Retrieval Practice — Alternating Currents — Alternating Currents

Q1. State the difference between AC and DC.

In DC, current flows in one direction only.
In AC, the current periodically reverses direction.
DC can vary in magnitude but not direction; AC alternates between positive and negative.

Q2. State the frequency and RMS voltage of the UK mains supply.

230 V RMS, 50 Hz.

Q3. Calculate the peak voltage 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 power 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.

Q6. Explain why the average value of AC over a complete cycle is zero, but the RMS value is not.

The positive and negative halves of the sine wave cancel, giving a mean of zero.
But power depends on I² (or V²), which is always positive.
RMS squares the values first (removing negatives), averages, then square roots — giving a non-zero value that represents the true power capability.

Q7. Write the equation for the mean power dissipated in an AC circuit, in terms of (a) RMS values and (b) peak values.

(a) P_mean = I_rms × V_rms. (b) P_mean = I₀V₀/2.

Q8. A sinusoidal AC supply has a peak current of 4.0 A. Calculate the RMS current.

I_rms = I₀/√2 = 4.0/√2 = 2.83 A ≈ 2.8 A.

Q9. Describe how to measure the peak voltage and frequency of an AC signal using an oscilloscope.

Peak voltage: count divisions from the centre line to the peak, multiply by Y-gain (V/div).
Frequency: count divisions for one complete cycle, multiply by time base (s/div) to get period T, then f = 1/T.

Q10. State the relationship between peak power and mean power for a sinusoidal AC signal.