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The rectification efficiency of a single phase Half Wave Controlled Rectifier having a resistive load and the delay angle of π/2 is _______%. (correct upto two decimal places
Correct answer is '20.23'. Can you explain this answer?
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The rectification efficiency of a single phase Half Wave Controlled R...
The rectification efficiency of a single-phase half-wave controlled rectifier can be calculated using the following formula:
η = (Vdc / Vac) × 100
Where:
η is the rectification efficiency
Vdc is the DC output voltage
Vac is the AC input voltage
In a half-wave controlled rectifier, the output voltage is given by the equation:
Vdc = (2 / π) * Vac * (1 - cos(α))
Where:
α is the delay angle
To calculate the rectification efficiency, we need to substitute the values into the equations and solve for η.
Given:
Delay angle (α) = π/2
1. Calculate the DC output voltage (Vdc):
Vdc = (2 / π) * Vac * (1 - cos(π/2))
= (2 / π) * Vac * (1 - 0)
= (2 / π) * Vac
2. Calculate the rectification efficiency (η):
η = (Vdc / Vac) × 100
= [(2 / π) * Vac / Vac] × 100
= (2 / π) × 100
≈ 20.23%
Therefore, the rectification efficiency of the single-phase half-wave controlled rectifier with a resistive load and a delay angle of π/2 is approximately 20.23%.
Note: The rectification efficiency represents the percentage of AC power converted into DC power. In this case, since we are using a half-wave rectifier, only half of the AC input waveform is utilized, resulting in a lower rectification efficiency compared to full-wave rectifiers. The delay angle determines the point in the AC cycle at which the rectifier starts conducting, and a delay angle of π/2 means the rectifier conducts from the peak of the AC waveform.
η = (Vdc / Vac) × 100
Where:
η is the rectification efficiency
Vdc is the DC output voltage
Vac is the AC input voltage
In a half-wave controlled rectifier, the output voltage is given by the equation:
Vdc = (2 / π) * Vac * (1 - cos(α))
Where:
α is the delay angle
To calculate the rectification efficiency, we need to substitute the values into the equations and solve for η.
Given:
Delay angle (α) = π/2
1. Calculate the DC output voltage (Vdc):
Vdc = (2 / π) * Vac * (1 - cos(π/2))
= (2 / π) * Vac * (1 - 0)
= (2 / π) * Vac
2. Calculate the rectification efficiency (η):
η = (Vdc / Vac) × 100
= [(2 / π) * Vac / Vac] × 100
= (2 / π) × 100
≈ 20.23%
Therefore, the rectification efficiency of the single-phase half-wave controlled rectifier with a resistive load and a delay angle of π/2 is approximately 20.23%.
Note: The rectification efficiency represents the percentage of AC power converted into DC power. In this case, since we are using a half-wave rectifier, only half of the AC input waveform is utilized, resulting in a lower rectification efficiency compared to full-wave rectifiers. The delay angle determines the point in the AC cycle at which the rectifier starts conducting, and a delay angle of π/2 means the rectifier conducts from the peak of the AC waveform.
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Community Answer
The rectification efficiency of a single phase Half Wave Controlled R...
Given
,
Single phase Half Wave Controlled Rectifier
Resistive load
∝= π/2
Rectification efficiency 
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The rectification efficiency of a single phase Half Wave Controlled Rectifier having a resistive load and the delay angle of π/2 is _______%. (correct upto two decimal placesCorrect answer is '20.23'. Can you explain this answer?
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