Alternating Current
Currents that swing back and forth, and the machines that make and use them.
AC & the Generator
A spinning coil makes alternating emf.
The electricity in your walls doesn’t flow one way: it surges back and forth fifty or sixty times a second. That alternating current comes straight from Faraday’s law applied to a spinning coil.
Alternating current
A battery drives current one way at a nearly steady size. A current whose direction keeps reversing is an alternating current (AC). We study the sinusoidal kind:
The current is positive for half of each period and negative for the other half. To drive it we need a source of sinusoidal emf: an AC generator (AC dynamo), which turns mechanical energy into electrical energy.
How a generator works
- Magnet: permanent or electro-, with facing poles giving a strong uniform field B.
- Armature: a coil, usually on a soft-iron core to strengthen the field, turning about an axis in its own plane and perpendicular to B. Its ends go to two slip rings.
- Brushes: two graphite brushes press on the rings and connect to the outer terminals, keeping contact as the rings turn.
Let the coil (N turns, area A) turn at constant ω, with its plane perpendicular to B at t = 0. The flux through each turn is BA cos ωt, so each turn has emf −dΦ/dt = BAω sin ωt, and the whole coil:
A 100-turn coil of area 200 cm² spins at 50 revolutions per second in a 0.25 T field. Find its peak emf.
- 1AC reverses direction periodically: i = i₀ sin(ωt + φ) (39.1).
- 2Generator: magnet, armature on slip rings, brushes.
- 3A coil spinning in B gives ℰ = NBAω sin ωt (39.2).
Check yourself
RMS Values
The size of a current that averages to zero.
An alternating current averages to zero, yet it lights bulbs perfectly well. The right way to describe its size is the root-mean-square value.
The average is zero
For i = i₀ sin(ωt + φ), the average over a time t is
Over a whole period this is exactly zero, and over a long time it tends to zero (the bracket stays between −2 and 2 while ωt grows).
Root mean square
But i² is never negative. Its average over a period:
Why rms is the useful value
A current i = i₀ sin(ωt + φ) through a resistor R produces heat i²R dt in each instant; over a period,
So a steady current equal to irms heats a resistor exactly as much as the alternating current does. That’s why AC voltages and currents are quoted as rms values: “230 V AC” means Vrms = 230 V, and the peak is 230√2 ≈ 325 V. (For 220 V supplies the peak is 311 V.)
An alternating current has a peak of 4.0 A and a frequency of 50 Hz. Find its rms value and the time it takes to rise from zero to its peak.
A 46 Ω resistor is connected to a 230 V, 60 Hz supply. Find (a) the rms current, (b) the peak current, (c) how long the current takes to fall from its peak to its rms value.
A current grows steadily as i = i₀(t/τ) from t = 0 to τ. Find its rms value over that interval.
- 1Mean of a sinusoidal current over a cycle (or long time) is zero.
- 2Mean square i₀²/2; rms = i₀/√2 (39.3), the heating-equivalent steady current.
- 3AC ratings are rms values: 230 V rms ↔ 325 V peak.
Check yourself
Videos, simulations & further reading (1)
Resistors, Capacitors & Inductors on AC
Each element limits and shifts the current differently.
Put a resistor, a capacitor or an inductor alone across an AC source. Each limits the current in its own way, and each shifts the current’s timing relative to the emf differently.
Resistor only
With ℰ = ℰ₀ sin ωt across R, the loop law gives ℰ₀ sin ωt = Ri:
Capacitor only
With no resistance, the capacitor’s voltage equals the source’s: q/C = ℰ₀ sin ωt, so q = Cℰ₀ sin ωt and
- On a battery a capacitor takes current only briefly while charging. On AC the charge must keep changing, so current flows all the time: a capacitor blocks DC but passes AC.
- XC, the capacitive reactance, acts like a resistance (in ohms). It is large at low frequency and small at high frequency; for DC (ω = 0) it is infinite.
- When the emf is zero the current is at its largest, and vice versa.
Find the reactance of a 50 µF capacitor at 20 Hz, 50 Hz and 1000 Hz.
Inductor only
For a pure inductor, ℰ₀ sin ωt − L di/dt = 0, so di/dt = (ℰ₀/L) sin ωt and i = −(ℰ₀/ωL) cos ωt + c. A sinusoidal current must average to zero, so c = 0:
The inductive reactance XL is zero for DC and grows with frequency.
A 0.30 H inductor is connected to a source of peak emf 150 V at 50 Hz. Find the peak current, and the source voltage at the instant the current is at its peak.
