Concept:Calculate the total secondary RMS voltage, divide by 2 for the center-tapped half-winding voltage, and multiply by \( \sqrt{2} \) to find the peak voltage \( V_m \).
Formula:$$V_{s,\text{total(rms)}} = \frac{N_s}{N_p} V_p = \frac{1}{10} \times 220\text{ V} = 22.0\text{ V}$$
$$V_{\text{half(rms)}} = \frac{22.0\text{ V}}{2} = 11.0\text{ V} \implies V_m = V_{\text{half(rms)}} \times \sqrt{2} = 11.0 \times 1.414 \approx 15.56\text{ V}$$
Solution:- Total secondary RMS voltage: \( V_{s,\text{total}} = 220 / 10 = 22\text{ V} \).
- RMS voltage across each half-winding: \( 22 / 2 = 11\text{ V} \).
- Peak voltage across each half-winding: \( V_m = 11 \times \sqrt{2} \approx 15.56\text{ V} \).
Why other options are incorrect:- Option B: 31.11 V is the peak voltage across the entire secondary winding (\( 22 \times \sqrt{2} \)).
- Option C: 11.00 V is the RMS voltage of the half-winding, not the peak voltage.
- Option D: 22.00 V is the total secondary RMS voltage.
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