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Full Wave Bridge Rectifier Circuit

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gurse

Electrical
Jun 27, 2019
17
I have a full wave rectifier thyristor control circuit on a 10 Ohms resistor coil which produced the following results with 240V AC rms.

DC Current through coil (A) DC Voltage across Coil (V) AC Voltage across coil (V rms)

1.15 10.8 92.0
2.43 22.5 153.9
3.44 31.9 189.7
4.63 43.4 211.4


If I reduce the mains input from 240V rms to 110V rms to thyristor full wave rectifier circuit, will I still be able to pull 4.63 Amp DC as the output of a full rectifier circuit is Vrms = Vp / √2, which works out as 110 * 1.414 / √2 = 110 V rms.
 
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What is based on well established formulas? If you already have the answer, then there isn't a question.

You specifically asked:

"If I reduce the mains input from 240V rms to 110V rms to thyristor full wave rectifier circuit, will I still be able to pull 4.63 Amp DC"

That isn't a formula, that is a speculation that you will be able to pull 4.63 Amp DC at 110V rms.

Since you are dealing with active electronics, for which you have not specified electrical characteristics, maybe you can. Or maybe you cannot.
 
Although the question was put in that way, later in the discussion I pointed out that the main interest was the AC and DC voltages that I was measuring across the coil.
But Waross pointed out that it could be due to metering errors.
 
A lot depends on your meter. The old analogue meters were not accurate when the wave form was not a sine wave.
Some of the first digital meters looked at the peak value and then indicated a value that assumed that the measured value was a sine wave. If you have one of these maters then AC readings of a chopped DC voltage may be a long way away from reality.
Please describe your meter so that we can confirm or eliminate this possible factor.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
So, you are still indicating a phase controlled rectifier feeding a coil (inductor). See my previous answers on how that works.

 
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