Characteristics of alternating currents

1. Overview

Alternating current (AC) is a flow of electric charge that periodically reverses its direction and changes its magnitude continuously with time. Unlike direct current (DC), which flows in a single constant direction, AC oscillates sinusoidally in most practical applications, including the national grid and domestic power supplies.

The fundamental physics of AC revolves around the fact that the instantaneous value of current and voltage varies over time. Because the average value of a sinusoidal current over a complete cycle is zero, we cannot use simple averages to describe the "strength" of the current. Instead, we use the root-mean-square (r.m.s.) value, which allows us to relate the heating effect of an alternating current to that of a direct current. Understanding the relationship between peak values, r.m.s. values, and mean power is essential for analyzing any AC circuit.

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Key Definitions

• Alternating Current (AC): A current which periodically reverses its direction and has a continuously changing magnitude.
• Peak Value ($I_0$ or $V_0$): The maximum value of the current or voltage in either direction, measured from the zero-line to the crest or trough.
• Peak-to-Peak Value ($V_{p-p}$): The total vertical distance between the positive peak and the negative peak ($V_{p-p} = 2V_0$).
• Period ($T$): The time taken for the alternating quantity to complete one full cycle. Measured in seconds (s).
• Frequency ($f$): The number of complete cycles per unit time. Measured in Hertz (Hz), where $1 \text{ Hz} = 1 \text{ s}^{-1}$.
• Angular Frequency ($\omega$): The rate of change of the phase angle of the sinusoidal waveform, defined as the product of $2\pi$ and the frequency. Measured in radians per second ($\text{rad s}^{-1}$).
• Root-mean-square (r.m.s.) value: The value of steady direct current (DC) that would dissipate the same average power in a given resistor as the alternating current.

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Content

3.1 Sinusoidal Equations

In Cambridge A-Level Physics, we model alternating currents and voltages as sine (or cosine) functions of time. This is because the rotation of a coil in a uniform magnetic field (the basis of an AC generator) produces an e.m.f. that varies sinusoidally.

The instantaneous value $x$ at any time $t$ is given by:

$$x = x_0 \sin(\omega t)$$

Where:

• $x$ is the instantaneous current ($I$) or voltage ($V$).
• $x_0$ is the peak value ($I_0$ or $V_0$).
• $\omega$ is the angular frequency, related to period and frequency by: $$\omega = 2\pi f = \frac{2\pi}{T}$$
• $t$ is the time elapsed since the start of the cycle.

Note on Phase: If the timing starts when the voltage is at its maximum, the equation $V = V_0 \cos(\omega t)$ is used. However, the syllabus focuses on the sine form.

3.2 Power in AC Circuits

When an alternating current flows through a resistor $R$, the electrical energy is converted into thermal energy (Joule heating). Because the current is constantly changing, the rate of energy transfer (power) is also constantly changing.

Instantaneous Power ($P$): Using $P = I^2 R$, and substituting $I = I_0 \sin(\omega t)$: $$P = (I_0 \sin(\omega t))^2 R = I_0^2 R \sin^2(\omega t)$$

Maximum (Peak) Power ($P_{max}$): The maximum power occurs when the current is at its peak ($I_0$). At this point, $\sin^2(\omega t) = 1$. $$P_{max} = I_0^2 R$$

Mean Power ($P_{mean}$): The mean power is the average value of the instantaneous power over one or more full cycles. Mathematically, the average value of $\sin^2(\theta)$ over a full cycle is exactly $0.5$. This leads to the fundamental relationship:

$$P_{mean} = \frac{1}{2} I_0^2 R$$ $$P_{mean} = \frac{1}{2} P_{max}$$

The Power Graph: If you plot Power ($P$) against time ($t$):

1. The graph is always positive (since $I^2$ is always positive).
2. The graph oscillates between $0$ and $P_{max}$.
3. The frequency of the power oscillations is twice the frequency of the current/voltage oscillations.

3.3 Root-Mean-Square (r.m.s.) Values

The r.m.s. value is the "effective" value of an AC supply. It allows us to use standard DC power formulas ($P = IV$, $P = I^2 R$, $P = V^2/R$) for AC circuits, provided we use the r.m.s. values.

The Relationship for Sinusoidal AC: For a perfectly sinusoidal wave, the r.m.s. value is related to the peak value by the factor $\frac{1}{\sqrt{2}}$:

$$I_{r.m.s.} = \frac{I_0}{\sqrt{2}}$$ $$V_{r.m.s.} = \frac{V_0}{\sqrt{2}}$$

(Note: $\frac{1}{\sqrt{2}} \approx 0.707$, meaning the r.m.s. value is roughly $71%$ of the peak value.)

