Full Wave Rectifier MCQ Quiz - Objective Question with Answer for Full Wave Rectifier - Download Free PDF

Concept:

In a full wave rectifier, the output frequency is twice the input frequency. This is because during each cycle of the input AC signal, the output signal completes two cycles.

Explanation: 

If the input frequency to a full wave rectifier is 50 Hz, the output frequency will be:

\( f_{out} = 2 \times f_{in} \)

Given that the input frequency \( f_{in} \) is 50 Hz,

\( f_{out} = 2 \times 50 \, \text{Hz} = 100 \, \text{Hz} \)

Therefore, the output frequency of the full wave rectifier will be 100 Hz.

The correct option is (2).

Explanation:

Average Value of Full-Wave Rectified Voltage

Definition: The average value of a full-wave rectified voltage is the mean of all the instantaneous values of the rectified voltage over one complete cycle of the input AC signal. It is a measure of the DC component of the rectified output.

Working Principle: In a full-wave rectifier, both the positive and negative halves of the AC signal are converted to positive voltage. This means that the output voltage is always positive, regardless of the input voltage polarity. The average value of this rectified voltage can be determined by integrating the rectified signal over one complete cycle and then dividing by the period of the cycle.

Formula: The average value (V_avg) of a full-wave rectified voltage with a peak value (V_m) can be calculated using the following formula:

V_avg = (2 * V_m) / π

Where:

• V_m is the peak value of the input AC voltage.
• π is a mathematical constant approximately equal to 3.14159.

Calculation:

Given that the peak value (V_m) is 66 V, we can substitute this value into the formula to find the average value:

V_avg = (2 * 66 V) / π

V_avg = 132 V / 3.14159

V_avg ≈ 42 V

Correct Option Analysis:

The correct option is:

Option 2: 42 V

This option correctly represents the average value of the full-wave rectified voltage with a peak value of 66 V, as calculated using the formula for the average value of a full-wave rectified signal.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 66 V

This option incorrectly suggests that the average value is equal to the peak value of the input voltage. However, the average value of a full-wave rectified signal is always less than the peak value. The peak value represents the maximum instantaneous voltage, not the average.

Option 3: 33 V

This option may confuse the average value with half the peak value. However, the correct average value is derived from the formula (2 * V_m) / π, which does not result in half the peak value.

Option 4: 88 V

This option may stem from a misunderstanding of the rectification process. The value 88 V is not related to the average calculation and is not derived from the provided peak value using the correct formula.

Conclusion:

Understanding the calculation of the average value of a full-wave rectified voltage is crucial in analyzing rectifier circuits. The correct average value for a full-wave rectified voltage with a peak value of 66 V is approximately 42 V, as determined by the formula (2 * V_m) / π. This understanding helps in designing and analyzing power supplies and other electronic circuits that utilize rectification to convert AC to DC.

Concept

The ripple voltage of a full-wave rectifier is given by:

\(V_R={2.4V_{dc}\over R_LC}\)

∵ \({V_{dc}\over R_L}=I_L\)

\(V_R={2.4I_{L}\over C}\)

where, VR = Ripple voltage

IL = Load current

C = Capacitor

Calculation

Given, C = 100 μF

IL = 50 mA

\(V_R={2.4\times 50\times 10^{-3}\over 100\times 10^{-6}}\)

VR = 1.2 kV

Concept:

Full-Wave Rectifier:

• A full-wave rectifier is used to convert alternating current (AC) into direct current (DC) by using two diodes, D1 and D2.
• In a full-wave rectifier, both diodes conduct during each half-cycle of the AC input.
• If the AC signal has a frequency of 50 Hz, this means the alternating voltage changes direction 50 times per second.
• Since both diodes conduct during each half-cycle, each diode will conduct twice per cycle, which means it will conduct 100 times per second.

Calculation:

Given,

Frequency of AC input = 50 Hz.

Since each cycle consists of two half-cycles, each diode will conduct twice per cycle.

Thus, the diode D1 conducts 100 times in one second.

∴ The correct answer is: 100.

The correct answer is option 1):(0.693)

Concept:

TUF:

• The transformer utilization factor (TUF) is the measure of the merit of a rectifier circuit. It is defined as the ratio of dc power output to the transformer Volt-Ampere (VA) rating required by the secondary winding.
• TUF = \(DC\: power\: output\over Effective\: VA \:rating\: of \:the\: transformer\)
• This factor indicates the effectiveness of transformer usage by the rectifier.

Ripple factor:

The amount of AC present in the output of the signal is called as ripple.

The ripple factor indicates the number of ripples present in the DC output.

The output of the power supply is given by

\({\rm{Ripple\;factor}} = \frac{{{I_{rms}}\;of\;AC\;component}}{{{I_{DC}}\;component}}\)

It is given as:

\(\gamma = \sqrt {{{\left( {\frac{{{V_{rms}}}}{{{V_{DC}}}}} \right)}^2} - 1} \)

Thus if the ripple factor is less, the power supply has less AC components and power supply output is purer (i.e more DC without much fluctuations)

Thus ripple factor is an indication of the purity of output of the power supply. 

