resistor cos φ = 1 and P = U · I; for an ideal coil or capacitor cos φ = 0 and P = 0.

Example 6: a series RLC circuit

A 30 Ω resistor, a coil with an inductive reactance of 80 Ω and a capacitor with a capacitive reactance of 40 Ω are connected in series to an effective voltage U = 100 V. Find the impedance, the current, the voltage across each element and the power of the circuit.

Show solution

Z = √(30² + (80 − 40)²) = √(900 + 1600) = 50 Ω; I = U / Z = 100 / 50 = 2 A.
U_R = I · R = 60 V, U_L = I · X_L = 160 V, U_C = I · X_C = 80 V. They add up to 300 V, not 100 V! Check: √(60² + (160 − 80)²) = √(3600 + 6400) = 100 V. The voltage across the coil can be larger than the source voltage.
Power: P = I² · R = 4 · 30 = 120 W; cos φ = R / Z = 0.6 → P = U · I · cos φ = 100 · 2 · 0.6 = 120 W.

The transformer and power transmission

A transformer is a closed steel core with two windings on it. The core is made of thin plates insulated from each other, so that eddy currents do not heat it. The primary winding (N₁ turns) is connected to the AC mains and the secondary (N₂ turns) to the load. The alternating current in the primary creates an alternating magnetic flux in the core; this flux passes through every turn of both windings and induces the same EMF in each turn. So the EMFs of the windings are proportional to their numbers of turns; at no load, U₁ / U₂ = N₁ / N₂.

k = U₁ / U₂ = N₁ / N₂k = U₁ / U₂ = N₁ / N₂

where:

kturns ratio (transformation ratio): k > 1 — step-down, k < 1 — step-up transformer
U₁, N₁voltage and number of turns of the primary winding
U₂, N₂voltage and number of turns of the secondary winding

A transformer works only with alternating current: direct current gives a constant flux, and no EMF appears in the secondary.

η = P₂ / P₁ = U₂ · I₂ / (U₁ · I₁)η = P₂ / P₁ = U₂ · I₂ / (U₁ · I₁)

where:

ηefficiency of the transformer
P₁power taken from the mains, in W
P₂power delivered to the load, in W

Large transformers reach an efficiency of 99 %. In an ideal transformer (η = 1), U₁I₁ = U₂I₂, so I₁ / I₂ = N₂ / N₁: it lowers the current by the same factor as it raises the voltage.

Example 7: three transformer tasks

1) A transformer has 2000 turns on the primary and 100 on the secondary, U₁ = 220 V. Find the turns ratio and U₂ at no load.
2) A transformer takes 0.5 A from the 220 V mains, and the secondary gives 8.8 A at 12 V. Find its efficiency.
3) The current in the secondary of an ideal transformer with k = 20 is 5 A. Find the current in the primary.

Show solution

1) k = N₁ / N₂ = 2000 / 100 = 20 (step-down); U₂ = U₁ / k = 220 / 20 = 11 V.
2) P₁ = U₁I₁ = 220 · 0.5 = 110 W; P₂ = U₂I₂ = 12 · 8.8 = 105.6 W; η = 105.6 / 110 = 0.96 = 96 %.
3) I₁ / I₂ = N₂ / N₁ = 1 / k → I₁ = 5 / 20 = 0.25 A. A transformer that lowers the voltage raises the current.

Example 8 (coded answer): the number of turns

The primary of a step-up transformer has 150 turns and is connected to 120 V. How many turns must the secondary have to give 3 kV at no load?

Show solution

U₁ / U₂ = N₁ / N₂ → N₂ = N₁ · U₂ / U₁ = 150 · 3000 / 120 = 3750.
Answer code: 3750. The most common slip is forgetting to convert kilovolts to volts (150 · 3 / 120 ≈ 3.75).

When power is sent along a line, the wires heat up and part of the energy is lost: ΔP = I²R. To send the same power P at voltage U, the current must be I = P / U (we take cos φ = 1). So the loss is inversely proportional to the square of the voltage:

ΔP = I² · R = P² · R / U²ΔP = I² · R = P² · R / U²

where:

ΔPpower lost in the line, in W
Ppower transmitted, in W
Utransmission voltage, in V
Rresistance of the line wires, in Ω

Raising the voltage n times cuts the loss n² times. That is why a step-up transformer raises the voltage at the power station and step-down transformers lower it again near the consumers.

Example 9: why high voltage?

1) A power of 1 MW is sent along a line with a resistance of 10 Ω. Find the power lost in the line if the voltage is a) 10 kV, b) 100 kV.
2) By what factor must the voltage be raised to cut the loss 4 times, with the same line?

Show solution

1a) I = P / U = 10⁶ / 10⁴ = 100 A; ΔP = 100² · 10 = 100 kW — 10 % of the power.
1b) I = 10⁶ / 10⁵ = 10 A; ΔP = 10² · 10 = 1 kW — only 0.1 %.
2) ΔP ∝ 1 / U²: n² = 4 → 2 times.

Key points

In a generator e = ℰ_m · sin ωt, ℰ_m = NBSω; the mains has f = 50 Hz, ω ≈ 314 rad/s, T = 0.02 s.
Effective value: I = I_m / √2, U = U_m / √2 — this is what meters show (220 V → U_m ≈ 311 V).
X_L = ωL grows with frequency, X_C = 1/(ωC) falls with frequency, and R does not depend on it.
Ohm’s law: I = U / Z, Z = √(R² + (X_L − X_C)²); power P = I²R = UI · cos φ.
Transformer: k = U₁ / U₂ = N₁ / N₂, η = P₂ / P₁; ideally I₁ / I₂ = N₂ / N₁.
Line loss ΔP = P²R / U²: raising the voltage n times cuts the loss n² times.

Check yourself

12 questions. Every correct answer earns XP.

1 / 12

The effective mains voltage is 220 V. What is its amplitude, approximately?