Question

A single-phase semi-converter is connected to a 230 V source and is feeding a load R = 10 Ω in series with a large inductance that makes the load current ripple free. For α = 45°, find the rectification efficiency. The RMS value of output voltage is 219.3 V

(a) 96.54 %

(b) 75.25 %

(c) 89.45 %

(d) 80.58 %

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This interesting question is from Single Phase Semi-Converters-2 topic in portion Converters of Power Electronics

Answer 1

The correct answer is (d) 80.58 %

Explanation: Vo =( Vm/π) (1 + cosα) = 176.72V

Io = Vo/R

Irms = Io = Vo/R = 17.67 A

Vrms = 219.3 V (given)

η = (Vo x Io) / (Vrms X Irms) = 0.8058 = 80.58 %.