Alternating current: peak and rms values

What alternating current is

An alternating voltage changes its size and direction regularly with time, following a sine curve: v = vₘ sin ωt. Here vₘ is the peak value (amplitude) and ω is the angular frequency, ω = 2πν, where ν (nu) is the frequency. The current it drives is an alternating current (AC). In a pure resistor R, i = iₘ sin ωt with iₘ = vₘ/R. The current and the voltage rise and fall together: they are in phase. Most devices in homes and offices run on AC. One reason is that an AC voltage is easy to step up or down with a transformer. In your textbook, small letters v and i are instantaneous values, vₘ and iₘ are peak values, and capital V and I are rms values.

Why the rms value is iₘ/√2

1. 1
   In a resistor, i = iₘ sin ωt, so the power is p = i²R = iₘ²R sin² ωt.
2. 2
   sin² ωt = ½(1 − cos 2ωt), and the average of cos 2ωt over a full cycle is zero. So the average of sin² ωt is ½.
3. 3
   Average power = ½iₘ²R.
4. 4
   The rms (root mean square) current I is the steady current that gives the same average power in R: I²R = ½iₘ²R.
5. 5★
   So, I = iₘ/√2 ≈ 0.707 iₘ. In the same way, V = vₘ/√2.

★ key step

Points to remember

• The average of i over a full cycle is zero, because the positive and negative halves cancel.
• AC ammeters and voltmeters read rms values.
• When we say the mains is 220 V, we mean the rms value. The peak is √2 × 220 V ≈ 311 V.
• With rms values, the DC formulas work for a resistor: P = VI = I²R = V²/R.
• The rms value is also called the effective value.

Worked example: an electric iron

Activity: frequency of AC mains with a sonometer

You need a sonometer with a thin metal wire, a step-down transformer, a key, a strong horseshoe magnet, a hanger with slotted masses and a metre scale.

1. 1
   Stretch the wire over the two bridges of the sonometer. Hang a known mass M on the hanger, so the tension is T = Mg.
2. 2
   Join the two ends of the wire, through a key, to the secondary of a step-down transformer connected to the AC mains.
3. 3
   Place the horseshoe magnet at the middle of the wire, with one pole on each side, so its field is horizontal and perpendicular to the wire.
4. 4
   Close the key. The alternating current in the wire feels a force IlB that reverses every half cycle, so the wire is forced to vibrate at the mains frequency.

   If an AC electromagnet pulls on a steel wire instead, the pull comes twice in each cycle, so the wire vibrates at 2ν.
5. 5
   Move the bridges, keeping the magnet midway between them, until the wire vibrates with the largest amplitude in one loop. This is resonance.
6. 6
   Record the length l between the bridges. Repeat with three or four different masses.
7. 7
   For each reading, calculate ν = (1/2l)√(T/m), where m is the mass per unit length of the wire. Take the mean.
8. 8★
   Result: the mean value comes out close to 50 Hz, the frequency of the Indian mains.

★ key step

Worked example: a sonometer reading