
The efficiency of a full wave rectifier is?
A. Half of half wave rectifier
B. 3 times of half wave rectifier
C. Double of half wave rectifier
D. Same as half wave rectifier
Answer
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Hint: A rectifier is an electrical device that converts alternating current (AC) which periodically reverses its direction to direct current (DC) which flows in only one direction.
Complete step by step solution:
Type of rectifier:
Half wave rectifier: In half-wave rectifier, rectification of a single-phase supply i.e. either the positive or negative half of the AC wave is passed, while the other half is blocked. Half-wave rectification requires a single diode in a single-phase supply, or three in a three-phase supply.
Full wave rectifier: A full-wave rectifier rectifies the whole of the input waveform to one of constant polarity (positive or negative) at its output. Full-wave rectification requires two diodes and a centre tapped transformer, or four diodes in a bridge configuration and any AC source.
Efficiency signifies a peak level of performance that uses the least amount of inputs to achieve the highest amount of output. It minimizes the waste of resources such as physical materials, energy, and time while accomplishing the desired output.
We know that,
\[Efficiency{\text{ }}\left( \eta \right){\text{ }} = {\text{ }}DC{\text{ }}Power{\text{ }}Output{\text{ }}/{\text{ }}AC{\text{ }}power{\text{ }}input\]
Therefore, the efficiency of a full wave rectifier \[\left( {{n_f}} \right)\] is \[81.2\% \](calculated)
And, efficiency of a half wave rectifier \[\left( {{n_h}} \right)\] is \[40.6\% \] (calculated)
$\dfrac{{{\eta _h}}}{{{\eta _f}}} = \dfrac{{40.6}}{{81.2}}$
$ = \dfrac{1}{2}$
$\therefore \,{\eta _f} = 2{\eta _h}$
Hence, the efficiency of a full wave rectifier is double of half wave-rectifier.
Note: Rectification is the process of converting bidirectional current flow to unidirectional current flow. The process is of vital importance in many areas of circuit design, including radio communication and AC to DC power conversion.
Complete step by step solution:
Type of rectifier:
Half wave rectifier: In half-wave rectifier, rectification of a single-phase supply i.e. either the positive or negative half of the AC wave is passed, while the other half is blocked. Half-wave rectification requires a single diode in a single-phase supply, or three in a three-phase supply.
Full wave rectifier: A full-wave rectifier rectifies the whole of the input waveform to one of constant polarity (positive or negative) at its output. Full-wave rectification requires two diodes and a centre tapped transformer, or four diodes in a bridge configuration and any AC source.
Efficiency signifies a peak level of performance that uses the least amount of inputs to achieve the highest amount of output. It minimizes the waste of resources such as physical materials, energy, and time while accomplishing the desired output.
We know that,
\[Efficiency{\text{ }}\left( \eta \right){\text{ }} = {\text{ }}DC{\text{ }}Power{\text{ }}Output{\text{ }}/{\text{ }}AC{\text{ }}power{\text{ }}input\]
Therefore, the efficiency of a full wave rectifier \[\left( {{n_f}} \right)\] is \[81.2\% \](calculated)
And, efficiency of a half wave rectifier \[\left( {{n_h}} \right)\] is \[40.6\% \] (calculated)
$\dfrac{{{\eta _h}}}{{{\eta _f}}} = \dfrac{{40.6}}{{81.2}}$
$ = \dfrac{1}{2}$
$\therefore \,{\eta _f} = 2{\eta _h}$
Hence, the efficiency of a full wave rectifier is double of half wave-rectifier.
Note: Rectification is the process of converting bidirectional current flow to unidirectional current flow. The process is of vital importance in many areas of circuit design, including radio communication and AC to DC power conversion.
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