Cambridge AS & A Level9702

Alternating currents

Physics 9702 Chapter Notes

What this chapter covers

Alternating currents - Characteristics of alternating currentsAlternating currents - Rectification and smoothing
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In this chapter

  1. 1Describing Alternating Currents
  2. 2RMS Values and Power in AC Circuits
  3. 3Rectification: Converting AC to DC
  4. 4Smoothing Rectified Current

1. Describing Alternating Currents

Unlike direct current (DC) which flows in one direction, alternating current (AC) periodically reverses the direction of charge flow. For your exams, we focus on sinusoidal AC, where the current and voltage vary smoothly like a sine wave. Key characteristics are used to describe this variation. The peak value (I₀ or V₀) is the maximum current or voltage achieved during a cycle. The period (T) is the time taken to complete one full cycle of alternation, measured in seconds. The frequency(f) is the number of complete cycles that occur per second, measured in Hertz (Hz). Frequency and period are reciprocals: f = 1/T. The entire wave can be described mathematically. The instantaneous current I at any time t is given by I = I₀ sin(ωt), where ω is the angular frequency. Angular frequency relates the rate of cycling to an angle in radians, and is calculated by ω = 2πf. A similar equation, V = V₀ sin(ωt), describes the alternating voltage.

I = I₀ sin(ωt)

V = V₀ sin(ωt)

ω = 2πf

f = 1/T

Key term

Peak Value (I₀ or V₀): The maximum magnitude of current or voltage reached during an alternating cycle.

Examiner insight

Examiners expect you to be able to read period and peak values directly from a voltage-time or current-time graph and use them to calculate frequency and angular frequency.

Common pitfall

Forgetting to set your calculator to radians mode when using sinusoidal equations like I = I₀ sin(ωt) to find the current at a specific time t.

Worked example 14 marks

An alternating current is represented by the equation I = 3.5 sin(100πt), where I is in amperes and t is in seconds. Determine:(a) the peak current,(b) the angular frequency,(c) the frequency, and(d) the period of the current.

  1. 1

    Step 1: Compare the given equation I = 3.5 sin(100πt) with the general form I = I₀ sin(ωt).

  2. 2

    Step 2:(a) By direct comparison, the peak current I₀ is the amplitude of the sine function. So, I₀ = 3.5 A.

  3. 3

    Step 3:(b) By comparison, the angular frequency ω is the coefficient of t. So, ω = 100π rad s⁻¹.

  4. 4

    Step 4:(c) Use the relationship ω = 2πf to find the frequency f. f = ω / 2π = 100π / 2π = 50 Hz.

  5. 5

    Step 5:(d) Use the relationship T = 1/f to find the period T. T = 1 / 50 = 0.020 s.

Worked example 23 marks

The graph shows the variation of an alternating voltage V with time t. From the graph, determine:(a) the peak voltage,(b) the period, and(c) the frequency. (Assume the peak is at 6V and one full cycle completes at 40ms).

  1. 1

    Step 1:(a) The peak voltage (V₀) is the maximum value on the vertical axis. From the graph, the peak is at 6 V. So, V₀ = 6.0 V.

  2. 2

    Step 2:(b) The period (T) is the time for one complete cycle. From the graph, one cycle is completed at t = 40 ms. So, T = 40 ms = 0.040 s.

  3. 3

    Step 3:(c) The frequency(f) is the reciprocal of the period. f = 1/T = 1 / 0.040 s = 25 Hz.

Recap

  • Alternating current (AC) periodically reverses direction in a sinusoidal pattern.
  • Peak value (I₀, V₀) is the maximum amplitude of the current or voltage.
  • Period (T) is the time for one complete cycle, while frequency (f) is the number of cycles per second (f=1/T).
  • The instantaneous current is given by I = I₀ sin(ωt), where ω = 2πf is the angular frequency.
  • Always set your calculator to radians for calculations involving the term ωt.

