This solution explains how to calculate the average output voltage for a full wave rectifier when the input AC supply voltage is given. We will analyze the relationship between the input AC voltage and the output DC voltage of the rectifier.
A full wave rectifier is an electronic circuit that converts an alternating current (AC) input into a direct current (DC) output. It utilizes both the positive and negative halves of the AC input waveform. For a sinusoidal AC supply, a full wave rectifier essentially takes the entire waveform, including the negative parts, and makes them positive, resulting in a pulsating DC output.
The problem states that the full wave rectifier is supplied from a 20 V AC supply. In electronics, when an AC voltage value is given without specifying whether it's peak or RMS, it is conventionally assumed to be the RMS (Root Mean Square) voltage. Therefore:
The relationship between the RMS voltage ($V_{rms}$) and the peak voltage ($V_m$) for a sinusoidal AC waveform is given by the formula:
$$ V_{rms} = \frac{V_m}{\sqrt{2}} $$
To find the peak voltage ($V_m$), we rearrange the formula:
$$ V_m = V_{rms} \times \sqrt{2} $$
Substituting the given $V_{rms}$ value:
$$ V_m = 20 \text{ V} \times \sqrt{2} $$
$$ V_m \approx 20 \text{ V} \times 1.4142 $$
$$ V_m \approx 28.284 \text{ V} $$
For an ideal full wave rectifier, the average value of the output voltage ($V_{avg}$) is related to the peak voltage ($V_m$) by the following formula:
$$ V_{avg} = \frac{2 V_m}{\pi} $$
This formula arises from integrating the rectified waveform over one full cycle and dividing by the period.
Now, we substitute the calculated peak voltage ($V_m \approx 28.284$ V) into the formula for $V_{avg}$:
$$ V_{avg} = \frac{2 \times (20 \sqrt{2})}{\pi} \text{ V} $$
$$ V_{avg} = \frac{40 \sqrt{2}}{\pi} \text{ V} $$
Let's calculate the numerical value:
$$ V_{avg} \approx \frac{40 \times 1.4142}{3.14159} $$
$$ V_{avg} \approx \frac{56.568}{3.14159} $$
$$ V_{avg} \approx 18.005 \text{ V} $$
The calculated average output voltage is approximately $18.005$ V. Let's compare this with the given options:
Our calculated value of $18.005$ V is very close to $18.01$ V.
Based on the standard formulas for AC RMS voltage and the average output of a full wave rectifier, the average output voltage for a 20 V RMS AC supply is approximately $18.01$ V.
What is the ripple factor of full-wave bridge rectifier?
The maximum efficiency of a half-wave rectifier is
For a full wave rectifier, the output frequency
A half wave rectifier requires -
For a Bridge rectifier circuit, the secondary voltage is given by V s= 50sinωt and the load resistance is R L= 800Ω. Calculate the rectification efficiency.