Alternating Currents and Transformers

SPEC 3.7.5.5 & 3.7.5.6

RMS values, oscilloscopes, transformer equations, and power transmission

Alternating Current (AC)

Definition

Alternating current (AC) is an electric current that periodically reverses direction, in contrast to direct current (DC) which flows in only one direction.

When a coil rotates in a magnetic field, an EMF is induced. From Lesson 7.06, this EMF follows:

Why AC is Sinusoidal

ε = B A N ω sin (ω t)

Because the equation contains sin(ωt), the induced EMF is alternating - it changes direction with time (shown by the changing sign).

UK Mains Supply

In the UK, mains electricity is supplied as AC at 230 V (RMS) and 50 Hz. This means the current reverses direction 50 times per second.

Sinusoidal AC Waveform

AC voltage and current vary sinusoidally with time. The diagram shows the four key measurements:

PointMeasurementDescription

①Peak Voltage (V₀ or V̂)Maximum amplitude from equilibrium (zero line)

②Time Period (T)Time for one complete cycle; frequency f = 1/T

③RMS Voltage (V_rms)= V₀/√2 ≈ 0.707 × V₀ (effective DC equivalent)

④Peak-to-Peak (V_p-p)= 2V₀ (total amplitude from min to max)

Root Mean Square (RMS) Values

Definition

The RMS value of an alternating current or voltage is the equivalent DC value that would produce the same power dissipation in a resistor. It is the average of all the squares of the possible voltages, giving the average value of voltage output by the supply (in either direction).

Why do we need RMS? The average value of a sinusoidal AC signal is zero (equal positive and negative halves). But power is proportional to V² or I², which is always positive.

Key Formulas

V_{r m s} = \frac{V _{0}}{2} \approx 0.707 \times V_{0}

I_{r m s} = \frac{I _{0}}{2} \approx 0.707 \times I_{0}

Where V₀ and I₀ are the peak values of voltage and current.

Important

When mains voltage is quoted as "230 V", this is the RMS value. The actual voltage oscillates between +325 V and -325 V!

01Worked Example - Peak to RMS

An AC power supply has a peak voltage of 12.0 V. Calculate:

(a) the RMS voltage

(b) the peak-to-peak voltage

a)

Using

V_{r m s} = \frac{V _{0}}{2}

•

V_{r m s} = \frac{12.0}{2} = \frac{12.0}{1.414} = 8.49 V

b)

Peak-to-peak voltage = 2 × V₀

•

V_{p - p} = 2 \times 12.0 = 24.0 V

02Worked Example - UK Mains

UK mains electricity is supplied at 230 V RMS and 50 Hz. Calculate:

(a) the peak voltage

(b) the peak-to-peak voltage

(c) the time period of the AC supply

a)

Rearranging

V_{r m s} = \frac{V _{0}}{2}

gives

V_{0} = V_{r m s} \times 2

•

V_{0} = 230 \times 2 = 230 \times 1.414 = 325 V

(to 2 s.f.: 330 V)

b)

V_{p - p} = 2 \times V_{0} = 2 \times 325 = 650 V

c)

T = \frac{1}{f} = \frac{1}{50} = 0.020 s

(or 20 ms)

Using an Oscilloscope

An oscilloscope displays how voltage varies with time. You can adjust the scale of both axes to make measurements easier:

Y-Gain Control

Volts per division on the vertical (Y) axis. Controls the height of the trace.

Time Base Control

Time per division on the horizontal (X) axis. Controls how "stretched" the waveform appears.

Taking measurements: Count the number of divisions and multiply by the setting:

Peak voltage: Divisions from centre to peak × Y-gain (V/div)

Time period: Divisions for one cycle × Time base (s/div)

Frequency: f = 1/T

Oscilloscope Traces: DC vs AC

The oscilloscope can display different signals. The time-base can be turned on or off. With time-base off, the trace shows all the possible voltages at any time in one area:

Straight line parallel to axis at output voltage level

Single dot at the output voltage level

Sinusoidal waveform showing voltage vs time

Vertical line showing all possible voltages

Why Turn Time Base Off?

With time base off, the trace shows all the possible voltages at any time in one area. This is useful for measuring the voltage range without needing to read a moving trace. For AC, the vertical line length = peak-to-peak voltage.

03Worked Example - Oscilloscope Reading

An oscilloscope displays an AC waveform. The Y-gain is set to 2.0 V per division and the time base is 5.0 ms per division. The trace shows a peak height of 3.5 divisions and one complete cycle takes 4 divisions.

