Effective or RMS value

  • Calculate the root-mean-square value of a waveform.

The effective or RMS value of an alternating quantity is that steady current (DC) which when flowing through a given resistance for a given time produces the same amount of heat produced by the alternating current flowing through the same resistance for the same time.

\[X_{\mathrm{eff}} =X_{\mathrm{RMS}}=\left [ \frac{1}{T}\int_{0}^{T}x^{2}(t)dt \right ]^{\frac{1}{2}}\] RMS value of a sinusoidal current =

\[\dfrac{I_{m}}{\sqrt{2}} = 0.707 I_{m}\]

Solved Examples

Solved Example:

20-3-01

The R.M.S. value and mean value is the same in the case of: (SSC JE EE Paper 8- Oct 2020 Morning)

Solution:
In case of DC, mean (average), RMS and instantaneous values are same.
In case of symmetrical square wave (square wave with equal positive and negative peaks) also, RMS value = mean value = peak value.

Correct Answer: C

Solved Example:

20-3-02

For the same peak value which of the following wave will have the highest R.M.S. value ? (SSC JE EE Paper 4- Jan 2018 Morning)

Solution:
To get a high RMS value, you want a waveform that is as high as possible most of the time, which happens in case of square wave as the instantaneous value is constant (and high) for half cycle. Since the values are squared while calculating RMS values, during negative half also, same effect takes place.

Correct Answer: A

Solved Example:

20-3-03

For a sine wave with peak value I$_{max}$ the R.M.S. value is:

Solution:
\begin{align*} i_{RMS} &= \left[\dfrac{1}{\pi}\int_{0}^{\pi}i^{2}(t)dt \right ]^{\frac{1}{2}}\\ &=\left[\dfrac{I_{max}^2}{\pi}\int_{0}^{\pi} \sin ^2 t\ dt \right ]^{\frac{1}{2}}\\ &= \dfrac{I_{max}}{\sqrt{2}\sqrt{\pi}}\left[\int_{0}^{\pi} (1- \cos 2t)\ dt \right ]^{\frac{1}{2}}\\ &= \dfrac{I_{max}}{\sqrt{2}\sqrt{\pi}}\left[ \left(t- \dfrac{\sin 2t}{2}\right)_0^\pi \right ]^{\frac{1}{2}}\\ &= \dfrac{I_{max}}{\sqrt{2}\sqrt{\pi}}\left[ (\pi - 0)-(0-0)\right]^{\frac{1}{2}}\\ &= \dfrac{I_{max}\sqrt{\pi}}{\sqrt{2}\sqrt{\pi}} = \dfrac{I_{max}}{\sqrt{2}} = 0.707I_{max} \end{align*}

Correct Answer: B

Solved Example:

20-3-04

A light bulb is rated at 100W for a 220 V supply. The peak voltage of the source is:

Correct Answer: C

Solved Example:

20-3-05

The RMS value of the periodic waveform shown in the following figure is:

20.3-05



Solution:
The RMS voltage (or current) of a periodic waveform is defined as the effective value equivalent of a steady DC (constant) value which gives the same effect. \begin{align*} V_{RMS} &= \left[\dfrac{1}{T}\int_{0}^{T}v^{2}(t)dt \right ]^{\frac{1}{2}}\\ &=\left[\frac{1}{6}\int_{0}^{2}v^{2}(t)dt + \int_{2}^{5}v^{2}(t)dt + \int_{5}^{6}v^{2}(t)dt\right]^{\frac{1}{2}}\\ &=\left[\dfrac{1}{6}\int_{0}^{2}(1.5t)^{2}dt + \int_{2}^{5}(3)^{2}dt + \int_{5}^{6}0^{2}dt\right]^{\frac{1}{2}}\\ &= \left[\dfrac{1}{6}\left(6 + 27 + 0\right) \right]^{\frac{1}{2}}\\ &= 2.35\ V \end{align*}

Correct Answer: B