Concept:In a center-tapped rectifier, each diode is fed by one half of the total secondary winding. The peak voltage is \( V_m = \sqrt{2} \times V_{\text{half(rms)}} \).
Formula:$$V_{\text{half(rms)}} = \frac{V_{\text{total(rms)}}}{2} = \frac{24.0\text{ V}}{2} = 12.0\text{ V}$$
$$V_m = \sqrt{2} \times V_{\text{half(rms)}} = 1.4142 \times 12.0\text{ V} \approx 16.97\text{ V}$$
Solution:- Total secondary RMS voltage is \( 24.0\text{ V} \), so each half-winding supplies \( 12.0\text{ V} \) RMS relative to the center tap.
- The peak voltage per half-cycle is \( V_m = 12.0 \times \sqrt{2} \approx 16.97\text{ V} \).
- For ideal diodes, the peak load voltage is \( 16.97\text{ V} \).
Why other options are incorrect:- Option A: 24.0 V is the total RMS secondary voltage.
- Option C: 33.94 V is the peak end-to-end voltage (\( 24 \times \sqrt{2} \)), which does not appear across the load.
- Option D: 12.0 V is the RMS voltage of the half-winding, not its peak value.
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