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RRB ALP CBT 2 Electronic Mechanic Previous Paper: Held on 23 Jan 2019 Shift 1

Option 4 : RMS × Square root of eight

RRB ALP 2018 Full Test (Stage II - Part A)

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__Concept__:

The RMS value of a sinusoidal wave is given by:

\(V_{rms}=\frac{V_m}{√2}\)

Where V_{m} is the peak voltage of the sine wave.

**Derivation:**

A general sinusoidal wave can be expressed as:

v(t) = V_{m }sin (ωt)

V_{m} = Peak Value

ω = Angular frequency

RMS value 'or' the effective value of an alternating quantity is calculated as:

\({V_{rms}} = √{\frac{1}{T}\mathop \smallint \limits_0^T {v^2}\left( t \right)dt} \)

T = Time period

\({V_{rms}} = √{\frac{1}{T}\mathop \smallint \limits_0^T {V_m^2~sin^2}\left( \omega t \right)dt} \)

\({V_{rms}} = √{\frac{1}{T}\mathop \smallint \limits_0^T \frac{V_m^2-V_m^2~cos(2\omega t)} {2} dt}\)

\({V_{rms}} = √{\frac{1}{T}(\frac{V_m^2} {2}) [t]_0^T}\)

\({V_{rms}} = √{\frac{V_m^2} {2}}\)

\({V_{rms}} = \frac{V_m} {√{{2}}}\)

Peak to peak value = 2V_{m}

2V_{m} = V_{rms} × 2√2

2V_{m} = V_{rms} × √8

The average value of sinusoidal waveform over one complete cycle is zero as two halves cancel each other, so the average value is taken over half a cycle.

\({V_{avg}} = \frac{1}{\pi }\mathop \smallint \nolimits_0^\pi {V_p}\sin \theta \;d\theta \)

\({V_{avg}} = \frac{{2{V_p}}}{\pi } = 0.637 × {V_p}\)