Question 6A310

In a center-tapped full-wave rectifier, the Peak Inverse Voltage (PIV) that appears across each non-conducting tube is equal to twice the peak voltage from one end of the transformer secondary to the center tap. Let $V_{RMS_{total}}$ be the total RMS voltage of the transformer's secondary winding (end-to-end). The RMS voltage from one end to the center tap is $V_{RMS_{half}} = \frac{V_{RMS_{total}}}{2}$. The peak voltage for one half of the winding ($V_{peak_{half}}$) is $V_{RMS_{half}} \times \sqrt{2} = (\frac{V_{RMS_{total}}}{2}) \times \sqrt{2}$. The PIV across each tube is $2 \times V_{peak_{half}}$. Substituting the expression for $V_{peak_{half}}$: $PIV = 2 \times (\frac{V_{RMS_{total}}}{2} \times \sqrt{2}) = V_{RMS_{total}} \times \sqrt{2}$. Given the PIV rating of the tubes is 10,000 V: $10,000 V = V_{RMS_{total}} \times \sqrt{2}$ $V_{RMS_{total}} = \frac{10,000 V}{\sqrt{2}} = \frac{10,000 V}{1.414} \approx 7071 V$. Therefore, the maximum allowable total secondary RMS voltage for the transformer is approximately 7070 V. Option A is correct because it represents the total RMS secondary voltage that results in a PIV equal to the tube's rating. Options B, C, and D would either exceed the tube's PIV rating or provide a lower PIV than the tube could safely handle if interpreted as total RMS secondary voltage.