Calculate: (a) the peak voltage (b) the RMS voltage (c) the frequency

a)

Peak voltage = divisions × Y-gain = 3.5 × 2.0 = 7.0 V

b)

V_{r m s} = \frac{V _{0}}{2} = \frac{7.0}{1.414} = 4.95 V

c)

Period T = divisions × time base = 4 × 5.0 ms = 20 ms = 0.020 s

•

f = \frac{1}{T} = \frac{1}{0.020} = 50 Hz

Transformer Principles

Definition

A transformer is used with alternating currents to change the size of their voltage. It consists of a primary coil (attached to input), secondary coil (connected to output), and an iron core.

How It Works

The primary coil provides a changing magnetic field (because AC changes direction). This passes through the iron core and interacts with the secondary coil, inducing a voltage in the secondary (by Faraday's law).

Transformer Equations

By using Faraday's law, we can show that the ratio of voltages equals the ratio of turns:

Transformer Equation

\frac{N _{s}}{N _{p}} = \frac{V _{s}}{V _{p}}

Nₛ = secondary turns, Nₚ = primary turns

Vₛ = secondary voltage, Vₚ = primary voltage

Step-Up Transformer

Nₛ > Nₚ (more turns on secondary)

Vₛ > Vₚ (voltage increases)

Iₛ < Iₚ (current decreases)

Used at power stations

Step-Down Transformer

Nₛ < Nₚ (fewer turns on secondary)

Vₛ < Vₚ (voltage decreases)

Iₛ > Iₚ (current increases)

Used for consumers

Ideal Transformer (100% Efficient)

I_{p} V_{p} = I_{s} V_{s}

Power in = Power out (no energy losses)

Important

Transformers only work with AC! DC produces a constant flux, so no EMF is induced in the secondary.

04Worked Example - Transformer Turns Ratio

A step-down transformer has 2400 turns on the primary coil and is connected to the 230 V mains. The secondary coil has 120 turns.

Calculate: (a) the secondary voltage (b) the current in the primary if the secondary current is 3.0 A (assume ideal transformer)

a)

Using

\frac{N _{s}}{N _{p}} = \frac{V _{s}}{V _{p}}

, rearrange for Vₛ:

•

V_{s} = V_{p} \times \frac{N _{s}}{N _{p}} = 230 \times \frac{120}{2400} = 230 \times 0.05 = 11.5 V

b)

For ideal transformer: IₚVₚ = IₛVₛ

•

I_{p} = \frac{I _{s} V _{s}}{V _{p}} = \frac{3.0 \times 11.5}{230} = \frac{34.5}{230} = 0.15 A

Check: Step-down gives lower voltage but higher current. Vₛ < Vₚ ✓, Iₛ > Iₚ ✓

Worked Example - Step-Up Transformer Design05

A transformer is required to step up the 230 V RMS mains supply to 5.0 kV RMS for a neon sign. The secondary coil has 900 turns. How many turns are needed on the primary coil?

1

Convert secondary voltage: Vₛ = 5.0 kV = 5000 V

2

Using

\frac{N _{s}}{N _{p}} = \frac{V _{s}}{V _{p}}

, rearrange for Nₚ:

•

N_{p} = N_{s} \times \frac{V _{p}}{V _{s}} = 900 \times \frac{230}{5000}

•

N_{p} = 900 \times 0.046 = 41.4 turns \approx 41 turns

Check: For step-up, Nₛ > Nₚ. Indeed, 900 > 41 ✓

Transformer Efficiency

Real transformers are not 100% efficient. The efficiency can be found using:

Efficiency Formula

Transformer efficiency = \frac{I _{s} V _{s}}{I _{p} V _{p}} \times 100%

One of the main causes of energy loss is the production of eddy currents:

Eddy Currents and Lenz's Law

Eddy currents are induced by the alternating magnetic field in the primary coil, and form loops in the iron core. Due to Lenz's law, they oppose the field that produced them, reducing the field's flux density. They also generate heat, which causes energy to be lost.

How Laminations Reduce Losses

The core is made of layers of iron separated by insulator. The eddy currents cannot pass through the insulator, so their amplitude is reduced. This significantly decreases the I²R heating losses. Eddy currents can also be reduced by using a high resistivity core material.

Causes of Transformer Inefficiency

Energy LossCauseHow to Reduce

Eddy CurrentsChanging flux induces current loops in iron core → I²R heatingLaminated core (layers separated by insulator) or high resistivity core material

Resistance HeatingCurrent through coil windings causes I²R lossesUse thick wire with low resistance

HysteresisEnergy needed to magnetise/demagnetise core each cycle (if core is not easily magnetised)Use soft iron core - easy magnetisation and demagnetisation

Flux LeakageNot all flux from primary links to secondaryWind coils close together; good core design

Typical Efficiency

Large power transformers can be over 99% efficient. The main losses are eddy currents (reduced by laminations) and resistance heating in the coils.

