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Electronic Devices and Circuits

Q1: If an additional two diodes were used to connect the 1 kW load across a bridge rectifier circuits, utilizing the full secondary of the transformer, how much d.c. power could be delivered using a transformer with the rating of 105 VA?

A 35 W

B 60 W

C 85 W

D 100 W

ANS:C - 85 W

Efficiency = dc load power/ac load power.
Efficiency = 80.2% for full wave.
So 0.802 = dc load power/105.
Then, dc load power =105 * 0.802 = 85 W.



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