Question
A single-phase full-bridge diode rectifier feeds a resistive load of 50 Ω from a 200 V, 50 Hz single phase AC supply. If the diodes are ideal, then the active power, in watts, drawn by the load is ______. (round off to nearest integer)
Answer :
800 (Range : 795 to 805)
Given :
(i) 1-∅ full bridge rectifier.
(ii) Resistive load, R = 50Ω
(iii) Supply voltage, Vs (rms) = 200 V
(iv) Frequency, f 50 Hz
Peak voltage, V m Vs (rms) × √2 = 200 √2 Volts
The active power drawn by the load in watts.
We know, average output power for the resistive load is given by,
P0(avg) = V20(avg)/ R
For 1-∅ full bridge diode rectifier, in case of resistive load,
V0(rms)=Vs(rms)= 200V
P0(avg) = 200 2/500 = 800 W
Hence, the correct answer is 800 W.
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