In case of full wave bridge rectifiers, maximum efficiency is ______ and ripple factor is ______.

Correct option is C

$$\bullet \ \text{The rectification efficiency of full wave rectifiers is double that of half wave rectifiers. The efficiency of half wave rectifiers is } 40.6\% \text{ while the rectification efficiency of full wave rectifiers is } 81.2\%. \\[6pt]\bullet \ \text{The ripple factor in full wave rectifiers is low hence a simple filter is required. The value of ripple factor in full wave rectifier is } 0.482 \text{ while in half wave rectifier it is about } 1.21. \\[10pt]\bullet \ \text{Form Factor} \\[4pt]\text{The form factor of the full wave rectifier is calculated using the formula:} \\[4pt]K_f= \frac{\text{RMS value of current}}{\text{Average value of current}}= \frac{I_{\mathrm{rms}}}{I_{\mathrm{dc}}}= \frac{I_{\max}/\sqrt{2}}{2I_{\max}/\pi}= \frac{\pi}{2\sqrt{2}}= 1.11 \\[10pt]\bullet \ \text{Ripple Factor} \\[4pt]K_r= \sqrt{(\text{Form factor})^2 - 1}= \sqrt{(1.11)^2 - 1}= 0.4817 \\[10pt]\bullet \ \text{Peak Factor} \\[4pt]\text{The following formula gives the peak factor of the full wave rectifier:} \\[4pt]K_p= \frac{\text{Peak value of current}}{\text{RMS value of current}}= \frac{I_{\max}}{I_{\max}/\sqrt{2}}= \sqrt{2} \\[10pt]\bullet \ \text{Maximum Efficiency} \\[4pt]\text{The maximum efficiency of the full-wave rectifier can be obtained using the following formula:} \\[4pt]\eta= \frac{\text{DC Output Power}}{\text{AC Output Power}}= \frac{(2V_m/\pi)}{(V_m/\sqrt{2})}= 0.812 \\[10pt]\bullet \ \text{The efficiency of the full wave rectifiers is } 81.2\%.$$​​