Question
Download Solution PDFA 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?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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:
- Determine Secondary Voltage:
Since the turns ratio is 1:1, the secondary voltage equals the primary voltage:
\(V_s = V_p = 240 \sin(\omega t) \text{ V (peak)}\)
- 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
Last updated on Jun 7, 2025
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