Quick check

  1. An AC supply has a frequency of 60 Hz. What is its period?1 mark
  2. What is the angular frequency of a 60 Hz AC supply? Give your answer in terms of π.1 mark

2. RMS Values and Power in AC Circuits

How do we describe the 'average' power of an AC supply if the current is constantly changing and its average value is zero? We use the Root-Mean-Square (r.m.s.) value. The r.m.s. value of an AC is the value of a steady DC that would dissipate heat at the same average rate in a given resistor. It's the 'effective' value. Because power is proportional to I² (P=I²R), the power is always positive, even when the current is negative. For a sinusoidal current, the average power is exactly half the maximum power: P_mean = P_max / 2. This leads to the crucial relationships between r.m.s. and peak values: I_rms = I₀/√2 and V_rms = V₀/√2. The great thing about r.m.s. values is that you can use them in all the familiar DC power equations (P=IV, P=I²R, P=V²/R) to find the average power in an AC circuit.

I_rms = I₀ / √2

V_rms = V₀ / √2

P_mean = P_max / 2

P_mean = I_rms * V_rms

P_mean = (I_rms)² * R

P_mean = (V_rms)² / R

Key term

Root-Mean-Square (r.m.s.) Value: The r.m.s. value of an alternating current is the value of the direct current that would produce the same heating effect (average power) in the same resistor.

Examiner insight

Questions often require you to convert between peak and r.m.s. values. Remember that mains voltage (e.g., 230 V) is always quoted as an r.m.s. value.

Common pitfall

Using peak values (I₀, V₀) in standard power formulas like P=V²/R, which overestimates the average power by a factor of 2. Always use r.m.s. values for average power calculations.

Fun fact

Your 230 V mains socket delivers a peak voltage of about 325 V! The r.m.s. value is used because it directly relates to the power you get and pay for.

Worked example 14 marks

The mains electricity supply in the UK is 230 V, 50 Hz. This voltage is the r.m.s. value.(a) What is the peak voltage of the UK mains?(b) A 100 W lamp is connected to this supply. What is the r.m.s. current it draws?

  1. 1

    Step 1:(a) Recall the relationship between r.m.s. and peak voltage: V_rms = V₀ / √2.

  2. 2

    Step 2: Rearrange for the peak voltage: V₀ = V_rms × √2.

  3. 3

    Step 3: Substitute the values: V₀ = 230 V × √2 = 325.27 V. So, the peak voltage is approximately 325 V.

  4. 4

    Step 4:(b) Use the power equation P = V_rms × I_rms. The power given for an appliance is its average power.

  5. 5

    Step 5: Rearrange for the r.m.s. current: I_rms = P / V_rms.

  6. 6

    Step 6: Substitute the values: I_rms = 100 W / 230 V = 0.435 A.

Worked example 24 marks

An alternating current in a 10 Ω resistor has a peak value of 2.0 A. Calculate(a) the r.m.s. current,(b) the maximum power dissipated, and(c) the average power dissipated.

  1. 1

    Step 1:(a) Calculate the r.m.s. current using I_rms = I₀ / √2. I_rms = 2.0 A / √2 = 1.41 A.

  2. 2

    Step 2:(b) The maximum power (P_max) occurs when the current is at its peak (I₀). P_max = (I₀)² × R = (2.0 A)² × 10 Ω = 40 W.

  3. 3

    Step 3:(c) The average power (P_mean) can be found in two ways. Method 1: P_mean = P_max / 2 = 40 W / 2 = 20 W. Method 2: P_mean = (I_rms)² × R = (1.41 A)² × 10 Ω ≈ 20 W. Both methods give the same result.

Recap

  • The r.m.s. value of an AC is its effective value for power calculations.
  • The r.m.s. current is the DC current that delivers the same average power to a resistor.
  • For a sinusoidal AC, I_rms = I₀/√2 and V_rms = V₀/√2.
  • Average power can be calculated using standard power formulas with r.m.s. values: P_mean = I_rms * V_rms.
  • The mean (average) power dissipated is half the maximum instantaneous power: P_mean = P_max / 2.
  • Mains voltage (e.g., 230 V) is always quoted as an r.m.s. value unless specified otherwise.

Quick check

  1. A mains voltage has a peak value of 170 V. What is its r.m.s. value?1 mark
  2. If the maximum power dissipated by a resistor in an AC circuit is 200 W, what is the average power dissipated?1 mark

3. Rectification: Converting AC to DC

Many electronic devices, like your phone charger, need a steady direct current (DC) to function, but the mains supply provides AC. The process of converting AC to DC is called rectification. This is achieved using diodes, which act as one-way gates for current. For half-wave rectification, a single diode is placed in series with the load. The diode only conducts during the positive half-cycles of the AC input, blocking the negative half-cycles. This results in a pulsating DC output where half the power is lost. For a more efficient conversion, full-wave rectification is used. This typically involves a 'bridge rectifier' arrangement of four diodes. This clever circuit flips the negative half-cycles of the AC input into positive ones. As a result, current always flows through the load in the same direction, producing a more continuous (but still pulsating) DC output.