For full-wave rectifier: γ = 0.48

For half-wave rectifier: γ = 1.21

Explanation:

• A half-wave rectifier conducts when the alternating current is positive, which happens during 180° of the A.C cycle, and blocks current during the other 180 degrees
• A full-wave rectifier conducts during both 180° periods of an A.C cycle; So you can see, it conducts current for twice the amount of time
• The full-wave rectifier has twice the output voltage of a half-wave rectifier is, that it utilizes both half-cycles of the input.

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Important:

Rectifier: 

A rectifier is an electrical device that converts alternating current (AC), which periodically reverses direction, to direct current (DC), which flows in only one direction.

Center-Tap Full Wave Rectifier:

It consists of two diodes as shown:

The RMS or the effective value of a waveform f(t) is given by:
\({f_{rms}} = \sqrt {\frac{1}{T}\mathop \smallint \limits_0^T {f^2}\left( t \right)dt}\)

Application:

The relation between the peak value and rms value of the waveform is given as:

\(V_p=\sqrt{2} V_{rms}\)

The relation between the peak value and the average value for a half and full-wave rectifier is:

\({V_{avg}} = \frac{{{V_m}}}{\pi }\) for half-wave rectifier

\({V_{avg}} = \frac{{2{V_m}}}{\pi }\) for full-wave rectifier

Solution

Concept

During the +ve cycle of input 

• Diode D₂ conducts and D₁ is off
• from the voltage division rule, the voltage across the 2kΩ resistance around which output is calculated is 
• V = 0.5 V_in

During the -ve cycle of input

• D₂ is off and D₁ conducts
• from voltage division rule
• V = -0.5 V_in

Hence during the full cycle of input,

we will observe the following output voltage waveform

We could observe that it is a full-wave rectified wave

Calculation

The maximum output voltage is the 

V_max = |0.5 V_in|

V_max = + 5volt

Hence the correct answer is + 5 volt

The correct option is 3

Concept:

• From the figure, it can be seen that two cycles of output occur for each input cycle of the input voltage.
• This is because the full-wave rectifier has inverted the negative half cycle of the input voltage.
• As a result, the output of a full-wave rectifier has a frequency double the input AC frequency.

Important specifications of different rectifiers are shown below:

• TUF: The transformer utilization factor (TUF) is the measure of merit of a rectifier circuit. It is defined as the ratio of dc power output to the transformer Volt-Ampere (VA) rating required by the secondary winding.

TUF = (DC power output) / (Effective VA rating of transformer)

\(TUF = K = \;\frac{{{P_{dc}}}}{{{V_s}{I_s}}}\)

• It is a quantitative indication of the utilization of VA Rating of Transformer. The more the value of TUF, the more will be the utilization. In other words, the VA rating of required transformer will be less if TUF is more and vice versa.
• NOTE: 1/TUF for full wave bridge rectifier signifies that transformer must be 1.23 times higher than that when it is used to deliver power from a pure ac voltage source.
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Points to Remember:

• TUF for Half-Wave rectifier is 28.6%
• TUF for Full-Wave centre tapped rectifier is 57.2%
• TUF for Full-Wave bridge rectifier is 81%

Full wave rectifier:

Case 1: During +ve half cycle

D₁ and D₃ conducts

V_o = V_s

Case 2: During -ve half cycle

D₂ and D₄ conducts

Vo = -V_s

The output waveform is given above:

The average output voltage is:

\(V_{o(avg)}={2V_m\over \pi}\)

Half wave rectifier:

RMS Value\(= \frac{{Max.\;value}}{2}\)

Average Value =\(\frac{{Max.\;value}}{\pi }\)

Full-wave rectifier:

\(rms\;value = \frac{{maximum\;value}}{{\sqrt 2 }}\)

\(Average\;value = 2\left( {\frac{{maximum\;value}}{\pi }} \right)\)

Calculation:

The RMS value of a half-wave rectified current is 10 Ampere

RMS Value \(= \frac{{Max.\;value}}{2}\)

10 = \(= \frac{{Max.\;value}}{2}\)

Maximum value (I_m) = 20 A

The RMS value for full-wave rectification

\(rms\;value = \frac{{maximum\;value}}{{\sqrt 2 }}\)

\(rms\;value = \frac{{20}}{{\sqrt 2 }}\)

RMS value = 14.14 A

The ripple factor indicates the number of ripples present in the DC output.

The output of the power supply is given by:

Mathematically, ripple factor (r) is defined as:

\(r= \frac{{{I_{rms}}\;of\;AC\;component}}{{{I_{DC}}\;component}}\)

Thus if the ripple factor is less, the power supply has less AC components and power supply output is more pure (i.e more DC without much fluctuations)

Thus ripple factor is an indication of the purity of output of power supply

Important Points:

For Half-wave rectifier, the ripple factor is = 1.21

For Full-wave rectifier, the ripple factor is = 0.48

For Bridge-wave rectifier, the ripple factor is = 0.48