use equations of the form x = x0 sin ωt representing a sinusoidally alternating current or voltage

1. Introduction to Alternating Current (AC)

In an alternating current the magnitude and direction of the current (or voltage) vary periodically with time. For the Cambridge AS & A‑Level syllabus the sinusoidal form is assumed because it is produced by the most common source – an AC generator.

1.1 Sinusoidal representation

The instantaneous value of any sinusoidally varying quantity x (current i, voltage v, etc.) is written as

\(x(t)=x_{0}\sin(\omega t+\phi)\)

  • \(x_{0}\) – peak (maximum) value, also called the amplitude.

  • \(\omega\) – angular frequency (rad s\(^{-1}\)).

  • t – time (s).

  • \(\phi\) – phase angle (rad) measured from a chosen reference waveform.

1.2 Relationship between angular frequency, frequency and period

\(\displaystyle \omega = 2\pi f = \frac{2\pi}{T}\)

  • f – frequency (Hz), the number of cycles per second.

  • T – period (s), the time for one complete cycle.

2. Key Quantities for a Sinusoidal AC

QuantitySymbolExpression (for \(x=x_{0}\sin\omega t\))Physical meaning
Peak (maximum) value\(x_{0}\)\(x_{0}\)Maximum magnitude reached during a cycle.
Peak‑to‑peak value\(x_{pp}\)\(2x_{0}\)Difference between the most positive and most negative values.
Root‑mean‑square (RMS) value\(x_{\rm rms}\)\(\displaystyle \frac{x_{0}}{\sqrt{2}}\)Effective DC equivalent; produces the same heating effect in a resistor as a DC current of the same magnitude.
Average (mean) value over a full cycle\(\overline{x}\)0For a pure sinusoid the algebraic average over one complete cycle is zero; therefore RMS, not average, is used for power calculations.
PeriodT\(\displaystyle \frac{2\pi}{\omega}\)Time taken for one complete sinusoidal cycle.
Frequencyf\(\displaystyle \frac{1}{T}= \frac{\omega}{2\pi}\)Number of cycles per second.

2.1 Example – UK mains supply

For a typical UK mains voltage described by \(v(t)=230\sin(2\pi\;50\,t)\) V:

  1. Peak voltage: \(v_{0}=230\;\text{V}\).

  2. RMS voltage: \(v_{\rm rms}= \dfrac{230}{\sqrt{2}}\approx 163\;\text{V}\).

  3. Period: \(T=\dfrac{1}{50}=0.020\;\text{s}\).

  4. Angular frequency: \(\omega = 2\pi f = 2\pi\times50\approx 314\;\text{rad s}^{-1}\).

3. Phase Relationships

Two sinusoidal quantities can be written as

\(x(t)=x{0}\sin(\omega t+\phi{x}),\qquad

y(t)=y{0}\sin(\omega t+\phi{y})\)

The phase difference is \(\Delta\phi = \phi{y}-\phi{x}\).

  • In‑phase: \(\Delta\phi = 0^{\circ}\) (or 0 rad) – maxima, minima and zero‑crossings occur simultaneously.

  • Out‑of‑phase (180°): \(\Delta\phi = \pi\) – the waveforms are exact opposites.

  • Quarter‑cycle shift (90°): \(\Delta\phi = \pm\frac{\pi}{2}\) – the peak of one coincides with the zero crossing of the other.

3.1 Phasor representation

A sinusoid can be represented by a rotating vector (phasor) of magnitude \(x_{0}\) making an angle \(\phi\) with the real axis. The instantaneous value is obtained by projecting the phasor onto the horizontal axis.

Phasor diagram showing two vectors with a phase difference \(\Delta\phi\). (Insert diagram)

4. AC in Basic Circuit Elements

4.1 Resistor (R)

\(v{R}(t)=i{R}(t)R\)

  • Voltage and current are in phase (\(\Delta\phi=0\)).

  • RMS relationship: \(V{\rm rms}=I{\rm rms}R\).

4.2 Inductor (L)

\(v{L}(t)=L\frac{di}{dt}=L\omega I{0}\cos\omega t

=\omega L I_{0}\sin\!\left(\omega t+\frac{\pi}{2}\right)\)

  • Voltage leads current by \(+90^{\circ}\).

  • Inductive reactance: \(X_{L}= \omega L\) (Ω).

  • RMS form: \(V{\rm rms}=I{\rm rms}X_{L}\).

4.3 Capacitor (C)

\(i{C}(t)=C\frac{dv}{dt}=C\omega V{0}\cos\omega t

=\omega C V_{0}\sin\!\left(\omega t-\frac{\pi}{2}\right)\)

  • Current leads voltage by \(+90^{\circ}\) (or voltage lags current by \(90^{\circ}\)).

  • Capacitive reactance: \(X_{C}= \dfrac{1}{\omega C}\) (Ω).

  • RMS form: \(I{\rm rms}=V{\rm rms}/X_{C}\).

