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The voltage regulation of a transformer at full-load 0.8 p.f leading is -2%. Its voltage regulation at full load 0.8 p.f lagging
- a)will be positive
- b)will be negative
- c)may be positive
- d)may be negative
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The voltage regulation of a transformer at full-load 0.8 p.f leading i...
The leading p.f. has negative v.r. and lagging p.f. has major portion of positive voltage regulation.
Most Upvoted Answer
The voltage regulation of a transformer at full-load 0.8 p.f leading i...
Explanation:
Voltage regulation is defined as the difference between the no-load voltage and full-load voltage, expressed as a percentage of the full-load voltage.
Given data:
Voltage regulation at full-load 0.8 p.f leading = -2%
Full-load power factor is lagging, i.e., cosθ = 0.8 lagging
To find:
Voltage regulation at full load 0.8 p.f lagging
Solution:
Voltage regulation is given by the formula:
% Voltage regulation = [(No-load voltage - Full-load voltage) / Full-load voltage] x 100
Let Vnl and Vfl be the no-load voltage and full-load voltage respectively.
Given, voltage regulation at full-load 0.8 p.f leading = -2%
Therefore, we have:
-2 = [(Vnl - Vfl) / Vfl] x 100
=> Vnl - Vfl = -0.02 Vfl
=> Vnl = (1 - 0.02) Vfl
=> Vnl = 0.98 Vfl
Now, for full-load 0.8 p.f lagging, we have:
Ifl = P / (Vfl x cosθ)
Ifl = P / (Vfl x 0.8)
Let Vfl' be the full-load voltage at 0.8 p.f lagging.
Then, we have:
Ifl = P / (Vfl' x cosθ')
Ifl = P / (Vfl' x 0.8)
Since the load is capacitive in nature, the voltage drop across the transformer winding will decrease, leading to an increase in the output voltage.
Hence, the voltage regulation at full load 0.8 p.f lagging will be positive.
Therefore, option A is the correct answer.
Voltage regulation is defined as the difference between the no-load voltage and full-load voltage, expressed as a percentage of the full-load voltage.
Given data:
Voltage regulation at full-load 0.8 p.f leading = -2%
Full-load power factor is lagging, i.e., cosθ = 0.8 lagging
To find:
Voltage regulation at full load 0.8 p.f lagging
Solution:
Voltage regulation is given by the formula:
% Voltage regulation = [(No-load voltage - Full-load voltage) / Full-load voltage] x 100
Let Vnl and Vfl be the no-load voltage and full-load voltage respectively.
Given, voltage regulation at full-load 0.8 p.f leading = -2%
Therefore, we have:
-2 = [(Vnl - Vfl) / Vfl] x 100
=> Vnl - Vfl = -0.02 Vfl
=> Vnl = (1 - 0.02) Vfl
=> Vnl = 0.98 Vfl
Now, for full-load 0.8 p.f lagging, we have:
Ifl = P / (Vfl x cosθ)
Ifl = P / (Vfl x 0.8)
Let Vfl' be the full-load voltage at 0.8 p.f lagging.
Then, we have:
Ifl = P / (Vfl' x cosθ')
Ifl = P / (Vfl' x 0.8)
Since the load is capacitive in nature, the voltage drop across the transformer winding will decrease, leading to an increase in the output voltage.
Hence, the voltage regulation at full load 0.8 p.f lagging will be positive.
Therefore, option A is the correct answer.
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The voltage regulation of a transformer at full-load 0.8 p.f leading is -2%. Its voltage regulation at full load 0.8 p.f lagginga)will be positiveb)will be negativec)may be positived)may be negativeCorrect answer is option 'A'. Can you explain this answer?
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