Impedance and phase
In all three cases the peak current is the peak emf divided by an effective resistance, the impedance Z:
Summary
| Element | Z | Phase φ of current (i = i₀ sin(ωt + φ)) | At high frequency |
|---|---|---|---|
| Resistor | R | 0 (in phase) | unchanged |
| Capacitor | 1/ωC | +π/2 (leads) | passes more |
| Inductor | ωL | −π/2 (lags) | passes less |
- 1R: in phase, i₀ = ℰ₀/R.
- 2C: current leads by 90°, X_C = 1/ωC; blocks DC, passes AC.
- 3L: current lags by 90°, X_L = ωL; zero for DC.
- 4i₀ = ℰ₀/Z (39.10); φ is the phase factor.
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Videos, simulations & further reading (2)
LCR Circuits & Resonance
Vector bookkeeping, resonance and tuning.
With several elements in series, a neat vector trick gives the impedance and phase at once. It reveals resonance: the frequency at which a circuit responds most strongly, the principle behind every radio tuner.
The vector (phasor) method
For a series circuit with ℰ = ℰ₀ sin ωt:
- Draw R as a vector along the x-axis.
- Draw the capacitive reactance 1/ωC along +y (π/2 ahead) and the inductive reactance ωL along −y (π/2 behind).
- Add them. The resultant’s length is the impedance Z; its angle with the x-axis is the phase factor φ.
On switching on, the current behaves in a complicated way, but soon settles into this steady state at the source’s frequency.
CR and LR circuits
CR: Z = √(R² + 1/ω²C²), tan φ = 1/ωCR, current leads. LR: Z = √(R² + ω²L²), tan φ = ωL/R, current lags.
The LCR circuit
The two reactances point in opposite directions, so the net reactance is X = 1/ωC − ωL:
- XC > XL: φ > 0, current leads (capacitive).
- XC < XL: φ < 0, current lags (inductive).
- XC = XL: net reactance zero, current in phase, Z = R.
L = 0.10 H, C = 25 µF and R = 30 Ω are in series with ℰ = (20 V) sin(400 t). Find the impedance, peak current, phase and resonant frequency.
A coil of resistance 40 Ω and inductance 0.40 H is connected to 230 V, 50 Hz. Find the rms current. Then a 60 µF capacitor is added in series: find the new rms current.
Resonance
Vary ω and the peak current i₀ = ℰ₀/Z is greatest when 1/ωC = ωL:
A small R makes the resonance sharp: the current is large only very close to ν₀. Such a circuit used near ν₀ is a resonant circuit.
LC oscillations
With R = 0 the current at resonance would be infinite for any emf: a finite current can flow with no source at all. Connect a charged capacitor to a pure inductor: it discharges through the inductor, whose emf then recharges it the other way, and so on, at ν = 1/2π√(LC). Energy swaps between the capacitor’s electric field and the inductor’s magnetic field.
- 1Phasors: R along x, 1/ωC up, ωL down; Z = √(R² + X²), tan φ = X/R (39.11).
- 2LCR: X = 1/ωC − ωL (39.12–39.13); leads if capacitive, lags if inductive.
- 3Resonance at ω₀ = 1/√(LC) (39.14): Z = R, i₀ maximal; sharper for small R; used in tuning.
- 4LC circuit: energy oscillates between C and L at 1/2π√(LC).
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Videos, simulations & further reading (9)
Power, Choke Coils & Hot-Wire Meters
Only the in-phase part of the current does work.
In an AC circuit, current times voltage isn’t the whole story: if the two are out of step, some energy just sloshes back and forth. The power factor measures how much is really used.
Average power
With ℰ = ℰ₀ sin ωt and i = i₀ sin(ωt + φ), the work done in time dt is ℰi dt = ℰ₀i₀(sin²ωt cos φ + sin ωt cos ωt sin φ) dt. Over a full cycle the second term averages to zero:
- Purely resistive: φ = 0, cos φ = 1, P = ℰrmsirms.
- Purely reactive (only L and/or C): φ = ±π/2, cos φ = 0, no power is absorbed, even though current flows.
A motor on a 230 V supply draws 2.0 A rms, lagging the voltage by 37°. What power does it take?
The choke coil
A choke is a coil of large inductance and small resistance. Old fluorescent tube fittings put one in series with the tube to cut down the voltage across it. Treating the tube as a resistance R and the choke as an ideal inductor, the circuit is an LR circuit with Z = √(R² + ω²L²), and the tube’s rms voltage is
The advantage over a series resistor: an ideal inductor consumes no power (its voltage and current are 90° apart), so no energy is wasted as heat.
