A half-wave rectifier is designed using a transformer and a diode. The primary winding of the transformer, with N1 turns, is connected to a 240 sin(ωt) V supply. The secondary winding has N2 turns. What is the rectified DC output voltage (Vdc) if N1/N2 = 1 : 1?

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  1. 240π
  2. 240/π
  3. 240
  4. 480/π

Answer (Detailed Solution Below)

Option 2 : 240/π
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Detailed Solution

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Concept:

A half-wave rectifier converts AC to DC by allowing only one half-cycle of the input waveform to pass. The DC output voltage is the average value of the rectified waveform.

The average DC voltage for a half-wave rectifier is given by:

\(V_{dc} = \frac{V_m}{\pi} \)

where Vm is the peak voltage of the input AC signal.

Given:

  • Primary voltage: \(V_p = 240 \sin(\omega t) \; V\)
  • Turns ratio \( \frac{N_1}{N_2} = 1:1 \)
  • Rectifier type: Half-wave

Calculation:

  1. Determine Secondary Voltage:

    Since the turns ratio is 1:1, the secondary voltage equals the primary voltage:

    \(V_s = V_p = 240 \sin(\omega t) \text{ V (peak)}\)

  2. Calculate DC Output Voltage:

    For a half-wave rectifier:
    \( V_{dc} = \frac{V_m}{\pi} = \frac{240}{\pi} \text{ V} \)

Thus the correct answer is option 2

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