Question

In a full wave rectifier the input is from 30-0-30V transformer. the load and diode resistance are 100ohm and 10ohm respectively. calculate the average voltage, rectification efficiency and percentage regulation​

Answer 1

Answer:

The correct answer to this question is $V_{avg} = 49.1V$

Explanation:

Given - A full wave rectifier the input is from 30-0-30V transformer and the load and diode resistance are 100ohm and 10ohm.

To Find - calculate the average voltage, rectification efficiency and percentage regulation​.

A full-wave rectifier with a $30 V$ input is present. The diode resistance is listed as $10$, whereas the load resistance is reported as $100$ .

Determine the average voltage, rectification efficiency, and percentage regulation.

This is calculated as follows:

$V_{rms} = \frac{V_{m} }{\sqrt{2} }$

$V_{avg} =\frac{2V_{m} }{\pi }$

$V_{m} = V_{rms}$ × $\sqrt{2}$

The RMS voltage of the rectifier is 30 V, and thus, the peak value of volatage is $30\sqrt{2}$ V.

$V_{avg} =\frac{2V_{m} }{\pi }$

$V_{avg} = \frac{2 . 54.545 . \sqrt{2} }{\pi }$

$V_{avg} = 49.1V$

Thus, the average voltage of the rectifier is given to be $V_{avg} = 49.1V$

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Answer 2

Answer:

The answer is $V_{avg} =49.1V$

percentage regulation​:

The power supply is more steady and reliable the lower the load regulation. Typical well-regulated power supplies have load regulations of less than 1%, which means that over the supply's load current range, the output voltage will only fluctuate by a maximum of 1%.

Explanation:

While RMS is utilised when the random variables supplied in the data are both positive and negative, like sinusoids, average is used to determine the central tendency of a given data set. 4. RMS is somewhat specialised in its practical application, whereas average is widely employed in every scientific and engineering sector you can think of. total regulation, rectification efficiency, and average voltage should be calculated

There is a full-wave rectifier here with a 30V input. The load resistance is claimed to be 100 while the diode resistance is given as 10. The average voltage, rectification efficiency, and percentage of regulation should be determined.

$V_{rms} =\frac{V_{m} }{\sqrt{2} }$

$V_{avg} =\frac{2V_{m} }{\pi }$

$V_{m} =V_{rms} *\sqrt{2}$

The rectifier's RMS voltage is 30 V, hence the voltage's peak value is

$30\sqrt{2} V$

$V_{avg} =\frac{2V_{m} }{\pi }$

$V_{avg} =\frac{2.54.545.\sqrt{2} }{\pi }$

$V_{avg} =49.1V$

Consequently, the rectifier's average voltage is given as $V_{avg} =49.1V$

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