PhysicsCore24 min read

Alternating Current

RMS values, reactance, resonance and why the grid uses a.c.

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01

What alternating means

Definition

Alternating current — A current that periodically reverses direction, usually varying sinusoidally with time.

Direct current flows one way and keeps its value. Alternating current reverses direction periodically, and in the mains supply it does so smoothly, following a sine wave.

Two numbers describe it. The frequency is how many complete cycles pass each second — 50 Hz in Pakistan, so the current reverses a hundred times a second and completes fifty full cycles. The peak value is the maximum it reaches in either direction.

The average value over a whole cycle is zero, because the negative half exactly cancels the positive. That makes the ordinary average useless for describing a.c., and it is why a different kind of average is needed.

02

RMS: the value that does the work

Definition

RMS value — The steady direct value that would deliver the same average power to a resistor as the alternating one does. For a sine wave, V_rms = V_peak / √2.

Power does not care which way the current flows. Heating goes as I²R, and squaring makes the negative half positive — so an alternating current heats a resistor perfectly well even though its average is zero.

The useful measure is therefore the root mean square: square the values, take the mean of those squares, then take the square root. That undoes the cancellation and leaves a number that predicts power correctly.

For a sine wave the arithmetic works out to peak divided by √2, about 0.707 of the peak. Every a.c. value you meet in ordinary life is an RMS value unless it says otherwise.

So Pakistan's 230 V mains actually peaks at about 325 V, and swings between +325 V and −325 V. Insulation has to withstand the peak; the heating is set by the RMS. That is the practical reason the distinction matters.

V_rms = V_peak / √2I_rms = I_peak / √2P_average = I_rms² R = V_rms I_rmsquoted mains voltages are always RMS; the peak is about 1.41 times larger
V_rms
root mean square voltageV
V_peak
peak voltageV
P
average powerW
Worked example 15 marks

A 230 V RMS supply is connected to a 60 Ω heater. Find the peak voltage, the RMS current and the average power.

  1. V_peak = V_rms × √2 = 230 × 1.414.Multiply going from RMS to peak; divide coming back.
  2. V_peak = 325 V.The value the insulation must withstand.
  3. I_rms = V_rms / R = 230 / 60.Use RMS with RMS throughout.
  4. I_rms = 3.83 A.
  5. P = V_rms × I_rms = 230 × 3.83 = 881 W.Using peak values here would overstate the power by a factor of two.

325 V peak, 3.83 A, 881 W

The dashed lines mark either the RMS or the peak. Notice the RMS line sits well inside the peak — at about 0.707 of it. It is not an average of the wave; it is the steady value that would heat a resistor at the same rate.

03

Capacitors and inductors in a.c.

A resistor behaves the same whether the current is steady or alternating. Two other components do not, and both oppose alternating current in a way that depends on frequency.

A capacitor blocks direct current completely — no charge crosses the gap between its plates. But with alternating current the plates charge and discharge repeatedly, so current flows in the circuit continuously. The opposition, called capacitive reactance, falls as frequency rises: the faster the reversals, the less time there is for charge to build up and oppose the flow.

An inductor is a coil, and it opposes any change in current by inducing a back e.m.f. — Lenz's law again. Its opposition, inductive reactance, rises with frequency, because faster changes induce a larger opposing e.m.f.

The two behave in exactly opposite ways, which is what makes the next section possible. It is also why a capacitor passes a treble signal while an inductor passes bass — the basis of the crossover in a loudspeaker.

Direct currentAs frequency rises
Resistoropposes normallyno change
Capacitorblocks completelyopposition falls
Inductorpasses freelyopposition rises
X_C = 1 / (2π f C)X_L = 2π f LZ = √(R² + (X_L − X_C)²)reactance is in ohms; capacitive falls with frequency, inductive rises
X_C
capacitive reactanceΩ
X_L
inductive reactanceΩ
C
capacitanceF
L
inductanceH
Z
impedanceΩ
04

Resonance, and why the grid uses a.c.

Put a resistor, a capacitor and an inductor in series and sweep the frequency. At one particular frequency the two reactances become equal and cancel, leaving only the resistance. The current then reaches its maximum, and the circuit is at resonance.

A resonant circuit responds strongly to one frequency and weakly to all others, which is exactly what tuning means. Turning the dial on a radio changes a capacitance, moving the resonant frequency until it matches the station you want; every other station is still arriving at the aerial but produces almost no current.

Finally, the reason the grid runs on alternating current at all. Transformers only work on a.c., because they depend on a changing magnetic field. Transformers allow the voltage to be stepped up for transmission, and since P = VI, a high voltage means a small current for the same power delivered.

