Alternating Currents

Peak values, RMS values, and why AC power calculations need special treatment.

Spec Points Covered
  • Distinguish between AC and DC in terms of currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). direction and time variation.
  • Describe AC using $V = V_{0} \sin(\omega t)$ and I = I₀ sin(ωt), identifying peak values and frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz)..
  • Define RMS values as the DC equivalent for powerThe rate of energy transfer. Measured in watts (W). dissipation, and apply $V_{\text{rms}} = V_{0}/\sqrt 2$ and $I_{\text{rms}} = I_{0}/\sqrt 2$.
  • Explain why the mean value of AC over a full cycle is zero, but RMS is not.
  • Calculate mean powerThe rate of energy transfer. Measured in watts (W). in AC circuits using $P = I_{\text{rms}} \times V_{\text{rms}}$.
  • Read peak voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). Informal term for potential difference. and frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). from an oscilloscope trace and convert to RMS.
Notes
01 AC reverses direction periodically — DC does not Direct current (DC) 3.7.5.5 → 02 AC is described by peak values, frequency, and sine equations $\omega = 2\pi f$ 3.7.5.5 → 03 RMS is the DC equivalent for power — not the average Root mean square (RMS) value 3.7.5.5 → 04 The average of AC is zero, but power is always positive $P = I^2 R$ 3.7.5.5 → 05 Mean power in AC circuits uses RMS values $P_{\text{mean}} = I_{\text{rms}} \times V_{\text{rms}}$ 3.7.5.5 → 06 Reading AC values from an oscilloscope trace 3.7.5.5 → 07 UK mains: 230 V means RMS, the peak is 325 V 3.7.5.5 → 08 Comparing AC and DC on an oscilloscope — what the traces look like 3.7.5.5 →
Σ Key Equations Full Reference →
On Data Sheet
Not on Data Sheet
RMS voltage
$$V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$$
  • Where:
    • $V_{\text{rms}}$ = RMS voltage (V)
    • $V_0$ = peak voltage (V)
The DC voltage that would give the same power dissipation. 1/√2 ≈ 0.707. Applies to sinusoidal AC only.
Instantaneous AC voltage
$$V = V_0 \sin(\omega t)$$
  • Where:
    • $V$ = instantaneous voltage (V)
    • $V_0$ = peak voltage (V)
    • $\omega$ = angular frequency (rad s⁻¹)
    • $t$ = time (s)
Assumes V = 0 at t = 0. ω = 2πf. Calculator must be in radians.
RMS current
$$I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$$
  • Where:
    • $I_{\text{rms}}$ = RMS current (A)
    • $I_0$ = peak current (A)
Same factor as RMS voltage. The DC current that would produce the same heating in a resistor.
Mean power (RMS form)
$$P_{\text{mean}} = I_{\text{rms}} V_{\text{rms}}$$
  • Where:
    • $P_{\text{mean}}$ = mean power dissipated (W)
    • $I_{\text{rms}}$ = RMS current (A)
    • $V_{\text{rms}}$ = RMS voltage (V)
Standard P = IV but using RMS values. Also valid: P = I_rms²R and P = V_rms²/R.
Angular frequency
$$\omega = 2\pi f$$
  • Where:
    • $\omega$ = angular frequency (rad s⁻¹)
    • $f$ = frequency (Hz)
Connects frequency to the sin(ωt) equations. Also used in SHM and circular motion.
Mean power (peak form)
$$P_{\text{mean}} = \frac{I_0 V_0}{2}$$
  • Where:
    • $P_{\text{mean}}$ = mean power dissipated (W)
    • $I_0$ = peak current (A)
    • $V_0$ = peak voltage (V)
Mean power is half the peak power. The factor of 2 comes from (1/√2) × (1/√2) = 1/2.
Q Retrieval Practice All 10 Questions →
Q1. State the difference between AC and DC.
  • In DC, currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flows in one direction only.
  • In AC, the currentThe rate of flow of charge. Measured in amperes (A). periodically reverses direction.
  • DC can vary in magnitude but not direction; AC alternates between positive and negative.
Q2. State the frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). and RMS voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference. of the UK mains supply.
230 V RMS, 50 Hz.
Q3. Calculate the peak voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference. of the UK mains supply.
V₀ = V_rms × √2 = 230 × √2 = 325 V.
Q4. Define the RMS value of an alternating current.
The value of direct current that would produce the same heating effect (same powerThe rate of energy transfer. Measured in watts (W). dissipation) in a given resistor.
Q5. State the equations relating RMS and peak values for a sinusoidal AC supply.
V_rms = V₀/√2 and I_rms = I₀/√2.