### Effective or RMS value

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

### 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.

### 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.

### 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*}

### Solved Example:

#### 20-3-04

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

### Solved Example:

#### 20-3-05

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