AC Voltage and Simple AC Circuits: Quick Revision

A brief note on AC Voltage and Simple AC Circuits. Read it before a test. Try each question at the end before you open its answer.

The alternating voltage

• The equation: $v = V_m \sin \omega t$, with $\omega = 2\pi f$ and $T = \dfrac{1}{f}$.
• The rms value: $V_{rms} = \dfrac{V_m}{\sqrt{2}} = 0.707\, V_m$, and $I_{rms} = \dfrac{I_m}{\sqrt{2}}$. Meters and ratings give rms values.
• Its meaning: the steady current that heats a resistor as much. For any wave, square, take the mean, take the root.
• The mean: zero over a full cycle, and $\dfrac{2}{\pi}\, I_m$ over half a cycle.

One part at a time

• Resistor: $I_m = \dfrac{V_m}{R}$. The current is in phase with the voltage. $P = V_{rms} I_{rms} = \dfrac{V_{rms}^2}{R}$.
• Coil: $X_L = \omega L = 2\pi f L$. The current lags by $90^\circ$. It passes direct current and holds back fast changes.
• Capacitor: $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$. The current leads by $90^\circ$. It blocks direct current and passes fast changes.
• Both reactances: the unit is the ohm, $I_{rms} = \dfrac{V_{rms}}{X}$, and the average power is zero.
• Memory aid: ELI the ICE man. In L the voltage (E) comes before the current (I). In C the current comes before the voltage.

Phasors

• A phasor is an arrow that turns at $\omega$. Its length is the peak value and its vertical part is the value now.
• The angle between two phasors is the phase difference $\varphi$. A phase is a delay $\Delta t = \dfrac{\varphi}{360^\circ}\, T$.
• Voltages in series add as arrows, tip to tail, not as numbers.

A worked example

A pure coil of $0.12$ H and a pure capacitor of $47\ \mu$F are joined, one at a time, to a source of $18$ V rms and $60$ Hz. Find each current.

1. The coil. $X_L = 2\pi \times 60\ \text{Hz} \times 0.12\ \text{H} = 45.2\ \Omega$. So $I_{rms} = \dfrac{18\ \text{V}}{45.2\ \Omega} = 0.398$ A, lagging the voltage.
2. The capacitor. $X_C = \dfrac{1}{2\pi \times 60\ \text{Hz} \times 47 \times 10^{-6}\ \text{F}} = 56.4\ \Omega$. So $I_{rms} = \dfrac{18\ \text{V}}{56.4\ \Omega} = 0.319$ A, leading the voltage.

Practise with new numbers

Check yourself

1. A voltmeter on an AC supply reads $12$ V. What is the peak voltage? Why does the meter not read the average?

2. The frequency of a source is made $4$ times larger. What happens to the reactance of a coil and of a capacitor?

3. At $50$ Hz, the current peak of a pure part comes $5$ ms before the voltage peak. Which part is it?