Alternating currents
| English | Chinese | Pinyin |
|---|---|---|
| alternating current | 交流电 | jiāo liú diàn |
| sine wave | 正弦波 | zhèng xián bō |
| peak value | 峰值 | fēng zhí |
| direct current | 直流电 | zhí liú diàn |
| root-mean-square | 均方根 | jūn fāng gēn |
Mains that flips 50 times a second
- Mains electricity is alternating current 交流电 — it reverses direction 50 times a second.
- Its value follows a sine wave 正弦波: $I = I_0\sin(\omega t)$.
- And "230 V mains" is not the peak — here's why.
The basics
- Period $T$, frequency $f = \dfrac{1}{T}$, angular frequency $\omega = 2\pi f$.
- Peak value 峰值 $I_0$ (the amplitude); peak-to-peak is $2I_0$.
A steady direct current 直流电 compared with a sinusoidal alternating current of peak $I_0$ and period $T$
Alternating current
I = a sin(bt)
AC is a sine wave — amplitude is the peak, b sets the frequency.
An alternating current:
a.c. keeps swapping direction — mains does so 50 (or 60) times a second.
Mains supply is $50\ \text{Hz}$. What is its period?
$T = \dfrac{1}{f} = \dfrac{1}{50} = 0.020\ \text{s}$.
Power in a resistor
- Instant power $P = I_0^{2}R\sin^{2}(\omega t)$ — always positive, peaking at $I_0^{2}R$.
- Since the average of $\sin^{2}$ is $\tfrac{1}{2}$, the average power is half the peak.
A substation's transformers step alternating voltage up or down
The average power of a.c. in a resistor is ____ the peak power.
The mean of $\sin^{2}$ over a cycle is $\tfrac{1}{2}$, so $\langle P\rangle = \tfrac{1}{2}P_{\text{peak}}$.
Root-mean-square 均方根 values
- The r.m.s. value is the steady DC that gives the same average power: $I_{\text{rms}} = \dfrac{I_0}{\sqrt{2}}$.
- Quoted AC values are r.m.s. "230 V mains" → peak $V_0 = 230\sqrt{2} \approx 325\ \text{V}$.
An oscilloscope shows how a voltage varies with time
An a.c. has a peak current of $2.0\ \text{A}$. What is the r.m.s. current?
$I_{\text{rms}} = \dfrac{I_0}{\sqrt{2}} = \dfrac{2.0}{\sqrt{2}} \approx 1.41\ \text{A}$.
A 230 V (r.m.s.) mains supply has a peak voltage of about 325 V.
$V_0 = V_{\text{rms}}\sqrt{2} = 230 \times \sqrt{2} \approx 325\ \text{V}$ — components must be rated for the peak.
You've got it
- a.c. is a sine wave: $I = I_0\sin(\omega t)$, peak $I_0$, frequency $f$
- average power in a resistor = half the peak ($\langle\sin^{2}\rangle = \tfrac{1}{2}$)
- r.m.s. = $\dfrac{\text{peak}}{\sqrt{2}}$ — the DC-equivalent value (what mains figures quote)