Question 18.10: A 230 V, 60 Hz voltage is applied to the primary of a 5:1 st...

The voltage across the two ends of secondary = \frac{230}{5}= 46 V

Voltage from center tapping to one end, V_{rms} = \frac{46}{2} = 23 V

(a) The d.c. voltage across the load, V_{d.c.} = \frac{2V_{m}}{\pi} = \frac{2 × 23 × \sqrt{2}}{\pi} = 20.7 V

(b) The d.c. current flowing through the load, I_{d.c.} =\frac{V_{d.c.}}{(r_{s} + r_{f} +R_{L})} = \frac{20.7}{1000} = 20.7 mA

(c) The d.c. power delivered to the load,

P_{d.c.} = (I_{d.c.})^{2} × R_{L} = (20.7 × 10^{– 3})^{2} × 900 = 0.386 W

(d) PIV across each diode =2V_{m} = 2 × 23 × \sqrt{2} = 65 V

(e) Ripple voltage, V_{r, rms} = \sqrt{(V_{rms})^{2} – (V_{d.c.})^{2}}

= \sqrt{(23)^{2} – (20.7)^{2}} = 10.05 V

Frequency of ripple voltage = 2 × 60 = 120 Hz

(f) Rectification efficiency, η = \frac{Pd.c.}{Pa.c.}=\frac{(V_{d.c.})^{2}/R_{L}}{(V_{rms})^{2}/R_{L}}=\frac{(V_{d.c.})^{2}}{(V_{rms})^{2}}

= \frac{(20.7)^{2}}{(23)^{2}}=\frac{428.49}{529} = 0.81

Therefore, percentage of efficiency = 81%