Worked Example - Transformer Efficiency06

A transformer has a primary coil connected to a 230 V supply and draws a current of 0.80 A. The secondary coil provides 160 W of power. Calculate the efficiency of the transformer.

1

Calculate input power: Pᵢₙ = IₚVₚ = 0.80 × 230 = 184 W

2

Output power is given: Pₒᵤₜ = 160 W

3

Calculate efficiency:

•

η = \frac{P _{o u t}}{P _{in}} \times 100% = \frac{160}{184} \times 100% = 87%

Energy lost: 184 - 160 = 24 W (lost as heat in core and windings)

High Voltage Power Transmission

When transferring electrical power, the power lost due to resistance is equal to I²R. Therefore it is vital to reduce the current to a minimum value to prevent unnecessary energy losses.

The Key Principle

As power = current × voltage, when the voltage is stepped up, the current will decrease (for the same power). This dramatically reduces I²R losses in the transmission cables.

Key Calculation

If voltage is increased by factor of 10, current decreases by factor of 10.

Power loss = I²R decreases by factor of 100!

This is why we use very high voltages (~400 kV in UK) for transmission.

07Worked Example - Power Transmission

Alternating current leaves a power station at 25 kV and enters a step-up transformer with 50 turns on its primary coil and 750 turns on its secondary coil.

(a) Find the output voltage of the transformer.

(b) 500 MW of power is transmitted along 225 km of wire with resistance 0.2 Ω per km. Find the power wasted.

a)

Using

\frac{N _{s}}{N _{p}} = \frac{V _{s}}{V _{p}}

, so

V_{s} = V_{p} \times \frac{N _{s}}{N _{p}}

•

V_{s} = 25000 \times \frac{750}{50} = 25000 \times 15 = 375 kV

(or 3.75 × 10⁵ V)

b)

Total resistance = 225 × 0.2 = 45 Ω

•

Find current: I = P/V = (500 × 10⁶)/(3.75 × 10⁵) = 1333 A (to 3 s.f.)

•

Power wasted = I²R = (1333)² × 45 = 8 × 10⁷ W (80 MW)

Practice

Question 1:

(a) the RMS voltage

(b) the time period

(c) the peak-to-peak voltage

(a) V_rms = V₀/√2 = 18/1.414 = 12.7 V

(b) T = 1/f = 1/60 = 0.0167 s (16.7 ms)

(c) V_p-p = 2V₀ = 2 × 18 = 36 V

Practice

Question 2:

(a) the secondary voltage

(b) the power output

(c) the current in the primary coil

(a) V_s = V_p × (N_s/N_p) = 12 × (3000/500) = 12 × 6 = 72 V

(b) P_out = I_sV_s = 0.25 × 72 = 18 W

(c) η = P_out/Pᵢₙ, so Pᵢₙ = P_out/η = 18/0.92 = 19.6 W

I_p = Pᵢₙ/V_p = 19.6/12 = 1.63 A

Practice

Question 3:

(a) the peak voltage

(b) the RMS voltage

(c) the frequency of the AC supply

(a) V₀ = 2.4 div × 5 V/div = 12 V

(b) V_rms = V₀/√2 = 12/1.414 = 8.5 V

(c) T = 8 div × 2 ms/div = 16 ms = 0.016 s

f = 1/T = 1/0.016 = 62.5 Hz

Summary

Key Formulas

QuantityFormula

RMS Voltage

V_{r m s} = \frac{V _{0}}{2}

RMS Current

I_{r m s} = \frac{I _{0}}{2}

Frequency

f = \frac{1}{T}

Transformer Ratio

\frac{N _{s}}{N _{p}} = \frac{V _{s}}{V _{p}}

Ideal Transformer

I_{p} V_{p} = I_{s} V_{s}

Efficiency

η = \frac{I _{s} V _{s}}{I _{p} V _{p}} \times 100%

Power Loss

P_{l oss} = I^{2} R

Key Exam Skills

Convert between peak, peak-to-peak, and RMS values

Read voltage, period, and frequency from oscilloscope traces

Sketch oscilloscope traces for DC/AC with time-base on/off

Apply transformer equations and calculate efficiency

Explain why high voltage is used for power transmission (reduces I²R losses)

Describe causes of transformer inefficiency and how to reduce them