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CalibratePhysics

13.3

Average and RMS Values

AC Circuits · Maharashtra HSC

Theory

13.3 Average and RMS Values

Alternating voltages and currents pass through every value between zero and the peak value in one cycle, so some specific values must be defined before waveforms can be compared or measured.

Peak Value

The peak value of an alternating current (or emf) is the maximum value of the current (or emf) in either direction. It is denoted i0i_0 (or e0e_0).

(a) Average Value of AC

The average value of AC is the average of all the values of the current (or voltage) taken over one half cycle.

The average over a full cycle is always zero, because the average value of sin⁡ωt\sin\omega t over a complete cycle is zero — the positive half exactly cancels the negative half. The mean value of AC over a full cycle therefore has no significance, and the mean value of AC is defined as the average over half a cycle.

The average value of sin⁡θ\sin\theta for θ\theta in the range 00 to π\pi is

∫0πsin⁡θ dθ∫0πdθ=[−cos⁡θ]0ππ=2π=0.637\dfrac{\displaystyle\int_0^{\pi}\sin\theta\,d\theta}{\displaystyle\int_0^{\pi}d\theta} = \dfrac{\left[-\cos\theta\right]_0^{\pi}}{\pi} = \dfrac{2}{\pi} = 0.637

Hence the average value of an alternating current or emf is 0.6370.637 times its peak value:

iav=0.637 i0=2i0π,eav=0.637 e0=2e0πi_{av} = 0.637\,i_0 = \dfrac{2i_0}{\pi}, \qquad e_{av} = 0.637\,e_0 = \dfrac{2e_0}{\pi}

(b) RMS Value

A moving-coil ammeter and voltmeter measure the average value of the current and voltage respectively. It is obvious, therefore, that moving-coil instruments cannot be used to measure alternating current and voltage, since that average over a cycle is zero.

To measure these quantities we must use a property that does not depend on the direction of flow. The heating effect is such a property: heat depends on the square of the current, and a square is always positive.

Consider an alternating current of peak value i0i_0 flowing through a resistance RR, and let HH be the heat produced in time tt. The same quantity of heat HH can be produced in the same resistance RR in the same time tt by passing a steady current of constant magnitude through it. The value of that steady current is called the effective value, or virtual value, or rms value of the given alternating current, denoted irmsi_{rms}.

irms=∫0Ti02sin⁡2ωt dt∫0Tdt=i022=i02=0.707 i0i_{rms} = \sqrt{\dfrac{\displaystyle\int_0^T i_0^2\sin^2\omega t\,dt}{\displaystyle\int_0^T dt}} = \sqrt{\dfrac{i_0^2}{2}} = \dfrac{i_0}{\sqrt2} = 0.707\,i_0

using ∫0Tsin⁡2ωt dt=T2\displaystyle\int_0^T\sin^2\omega t\,dt = \dfrac{T}{2}. Similarly, for the emf,

erms=e02=0.707 e0e_{rms} = \dfrac{e_0}{\sqrt2} = 0.707\,e_0

Heat Produced Over a Complete Cycle

H=∫0Ti2R dt=i02R∫0Tsin⁡2ωt dt=i02R(T2)=irms2 R TH = \int_0^T i^2R\,dt = i_0^2R\int_0^T\sin^2\omega t\,dt = i_0^2R\left(\dfrac{T}{2}\right) = i_{rms}^2\,R\,T

This is the same as the heat produced by a direct current of magnitude irmsi_{rms} flowing for the same time — which is exactly what the rms value was defined to mean.

Standard Results

  • iav=0.637 i0i_{av} = 0.637\,i_0, eav=0.637 e0e_{av} = 0.637\,e_0 (half cycle); zero over a full cycle.
  • irms=0.707 i0i_{rms} = 0.707\,i_0, erms=0.707 e0e_{rms} = 0.707\,e_0; conversely i0=2 irms=1.414 irmsi_0 = \sqrt2\,i_{rms} = 1.414\,i_{rms}.
  • H=irms2RtH = i_{rms}^2Rt.
  • Every AC meter reading, and every quoted supply voltage such as 220 V, is an rms value.

Worked Pattern

For e=8sin⁡628.4 te = 8\sin 628.4\,t: comparing with e=e0sin⁡ωte = e_0\sin\omega t gives e0=8e_0 = 8 V and ω=628.4\omega = 628.4 rad/s, so f=ω/2π=628.4/(2×3.142)=100f = \omega/2\pi = 628.4/(2\times3.142) = 100 Hz. At t=10t = 10 ms =10−2= 10^{-2} s, e=8sin⁡(6.284)=8sin⁡(2π)=0e = 8\sin(6.284) = 8\sin(2\pi) = 0 V.

For e=6sin⁡(100πt)e = 6\sin(100\pi t): e0=6e_0 = 6 V, f=100π/2π=50f = 100\pi/2\pi = 50 Hz, T=1/f=0.02T = 1/f = 0.02 s, and at t=2t = 2 ms, e=6sin⁡(100π×2×10−3)=3.527e = 6\sin(100\pi\times2\times10^{-3}) = 3.527 V.

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

  1. 1

    An alternating voltage is given by e=8sin⁡628.4te=8\sin628.4t. Find i. peak value of e.m.f. ii. frequency of e.m.f. iii. instantaneous value of e.m.f. at time t = 10 ms.

    Maharashtra HSC board question — asked in February 2024Answer 1 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

  1. The alternating voltage is given by, e=8sin⁡628.4te=8\sin628.4t i. On comparing with e=e0sin⁡ωte=e_0\sin\omega t, we get e0=8e_0=8 V ii. ω=628.4\omega=628.4 rad/s ∴Frequency, f=ω2π=628.42×3.142=100 Hz\therefore\text{Frequency, }f=\dfrac{\omega}{2\pi}=\dfrac{628.4}{2\times3.142}=100\text{ Hz} iii. At t=10t=10 ms =10−2=10^{-2} s, e=8sin⁡(628.4×10−2)e=8\sin(628.4\times10^{-2}) =8sin⁡(200×π100)=8\sin\left(\dfrac{200\times\pi}{100}\right) =8sin⁡(2π)=8\sin(2\pi) =0 V=0\text{ V} Ans: i. The peak value of emf is 8 V. ii. The frequency of emf is 100 Hz. iii. The instantaneous value of emf at t = 10 ms is 0 V.