Key term

Rectification: The process of converting an alternating current (AC) into a direct current (DC).

Examiner insight

Be prepared to draw the output voltage graph for both half-wave and full-wave rectification and to trace the current path through a bridge rectifier for both halves of the AC cycle.

Common pitfall

Incorrectly tracing the path of current in a bridge rectifier, especially during the negative half-cycle. Remember the current must always flow through the load resistor in the same direction.

Worked example 12 marks

A sinusoidal AC voltage is applied to the input of a circuit containing a single diode and a resistor. Sketch the input voltage and the voltage across the resistor on the same axes, labelling each trace.

  1. 1

    Step 1: Draw a set of axes with Voltage on the y-axis and Time on the x-axis.

  2. 2

    Step 2: Sketch the input voltage as a continuous sine wave, showing at least two full cycles. Label this 'Input Voltage'.

  3. 3

    Step 3: On the same axes, sketch the output voltage. For every positive 'hump' of the input, draw an identical hump for the output. For every negative 'hump' of the input, draw a flat line at V=0.

  4. 4

    Step 4: Label the resulting trace of positive humps and flat lines as 'Output Voltage (Half-wave rectified)'.

Worked example 22 marks

The diagram shows a bridge rectifier with a load resistor R. For the instant when terminal X is positive and terminal Y is negative, trace the path of conventional current through the circuit.

  1. 1

    Step 1: Start at the positive terminal, X.

  2. 2

    Step 2: Follow the path to the junction of two diodes. Current can only pass through the forward-biased diode. Draw an arrow along this path.

  3. 3

    Step 3: The current then flows through the load resistor R. Draw an arrow through R (e.g., from top to bottom).

  4. 4

    Step 4: After the resistor, the current reaches another junction. It flows through the second forward-biased diode towards the negative terminal, Y.

  5. 5

    Step 5: The complete path is from X, through the first diode, through R, through the second diode, to Y.

Recap

  • Rectification is the process of converting AC to DC using diodes.
  • A diode is a semiconductor device that allows current to flow in only one direction.
  • A single diode in series with a load produces half-wave rectification, blocking all negative half-cycles.
  • A bridge rectifier with four diodes produces full-wave rectification by inverting the negative half-cycles.
  • Full-wave rectification is more efficient and provides a more continuous output than half-wave.

Quick check

  1. What is the name of the component used as a one-way gate for current in a rectifier circuit?1 mark
  2. Sketch the output voltage graph for a full-wave rectifier.2 marks

4. Smoothing Rectified Current

The output from a rectifier is a 'pulsating DC' – it flows in one direction but its value varies significantly. For most electronics, a much steadier voltage is required. This is achieved by 'smoothing'. The most common way to do this is to connect a large capacitor in parallel with the load resistor. Here's how it works: as the rectified voltage rises, the capacitor charges up, storing energy. As the rectified voltage begins to fall, the power supply can no longer provide the peak voltage. The capacitor then starts to discharge through the load resistor, supplying current and 'propping up' the voltage. It continues to do this until the next voltage pulse from the rectifier rises above the capacitor voltage and begins to recharge it. The result is a much smoother DC voltage with only a small variation, known as 'ripple'. The effectiveness of the smoothing depends on how slowly the capacitor discharges. This is governed by the time constant (τ = RC). A large time constant (achieved with a large capacitance C and/or a large load resistance R) leads to slower discharge and better smoothing (less ripple).

τ = RC

Key term

Smoothing: The process of reducing the variations (ripple) in a rectified DC voltage, typically using a capacitor.

Examiner insight

Examiners frequently ask you to explain how smoothing works and how the degree of smoothing is affected by the values of the capacitor and the load resistor. Clear, annotated graphs are an excellent way to support your explanation.

Common pitfall

Stating that a larger resistor improves smoothing without explaining why. The key is that a larger R increases the time constant (τ = RC), slowing the capacitor's discharge rate.

Worked example 14 marks

A full-wave rectifier output is smoothed by a capacitor. The initial circuit produces significant ripple. State and explain two changes to the circuit components that would reduce the ripple.

