CIE AS/A Level Physics 21.1 Characteristics of alternating currents Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 21.1 Characteristics of alternating currents Study Notes – New Syllabus
CIE AS/A Level Physics 21.1 Characteristics of alternating currents Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- understand and use the terms period, frequency and peak value as applied to an alternating current or voltage
- use equations of the form x = x0 sin ωt representing a sinusoidally alternating current or voltage
- recall and use the fact that the mean power in a resistive load is half the maximum power for a sinusoidal alternating current
- distinguish between root-mean-square (r.m.s.) and peak values and recall and use Ir.m.s. = I0 /√ 2 and Vr.m.s. = V0 /√ 2 for a sinusoidal alternating current
Period, Frequency, and Peak Value in Alternating Current (AC) and Voltage
Alternating current (AC) and alternating voltage vary continuously with time, usually in the form of a sine wave. To describe this variation, three important terms are used: period, frequency, and peak value.
Period (T)
The period is the time taken for one complete cycle of the alternating waveform.
- Unit: seconds (s)
- Measured from any point on the waveform to the corresponding point one cycle later.
Frequency (f)
Frequency is the number of complete cycles per second.
- Unit: hertz (Hz)
- It is the reciprocal of the period:
\( \mathrm{f = \dfrac{1}{T}} \)
Peak Value ( \( \mathrm{I_0} \) or \( \mathrm{V_0} \) )
The peak value is the maximum instantaneous value of current or voltage in one cycle.
- This is the highest point on the waveform (positive or negative).
- For a sine wave:
\( \mathrm{I(t) = I_0 \sin(\omega t)} \), \( \mathrm{V(t) = V_0 \sin(\omega t)} \)
Relationship with Angular Frequency:
\( \mathrm{\omega = 2\pi f = \dfrac{2\pi}{T}} \)
Example
An AC supply completes one cycle in \( \mathrm{0.02\ s} \). Find the frequency of the supply.
▶️ Answer / Explanation
Use:
\( \mathrm{f = \dfrac{1}{T}} \)
\( \mathrm{f = \dfrac{1}{0.02} = 50\ Hz} \)
Frequency = 50 Hz
Example
An alternating voltage is given by \( \mathrm{V(t) = 120 \sin(100\pi t)} \). Determine the peak voltage and the frequency.
▶️ Answer / Explanation
Peak voltage:
\( \mathrm{V_0 = 120\ V} \)
Frequency from angular frequency:
\( \mathrm{\omega = 100\pi} \)
\( \mathrm{f = \dfrac{\omega}{2\pi} = \dfrac{100\pi}{2\pi} = 50\ Hz} \)
Peak voltage = 120 V, Frequency = 50 Hz
Example
An AC current has a peak value of \( \mathrm{0.8\ A} \) and a frequency of 60 Hz. Write an expression for the instantaneous current \( \mathrm{I(t)} \) and determine the current at \( \mathrm{t = 2.0\ ms} \).
▶️ Answer / Explanation
Step 1: Find angular frequency
\( \mathrm{\omega = 2\pi f = 2\pi \times 60 = 120\pi\ rad\,s^{-1}} \)
Step 2: Write AC equation
\( \mathrm{I(t) = I_0 \sin(\omega t) = 0.8 \sin(120\pi t)} \)
Step 3: Calculate current at \( \mathrm{t = 2.0\ ms = 0.002\ s} \)
\( \mathrm{I = 0.8 \sin(120\pi \times 0.002)} \)
\( \mathrm{I = 0.8 \sin(0.24\pi)} \)
\( \mathrm{\sin(0.24\pi) \approx 0.684} \)
\( \mathrm{I = 0.8 \times 0.684 = 0.547\ A} \)
Instantaneous current ≈ \( \mathrm{0.55\ A} \)
Using Equations of the Form \( \mathrm{x = x_0 \sin(\omega t)} \) for AC Voltage and Current
Alternating current (AC) and alternating voltage vary sinusoidally with time. They are commonly expressed using the equation:
\( \mathrm{x = x_0 \sin(\omega t)} \)
- \( \mathrm{x} \) = instantaneous current or voltage
- \( \mathrm{x_0} \) = peak (maximum) value, such as \( \mathrm{I_0} \) or \( \mathrm{V_0} \)
- \( \mathrm{\omega} \) = angular frequency (rad s⁻¹)
- \( \mathrm{t} \) = time (s)
Angular Frequency Relationship:
\( \mathrm{\omega = 2\pi f = \dfrac{2\pi}{T}} \)
Why a Sine Curve?