4.4 Summary of phase and reactance

ElementVoltage–Current PhaseReactanceImpedance \(Z\)
Resistor (R)In phase (0°)—\(Z = R\)
Inductor (L)Voltage leads current (+90°)\(X_{L}= \omega L\)\(Z = jX_{L}\)
Capacitor (C)Current leads voltage (‑90°)\(X_{C}= \dfrac{1}{\omega C}\)\(Z = -jX_{C}\)

5. Power in Sinusoidal AC Circuits

The instantaneous power is the product of instantaneous voltage and current:

\(p(t)=v(t)i(t)\)

For a single sinusoid where the voltage and current have a phase difference \(\Delta\phi\), the average (real) power over one complete cycle is

\(P{\rm av}=V{\rm rms}I_{\rm rms}\cos\Delta\phi\)

The factor \(\cos\Delta\phi\) is the power factor.

5.1 Apparent, real and reactive power

  • Apparent power \(S = V{\rm rms}I{\rm rms}\) (unit VA).

  • Real (active) power \(P = S\cos\Delta\phi\) (unit W).

  • Reactive power \(Q = S\sin\Delta\phi\) (unit VAr).

5.2 Power triangle

S
/|
/ |
Q /  | P
/   |
/|
φ

Mathematically, \(S^{2}=P^{2}+Q^{2}\) and the angle between \(S\) and \(P\) is the phase difference \(\Delta\phi\). The triangle is a useful visual aid in exam questions.

6. Rectification and Smoothing (21.2 Syllabus)

6.1 Half‑wave rectifier

  • Only one half of the AC waveform (positive or negative) is allowed to pass.

  • Peak output voltage ≈ \(V{\rm pk}-V{D}\) (where \(V_{D}\) is the diode forward drop).

  • Average (DC) output voltage: \(\displaystyle V{\rm av}= \frac{V{\rm pk}}{\pi}\) for an ideal diode.

  • RMS value of the rectified output: \(\displaystyle V{\rm rms}= \frac{V{\rm pk}}{2}\).

6.2 Full‑wave (bridge) rectifier

  • Both halves of the AC waveform are utilised, giving twice the frequency of ripple.

  • Peak output voltage ≈ \(V{\rm pk}-2V{D}\).

  • Average (DC) output voltage: \(\displaystyle V{\rm av}= \frac{2V{\rm pk}}{\pi}\) (ideal diodes).

  • RMS value of the rectified output: \(\displaystyle V{\rm rms}= \frac{V{\rm pk}}{\sqrt{2}}\).

6.3 Smoothing with a capacitor

A filter capacitor placed across the load stores charge during the peaks of the rectified waveform and releases it when the waveform falls, reducing the ripple.

  • Ripple voltage (peak‑to‑peak) for a full‑wave rectifier:

    • \(\displaystyle V{r}\;\approx\;\frac{I{\rm load}}{f\,C}\)

  • For a half‑wave rectifier the same expression uses \(f\) (the line frequency) instead of \(2f\).

  • Increasing the capacitance \(C\) or the load resistance (reducing \(I_{\rm load}\)) lowers the ripple.

6.4 Example – 240 V mains, bridge rectifier, 100 µF filter

  1. Peak of the sinusoid: \(V_{\rm pk}= \sqrt{2}\times 240\approx 340\;\text{V}\).

  2. Assuming ideal diodes, \(V{\rm av}= \dfrac{2V{\rm pk}}{\pi}\approx 216\;\text{V}\).

  3. Load current \(I{\rm load}=0.5\;\text{A}\); ripple \(V{r}\approx \dfrac{0.5}{(2\times50)\times100\times10^{-6}}\approx 50\;\text{V}\).

  4. Choosing a larger capacitor (e.g., 1000 µF) would reduce the ripple to about 5 V.

7. Common Mistakes to Avoid (Exam Tips)

  1. Never forget the RMS factor \(\frac{1}{\sqrt{2}}\) when converting from peak to RMS values.

  2. Remember that the average value of a pure sinusoid over a full cycle is zero; therefore RMS, not average, is used for power calculations.

  3. When asked for a phase relationship, explicitly state which quantity leads and which lags (e.g., “voltage leads current by \(90^{\circ}\) in an inductor”).

  4. Use the correct sign for reactance in impedance: \(+jX{L}\) for inductors, \(-jX{C}\) for capacitors.

  5. In power‑factor problems, first determine \(\Delta\phi\) from the circuit (R, L, C combination) before applying \(P{\rm av}=V{\rm rms}I_{\rm rms}\cos\Delta\phi\).

  6. For rectifier questions, write down whether it is half‑wave or full‑wave, state the peak output voltage (accounting for diode drops), and use the appropriate formula for average voltage and ripple.

8. Quick Reference Checklist (for the exam)

  • Write the sinusoidal form \(x=x_{0}\sin(\omega t+\phi)\) and label each symbol.

  • Convert between \(\omega\), \(f\) and \(T\) using \(\omega = 2\pi f = 2\pi/T\).

  • Calculate RMS values: \(x{\rm rms}=x{0}/\sqrt{2}\).

  • State the phase relationship and reactance for R, L and C.

  • Apply \(P{\rm av}=V{\rm rms}I_{\rm rms}\cos\Delta\phi\) to find real power and identify the power factor.

  • Draw clear phasor diagrams and the power‑triangle (label units: W, VA, VAr).

  • For rectifiers: identify half‑wave or full‑wave, write the peak and average DC voltages, and use \(V{r}\approx I{\rm load}/(fC)\) (or \(I_{\rm load}/(2fC)\) for full‑wave) to estimate ripple.

  • Check units at every step – especially when mixing peak, RMS and average quantities.