Derivation of the r.m.s. Relationship:

1. By definition, $P_{mean} = I_{r.m.s.}^2 R$.
2. From the physics of the sine wave, $P_{mean} = \frac{1}{2} I_0^2 R$.
3. Equating the two: $I_{r.m.s.}^2 R = \frac{1}{2} I_0^2 R$.
4. Cancel $R$: $I_{r.m.s.}^2 = \frac{I_0^2}{2}$.
5. Taking the square root: $I_{r.m.s.} = \frac{I_0}{\sqrt{2}}$.

3.4 Interpreting Oscilloscope Traces

In the laboratory, AC characteristics are measured using a Cathode Ray Oscilloscope (CRO).

• Vertical Axis (Y-plates): Represents Voltage. The "Y-gain" or "Volts/div" setting allows you to calculate $V_0$.
  • $V_0 = (\text{number of divisions from center to peak}) \times (\text{Volts/div})$
• Horizontal Axis (X-plates): Represents Time. The "Time-base" or "Time/div" setting allows you to calculate $T$.
  • $T = (\text{number of divisions for one cycle}) \times (\text{Time/div})$
• Once $T$ is found, calculate $f = 1/T$.

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Worked Example 1 — Analyzing an AC Equation

A sinusoidal alternating current is described by the equation $I = 4.5 \sin(120\pi t)$, where $I$ is in Amperes and $t$ is in seconds. Calculate: (a) The peak current. (b) The frequency of the supply. (c) The r.m.s. current. (d) The instantaneous current at $t = 2.5 \text{ ms}$.

Solution:

(a) Peak Current ($I_0$): Compare $I = 4.5 \sin(120\pi t)$ to the standard form $I = I_0 \sin(\omega t)$. $I_0 = 4.5 \text{ A}$

(b) Frequency ($f$): Identify $\omega = 120\pi \text{ rad s}^{-1}$. Using $\omega = 2\pi f$: $120\pi = 2\pi f$ $f = \frac{120\pi}{2\pi} = 60 \text{ Hz}$

(c) r.m.s. Current ($I_{r.m.s.}$): $I_{r.m.s.} = \frac{I_0}{\sqrt{2}}$ $I_{r.m.s.} = \frac{4.5}{\sqrt{2}} = 3.18 \text{ A}$

(d) Instantaneous current at $t = 2.5 \times 10^{-3} \text{ s}$: $I = 4.5 \sin(120\pi \times 2.5 \times 10^{-3})$ $I = 4.5 \sin(0.3\pi)$ Ensure calculator is in RADIANS mode. $I = 4.5 \times 0.809 = 3.64 \text{ A}$

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Worked Example 2 — Power and Energy

A $150 \Omega$ heating element is connected to an AC mains supply quoted as $230 \text{ V}, 50 \text{ Hz}$. (a) Calculate the peak voltage of the supply. (b) Calculate the mean power dissipated by the heating element. (c) Calculate the energy dissipated in 10 minutes.

Solution:

(a) Peak Voltage ($V_0$): Mains values are quoted as r.m.s. unless stated otherwise. $V_{r.m.s.} = 230 \text{ V}$ $V_0 = V_{r.m.s.} \times \sqrt{2}$ $V_0 = 230 \times \sqrt{2} = 325.3 \text{ V}$ Answer: $325 \text{ V}$

(b) Mean Power ($P_{mean}$): $P_{mean} = \frac{V_{r.m.s.}^2}{R}$ $P_{mean} = \frac{230^2}{150} = \frac{52900}{150} = 352.67 \text{ W}$ Answer: $353 \text{ W}$

(c) Energy ($E$): $E = P_{mean} \times t$ $t = 10 \times 60 = 600 \text{ s}$ $E = 352.67 \times 600 = 211,602 \text{ J}$ Answer: $2.12 \times 10^5 \text{ J}$ (or $212 \text{ kJ}$)

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Worked Example 3 — CRO Trace Interpretation

An AC voltage is displayed on a CRO. The trace shows a complete cycle occupying $4.0$ horizontal divisions. The peak-to-peak height of the trace is $6.0$ vertical divisions. The time-base is set to $5.0 \text{ ms/div}$ and the Y-gain is set to $10 \text{ V/div}$. Determine the r.m.s. voltage and the frequency.