- Rotating coils in generators produce sinusoidal variation.
- The current and voltage rise smoothly from zero to a maximum, then reverse direction.
Graph Shape: A symmetric sine wave oscillating between \( +x_0 \) and \( -x_0 \).
Example
An AC voltage has a peak value of 15 V and a frequency of 25 Hz. Write an expression for the instantaneous voltage.
▶️ Answer / Explanation
Step 1: Calculate angular frequency
\( \mathrm{\omega = 2\pi f = 2\pi \times 25 = 50\pi\ rad\,s^{-1}} \)
Step 2: Write the sinusoidal equation
\( \mathrm{V(t) = 15 \sin(50\pi t)} \)
Instantaneous voltage: \( \mathrm{V(t) = 15 \sin(50\pi t)} \)
Example
An alternating current is described by \( \mathrm{I(t) = 6.0 \sin(200 t)} \). Determine the peak current and the frequency.
▶️ Answer / Explanation
Peak current:
\( \mathrm{I_0 = 6.0\ A} \)
Find angular frequency:
\( \mathrm{\omega = 200\ rad\,s^{-1}} \)
Use:
\( \mathrm{f = \dfrac{\omega}{2\pi} = \dfrac{200}{2\pi} \approx 31.8\ Hz} \)
Peak current = 6.0 A
Frequency ≈ 31.8 Hz
Example
A sinusoidal voltage is given by \( \mathrm{V(t) = 50 \sin(120\pi t)} \). Find the voltage at \( \mathrm{t = 3.0\ ms} \), and calculate the period \( \mathrm{T} \).
▶️ Answer / Explanation
Step 1: Voltage at \( \mathrm{t = 0.003\ s} \)
\( \mathrm{V = 50 \sin(120\pi \times 0.003)} \)
\( \mathrm{= 50 \sin(0.36\pi)} \)
\( \mathrm{\sin(0.36\pi) \approx 0.932} \)
\( \mathrm{V = 50 \times 0.932 = 46.6\ V} \)
Voltage at 3 ms ≈ 46.6 V
Step 2: Find the period
Given:
\( \mathrm{\omega = 120\pi} \)
Use:
\( \mathrm{T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{120\pi} = \dfrac{1}{60}} \)
\( \mathrm{T = 0.0167\ s = 16.7\ ms} \)
Period = 16.7 ms
Mean Power for a Sinusoidal Alternating Current in a Resistive Load
For a purely resistive load (e.g., a resistor connected to an AC supply), the instantaneous power varies with time because both the current and voltage are continuously changing.
For a sinusoidal current:
\( \mathrm{I(t) = I_0 \sin(\omega t)} \)
The instantaneous power is:
\( \mathrm{P(t) = I^2(t) R = I_0^2 R \sin^2(\omega t)} \)
Key Result:
\( \mathrm{P_{\text{mean}} = \dfrac{1}{2}P_{\text{max}}} \)
- Maximum power: \( \mathrm{P_{\text{max}} = I_0^2 R} \)
- Mean (average) power: \( \mathrm{P_{\text{mean}} = \tfrac{1}{2} I_0^2 R} \)
- Reason: \( \mathrm{\sin^2(\omega t)} \) has a time-average value of 0.5 over one full cycle.
Interpretation:
- The power oscillates between 0 and \( \mathrm{I_0^2 R} \).
- The average (mean) power supplied to the resistor is exactly half of the peak power.
- This relationship is why RMS values are used in AC circuits.
Example
An AC current has a peak value of \( \mathrm{I_0 = 4.0\ A} \) flowing through a \( \mathrm{10\ \Omega} \) resistor. Find the maximum power and the mean power.
