Transformer utilisation factor of a half wave rectifier is _________a)...
Transformer utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For a half wave rectifier, it’s low and equal to 0.287.
View all questions of this testTransformer utilisation factor of a half wave rectifier is _________a)...
Transformer Utilisation Factor (TUF) of a Half Wave Rectifier
The transformer utilisation factor (TUF) is the ratio of the DC power delivered to the load to the AC rating of the transformer. In a half wave rectifier, TUF is defined as the ratio of the DC output power to the product of the secondary voltage and the secondary current of the transformer.
TUF = DC output power / (V_s x I_s)
Where V_s is the secondary voltage and I_s is the secondary current.
Calculation of TUF for a Half Wave Rectifier
The DC output power of a half wave rectifier is given by:
P_dc = (V_m/π) x I_s x (1 - cos(α))
Where V_m is the maximum value of the secondary voltage and α is the firing angle or the angle between the secondary voltage and the rectified output voltage.
The AC rating of the transformer is given by:
P_ac = V_s x I_s
Therefore, the TUF of a half wave rectifier can be calculated as:
TUF = P_dc / P_ac
= (V_m/π) x (1 - cos(α)) / V_s
For a half wave rectifier, the maximum value of the secondary voltage is equal to the peak value of the input voltage. Therefore,
V_m = V_p
Where V_p is the peak input voltage.
The average value of the output voltage for a half wave rectifier is given by:
V_avg = (V_p / π) x (1 - cos(α))
Therefore, the TUF of a half wave rectifier can be expressed as:
TUF = (1/π) x (1 - cos(α)) / (1/2) x (1 - cos(α/2))
Simplifying this equation, we get:
TUF = 2 / (π x (2 - cos(α)))
For α = 0, TUF = 0
For α = π, TUF = 0.287
Therefore, the correct answer is option C, i.e., 0.287.