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 WB-60 WC-85 WD-100 W-C-85 W

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APTITUDE CRACK · Accepted Answer

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.

Electronic Devices and Circuits - Engineering

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?

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Electronic Devices and Circuits - Engineering

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?

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For help Students Orientation Mcqs Questions

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