▶️ Answer / Explanation
Maximum power:
\( \mathrm{P_{max} = I_0^2 R = (4.0)^2 \times 10 = 160\ W} \)
Mean power:
\( \mathrm{P_{mean} = \tfrac{1}{2}P_{max} = 80\ W} \)
Mean power = 80 W
Example
A sinusoidal AC source delivers a maximum power of 50 W to a resistor. Calculate the average power delivered.
▶️ Answer / Explanation
Use the key relationship:
\( \mathrm{P_{mean} = \tfrac{1}{2}P_{max}} \)
\( \mathrm{P_{mean} = \tfrac{1}{2}\times 50 = 25\ W} \)
Mean power = 25 W
Example
An AC circuit has a peak current of \( \mathrm{7.0\ A} \) through a resistor of \( \mathrm{5\ \Omega} \). Calculate the maximum power and verify that the mean power is half the maximum power.
▶️ Answer / Explanation
Step 1: Maximum power
\( \mathrm{P_{max} = I_0^2 R = 7.0^2 \times 5} \)
\( \mathrm{P_{max} = 49 \times 5 = 245\ W} \)
Step 2: Mean power
\( \mathrm{P_{mean} = \tfrac{1}{2}P_{max} = 122.5\ W} \)
Verification:
The average value of \( \sin^2(\omega t) \) over one cycle is 0.5 → the calculated mean power agrees with the theory.
Mean power = 122.5 W
Root-Mean-Square (r.m.s.) and Peak Values in AC Circuits
For sinusoidal alternating voltage or current, two values are commonly used:
- Peak value (\( \mathrm{I_0} \), \( \mathrm{V_0} \)) — the maximum instantaneous value reached in each cycle.
- Root-mean-square value (\( \mathrm{I_{\text{rms}}} \), \( \mathrm{V_{\text{rms}}} \)) — the effective steady value of DC that produces the same power in a resistor.
Definition:
The r.m.s. value is the square root of the mean of the square of the instantaneous values over one full cycle.
For a sinusoidal alternating current or voltage:
\( \mathrm{I_{\text{rms}} = \dfrac{I_0}{\sqrt{2}}} \)
\( \mathrm{V_{\text{rms}} = \dfrac{V_0}{\sqrt{2}}} \)
- This follows because the mean value of \( \mathrm{\sin^2(\omega t)} \) over one cycle is \( \tfrac{1}{2} \).
- Thus, r.m.s. value is ≈ 0.707 × peak value.
- R.m.s. values are used in AC mains ratings (e.g., “230 V AC”).
Distinguishing the two:
- Peak value — actual maximum magnitude reached.
- R.m.s. value — effective value for power calculations.
- For the same wave, \( \mathrm{I_0 = \sqrt{2} I_{\text{rms}}} \)
Example
A sinusoidal AC voltage has a peak value of 20 V. Find the r.m.s. voltage.
▶️ Answer / Explanation
\( \mathrm{V_{\text{rms}} = \dfrac{20}{\sqrt{2}} = 14.1\ V} \)
r.m.s. voltage ≈ 14.1 V
Example
The r.m.s. value of an AC current is 3.0 A. Find the peak current.
▶️ Answer / Explanation
Use:
\( \mathrm{I_0 = \sqrt{2}\, I_{\text{rms}}} \)
\( \mathrm{I_0 = 1.414 \times 3.0 = 4.24\ A} \)
Peak current = 4.24 A
Example
A voltage supply is described by \( \mathrm{V(t) = 50\sin(100\pi t)} \). Calculate the r.m.s. voltage and the mean power dissipated in a \( \mathrm{20\ \Omega} \) resistor.
▶️ Answer / Explanation
Step 1: Identify peak voltage
\( \mathrm{V_0 = 50\ V} \)
Step 2: Calculate r.m.s. voltage
\( \mathrm{V_{\text{rms}} = \dfrac{50}{\sqrt{2}} = 35.36\ V} \)
Step 3: Use AC power formula:
\( \mathrm{P = \dfrac{V_{\text{rms}}^2}{R}} \)
\( \mathrm{P = \dfrac{(35.36)^2}{20} = \dfrac{1250}{20} = 62.5\ W} \)
Mean power = 62.5 W