In the full-wave rectifier, relation between peak value and rms value is:

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ALP CBT 2 Electronic Mechanic Previous Paper: Held on 21 Jan 2019 Shift 2
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  1. \(V_p=\sqrt{4} V_{rms}\)
  2. \(V_p=\sqrt{3} V_{rms}\)
  3. \(V_p=\sqrt{2} V_{rms}\)
  4. \(V_p=\sqrt{1.2} V_{rms}\)

Answer (Detailed Solution Below)

Option 3 : \(V_p=\sqrt{2} V_{rms}\)
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The RMS or the effective value of a waveform f(t) is given by:
\({f_{rms}} = \sqrt {\frac{1}{T}\mathop \smallint \limits_0^T {f^2}\left( t \right)dt}\)

Application:

The relation between the peak value and rms value of the waveform is given as:

\(V_p=\sqrt{2} V_{rms}\)

26 June 1

The relation between the peak value and the average value for a half and full-wave rectifier is:

\({V_{avg}} = \frac{{{V_m}}}{\pi }\) for half-wave rectifier

\({V_{avg}} = \frac{{2{V_m}}}{\pi }\) for full-wave rectifier

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