A tube behaves like a 120 Ω resistor and should have 100 V rms across it, from a 230 V, 50 Hz supply. What choke inductance is needed?
A 20 µF capacitor on a 100 Hz supply carries 1.5 A rms. Find the rms voltage across it and the average energy stored.
Hot-wire instruments
A moving-coil meter can’t read AC: its torque reverses every half cycle and averages to zero. AC meters use an effect that depends on i², which is never negative. In a hot-wire ammeter, current heats a platinum–iridium wire AB stretched between fixed points; it expands and sags, and a spring pulls a thread wound round a small drum, turning a pointer. The rise in temperature, and so the deflection, depends on irms²; the scale is marked to read irms directly. A small shunt keeps the meter’s resistance low. A hot-wire voltmeter is the same but with a large series resistance instead of a shunt.
- 1P = ℰ_rms i_rms cos φ (39.15); cos φ is the power factor.
- 2Pure L or C circuits consume no average power.
- 3A choke (big L, small R) lowers voltage without wasting energy.
- 4Hot-wire meters respond to i² and read rms values.
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Videos, simulations & further reading (3)
Dynamos, Motors & Transformers
Making, using and shipping electrical power.
Three machines built on induction and magnetic force: the DC dynamo, the DC motor and the transformer, which together make it possible to generate, use and ship electrical power across a country.
The DC dynamo
Same magnet and armature as an AC dynamo, but the slip rings are replaced by a split ring: a cylinder cut into two halves C₁ and C₂, one joined to each end of the coil, with carbon brushes B₁ and B₂ pressing on them. Every half turn the gaps pass under the brushes and the connections to the outside swap. Arranged so that the swap happens just when the emf is zero, the external current always flows the same way.
This split ring with its brushes is the commutator. The output is one-way but pulsing; a second coil at right angles (its emf peaks when the first’s is zero) smooths it, and more coils smooth it further.
The DC motor
A DC motor is built just like a DC dynamo, but run in reverse: a battery sends current through the armature via the brushes, the field exerts a torque on the coil (largest when the coil’s plane lies along the field, zero when it faces it), and the commutator reverses the current every half turn so the torque keeps the same sense. The shaft turns the load.
As the coil turns it also acts as a generator, producing a back emf e that opposes the applied emf ℰ:
A motor with armature resistance 2.0 Ω runs on 12 V. At full speed its back emf is 10 V. Find the current at full speed and at the moment it starts.
The transformer
Two coils, a primary (N₁ turns, connected to the AC source) and a secondary (N₂ turns, feeding the load), wound on one laminated soft-iron core. The core’s magnetization dominates the field, so the same flux Φ passes through every turn of both coils.
Neglecting the primary’s resistance, ℰ₁ = N₁ dΦ/dt; the secondary has ℰ₂ = −N₂ dΦ/dt. Dividing:
Power transfer
With the secondary open, the primary is a pure inductor: a small current is flows, 90° out of phase with ℰ₁, and no power is drawn. Connect a load and a secondary current i₂ flows. It would change the core’s flux, so an extra primary current i₁ appears that exactly cancels its effect; i₁ is in phase with ℰ₁, and the source now delivers power ℰ₁i₁. With no losses this all reaches the load:
- Step-up (N₂ > N₁): higher voltage, lower current out; the primary is wound from thick wire.
- Step-down (N₂ < N₁): lower voltage, higher current out; the secondary needs thick wire.
- Efficiency η = output power/input power is lowered by coil resistance, hysteresis and eddy currents in the core, but 99 % is achievable.
A charger needs 12 V from the 230 V mains. The primary has 1150 turns. How many turns should the secondary have?
Long-distance transmission
A power station supplies a fixed power P = Vi. Sent through lines of resistance R, it loses i²R as heat. At high voltage the current, and so the loss, is small. So voltages are stepped up at the station, sent across country, and stepped down near the users.
1.0 MW is sent through lines of total resistance 4.0 Ω. Find the power lost at 11 kV and at 220 kV.
- 1DC dynamo: a commutator reverses connections each half turn; more coils smooth the output.
- 2DC motor: torque on a current loop; back emf e gives i = (ℰ − e)/R.
- 3Transformer: ℰ₂/ℰ₁ = −N₂/N₁ (39.16); i₂/i₁ = −N₁/N₂ (39.17); step-up or step-down.
- 4High-voltage transmission keeps i²R losses small.