Alternating current rises and falls and reverses many times a second. To compare it fairly with steady d.c. we use the root-mean-square value, the steady current that would deliver the same average power.
A sinusoidal current is i = I₀ sin ωt, with peak value I₀, angular frequency ω = 2πf and period T = 1/f. The power i²R is always positive and averages to half its peak, so the mean power is ½ the peak power. The r.m.s. value is then I₀/√2, and likewise V₀/√2 for voltage.
The top trace is the current, i = I₀ sinωt. The lower trace is the power, which never goes negative and oscillates at twice the frequency. Its mean sits at exactly half the peak, and that is the key to the r.m.s. value. Change the peak and the frequency and read the values off.
The r.m.s. value is built from the power, not the current.
The simple average of a sinusoidal current over a whole cycle is zero, which is why we use the r.m.s. value instead. The factor 1/√2 applies only to sinusoidal signals. The r.m.s. current is defined through power: it is the steady d.c. that dissipates the same mean power, so mean power = ½ peak power, not the peak power.
Four quick checks on peak, r.m.s. and power. Each correct answer earns XP and lights this skill on your star map.
The r.m.s. value of a sinusoidal current of peak value I₀ is:
For a sinusoidal supply, the mean power delivered to a resistor is:
The UK mains has an r.m.s. voltage of 230 V. The peak voltage is closest to:
A sinusoidal supply has a frequency of 50 Hz. Its period is:
Convert with care: to go from r.m.s. to peak multiply by √2; to go from peak to r.m.s. divide by √2. When finding mean power, use Iₕₘₛ and Vₕₘₛ (or ½ of the peak power), never the peak values directly.
This skill is now lit on your star map. Keep the chain going.