Solution:

Step 1: Determine the Period ($T$) $T = \text{divisions} \times \text{time-base}$ $T = 4.0 \text{ div} \times 5.0 \text{ ms/div} = 20 \text{ ms} = 0.020 \text{ s}$

Step 2: Calculate Frequency ($f$) $f = \frac{1}{T} = \frac{1}{0.020} = 50 \text{ Hz}$

Step 3: Determine Peak Voltage ($V_0$) The peak-to-peak height is $6.0$ divisions. Therefore, the peak height (from center) is $3.0$ divisions. $V_0 = \text{peak divisions} \times \text{Y-gain}$ $V_0 = 3.0 \text{ div} \times 10 \text{ V/div} = 30 \text{ V}$

Step 4: Calculate $V_{r.m.s.}$ $V_{r.m.s.} = \frac{V_0}{\sqrt{2}} = \frac{30}{\sqrt{2}} = 21.21 \text{ V}$

Answer: $f = 50 \text{ Hz}$, $V_{r.m.s.} = 21 \text{ V}$

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Key Equations

Quantity	Equation	Data Sheet?
Instantaneous Value	$x = x_0 \sin \omega t$	No
Angular Frequency	$\omega = 2\pi f = \frac{2\pi}{T}$	No
r.m.s. Current	$I_{r.m.s.} = \frac{I_0}{\sqrt{2}}$	Yes
r.m.s. Voltage	$V_{r.m.s.} = \frac{V_0}{\sqrt{2}}$	Yes
Mean Power (Resistive)	$P_{mean} = I_{r.m.s.}^2 R = \frac{V_{r.m.s.}^2}{R}$	No
Mean Power (General)	$P_{mean} = I_{r.m.s.} V_{r.m.s.} = \frac{1}{2} I_0 V_0$	No
Peak-to-Peak	$V_{p-p} = 2V_0$	No

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Common Mistakes to Avoid

• ❌ Confusing Peak and Peak-to-Peak: Students often use the full height of a CRO trace as $V_0$.
  • ✓ Right: $V_0$ is the distance from the zero-axis to the peak. Always divide the peak-to-peak value by 2 before using the r.m.s. formula.
• ❌ Calculator Mode Errors: Using Degrees mode for sinusoidal equations.
  • ✓ Right: The term $\omega t$ is in radians. Your calculator must be in RAD mode for any calculation involving $x = x_0 \sin(\omega t)$.
• ❌ Incorrect r.m.s. Definition: Defining $I_{r.m.s.}$ as "the average current."
  • ✓ Right: The average current of a sine wave is zero. $I_{r.m.s.}$ must be defined in terms of power equivalence to a steady direct current.
• ❌ Misinterpreting Mains Voltage: Assuming "230 V AC" means the peak voltage is 230 V.
  • ✓ Right: Standard AC ratings (like those on appliances) are always r.m.s. values. The peak voltage of a 230 V supply is actually $\approx 325 \text{ V}$.
• ❌ Power Formula Confusion: Using $P = I_0 V_0$ for average power.
  • ✓ Right: $I_0 V_0$ gives the peak power. The mean power is half of this: $P_{mean} = \frac{1}{2} I_0 V_0$.

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Exam Tips

1. The "Heating Effect" Definition: If an exam question asks "What is meant by the r.m.s. value of an alternating current?", you must mention:
   • It is a steady or direct current.
   • It produces the same mean power (or heating effect).
   • In the same resistor.
2. Sketching Power Graphs: If asked to sketch a graph of power against time for an AC circuit:
   • Ensure the graph never goes below the x-axis.
   • Ensure the peaks of the power graph align with both the peaks and the troughs of the current/voltage graph.
   • The power graph should look like a $\sin^2$ wave, which is always positive.
3. Unit Conversions: Be extremely careful with milliseconds (ms) on CRO time-bases. $1 \text{ ms} = 10^{-3} \text{ s}$. Forgetting this conversion will lead to a frequency error of a factor of 1000.
4. Significant Figures: In AC questions, values like $\sqrt{2}$ are constants. Your final answer should be rounded to the number of significant figures provided in the question data (usually 2 or 3 s.f.).
5. Non-Sinusoidal Waves: Note that the $I_{r.m.s.} = I_0 / \sqrt{2}$ relationship only applies to sinusoidal waves. If the exam provides a square wave or triangular wave, you must use the basic definition of r.m.s. (square the values, find the mean, then take the square root).