Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion
Discussion Forum : Electronic Devices and Circuits - Section 10 (Q.No. 14)
14.
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?
Discussion:
1 comments Page 1 of 1.
Ajit said:
9 years ago
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.
Efficiency = 80.2% for full wave.
So 0.802 = dc load power/105.
Then, dc load power =105 * 0.802 = 85 W.
(2)
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