That matters because the heat lost in the cables is I²R. Cutting the current by a factor of ten cuts the loss by a factor of a hundred. Transmitting at 500 kV instead of 230 V is the difference between losing a few per cent of the power and losing nearly all of it — and none of that is possible with d.c. and no transformers.

at resonance: X_L = X_Cf₀ = 1 / (2π √(L C))at f₀ the impedance is at its minimum and the current at its maximum
f₀
resonant frequencyHz
L
inductanceH
C
capacitanceF

Key points

  1. The average of a full a.c. cycle is zero; the RMS is not.
  2. V_rms = V_peak/√2 — quoted mains voltages are RMS.
  3. Capacitive reactance falls with frequency; inductive reactance rises.
  4. At resonance the two cancel and the current is greatest.
  5. The grid uses a.c. because transformers need a changing field, and high voltage means low current and small I²R losses.

Practice questions

6 questions · 24 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Explain why the average value of an alternating current over a complete cycle is zero, yet it still heats a resistor.
Model answer

The negative half of the cycle cancels the positive half, so the mean current is zero. Heating depends on I²R, and squaring makes the negative half positive, so energy is dissipated throughout.

Examiner tip. The squaring is the whole answer. It is also why the RMS rather than the mean is used.

SQ2[2 marks]
Define the RMS value of an alternating current.
Model answer

The steady direct current that would deliver the same average power to a resistor as the alternating current does.

Examiner tip. Define it by the power it delivers, not by the arithmetic of squaring and rooting.

SQ3[2 marks]
State how the reactance of a capacitor and of an inductor each change as the frequency increases.
Model answer

Capacitive reactance decreases, since X_C = 1/(2πfC). Inductive reactance increases, since X_L = 2πfL.

Examiner tip. They move in opposite directions — that opposition is what makes resonance possible.

Solved numericals

1 · 5 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[5 marks]
An a.c. supply has a peak voltage of 170 V and is connected to a 40 Ω resistor. Find the RMS voltage, the RMS current and the average power dissipated.
Full working
  1. Uses V_rms = V_peak/√2[1]
  2. V_rms = 170 / 1.414 = 120 V[1]
  3. I_rms = V_rms/R = 120/40[1]
  4. I_rms = 3.0 A[1]
  5. P = V_rms I_rms = 120 × 3.0 = 360 Wusing peak values would double this[1]

120 V, 3.0 A, 360 W

Examiner tip. Divide going peak → RMS, multiply going RMS → peak. Mixing peak and RMS in the same power calculation is the usual error.

Long questions

1 · 8 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[8 marks]
Electricity is generated at a power station and transmitted at high voltage.
  1. Explain why transformers only work with alternating current. [2]
  2. A station delivers 40 MW at 400 kV through cables of resistance 3.0 Ω. Calculate the current and the power lost as heat. [4]
  3. Calculate the power that would be lost if the same power were sent at 40 kV instead. [2]
Mark scheme
  1. A transformer works by a changing magnetic field inducing an e.m.f. in the secondary[1]
  2. Direct current gives a steady field, so nothing is induced[1]
  3. I = P/V = 40 × 10⁶ / 400 × 10³[1]
  4. I = 100 A[1]
  5. P_lost = I²R = 100² × 3.0[1]
  6. = 30 kW, less than 0.1% of the output[1]
  7. At 40 kV: I = 1000 Aten times the current[1]
  8. P_lost = 1000² × 3.0 = 3.0 MW — a hundred times more, and 7.5% of the output[1]

(b) 100 A, 30 kW (c) 3.0 MW

Examiner tip. Loss goes as the square of the current, so a tenfold voltage rise cuts the loss a hundredfold. That single fact is why the grid exists in the form it does.

Exam questions

1 · 5 marks

Multi-part questions with a full mark scheme.

Q1[5 marks]
A series circuit contains a resistor, a capacitor and an inductor connected to a variable-frequency supply.
  1. State the condition for resonance. [1]
  2. Describe what happens to the current at the resonant frequency, and explain why. [2]
  3. Give one practical use of a resonant circuit. [2]
Mark scheme
  1. The inductive and capacitive reactances are equal, X_L = X_C[1]
  2. The current reaches its maximum[1]
  3. The two reactances cancel, so the impedance falls to just the resistance — its minimum value[1]
  4. Tuning a radio receiver[1]
  5. Adjusting the capacitance shifts the resonant frequency to match one station, so only that signal produces a large current[1]

X_L = X_C; current is maximum; used for tuning a radio

Examiner tip. At resonance the impedance is at a minimum, not a maximum — which is why the current peaks.