  1. 1

    Change 1: Increase the capacitance of the smoothing capacitor.

  2. 2

    Explanation 1: A capacitor with a larger capacitance (C) can store more charge for a given voltage. This increases the time constant of the discharge circuit (τ = RC). A larger time constant means the capacitor discharges more slowly, so the voltage across the load drops less between peaks, reducing the ripple.

  3. 3

    Change 2: Increase the resistance of the load resistor (if possible).

  4. 4

    Explanation 2: A load with a larger resistance (R) draws less current for a given voltage. This also increases the time constant (τ = RC), causing the capacitor to discharge more slowly and thus reducing the ripple.

Worked example 23 marks

Sketch a graph showing the voltage across the load resistor in a smoothed full-wave rectified circuit. On the same axes, show the effect of using a capacitor with a smaller capacitance.

  1. 1

    Step 1: Draw axes for Voltage vs. Time. Sketch the unsmoothed full-wave rectified output (a series of connected positive humps) as a faint or dashed line for reference.

  2. 2

    Step 2: Draw the smoothed output for a large capacitor. The line should rise with the first hump to its peak, then slowly slope downwards until it meets the rising edge of the next hump. The variation (ripple) should be small.

  3. 3

    Step 3: Draw the output for a smaller capacitor. This line should also start at the peak, but it should slope downwards more steeply, resulting in a much lower voltage before the next hump recharges it. This shows a significantly larger ripple.

  4. 4

    Step 4: Label the two smoothed curves clearly as 'Large C (good smoothing)' and 'Small C (poor smoothing)'.

Recap

  • A capacitor connected in parallel with the load resistor is used to smooth a rectified voltage.
  • The capacitor charges when the input voltage is high and discharges through the load when the input voltage falls.
  • The effectiveness of smoothing is determined by the time constant, τ = RC.
  • A larger time constant (from a large C or large R) results in better smoothing and less ripple.
  • Full-wave rectification is easier to smooth than half-wave because the time between voltage peaks is shorter.

Quick check

  1. What is the term for the small remaining voltage variation after smoothing?1 mark
  2. To improve smoothing, should the time constant RC be large or small?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. An AC voltage is given by the equation V = 170 sin(100πt). Determine (a) the peak voltage, (b) the r.m.s. voltage, and (c) the frequency of the supply.3 marks
  2. Sketch the voltage-time graph for two cycles of a half-wave rectified sinusoidal AC supply. Label your axes.2 marks
  3. An alternating current has a frequency of 200 Hz and a peak value of 4.0 A. (a) Write an equation of the form I = I₀ sin(ωt) to represent this current. (b) Calculate its r.m.s. value.4 marks
  4. A 1200 W hairdryer is designed for use on a 230 V r.m.s. mains supply. Calculate (a) the r.m.s. current drawn by the hairdryer, and (b) the peak current.4 marks
  5. Explain, with the aid of a circuit diagram, the role of a single diode in producing half-wave rectification. Your explanation should include a sketch of the input and output voltage waveforms.5 marks
  6. The trace shown on an oscilloscope represents a voltage. The Y-gain is set to 2.0 V div⁻¹ and the time-base is 10 ms div⁻¹. The waveform has a peak amplitude of 2.5 divisions and one complete cycle has a length of 4 divisions. Determine (a) the peak voltage, (b) the r.m.s. voltage, (c) the period, and (d) the frequency of the signal.6 marks
  7. A student builds a full-wave bridge rectifier. Explain why the current flows through the load resistor in the same direction during both positive and negative half-cycles of the input AC. A diagram is not required.3 marks
  8. A rectified DC power supply includes a smoothing capacitor. (a) Explain the action of the capacitor in smoothing the output voltage. (b) The power supply is connected to a new device with a much lower resistance. Describe and explain the effect this will have on the smoothness of the output voltage.5 marks
  9. A 12 V DC supply and a 12 V r.m.s. AC supply are each connected to identical lamps. (a) Which lamp is brighter, or are they the same? Explain your reasoning. (b) Calculate the peak voltage of the AC supply.4 marks
  10. A device requires a reasonably smooth DC voltage of approximately 9 V from a 50 Hz AC supply which has a peak voltage of 15 V. Draw a circuit diagram for a power supply that could achieve this, labelling all the components. Sketch a graph to show the voltage across the final load resistor.6 marks

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