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A 400V / 100V, 10kVA two winding transformer is to be used as an autotransformer to supply a 400V circuit from 500V source when tested as two winding transformer at rated load .85 P.F. lag it’s efficiency is 97%. Determine efficiency as autotransformer at rated load and 0.85 power factor
  • a)
    95.5 %
  • b)
    96.9%
  • c)
    98.2 %
  • d)
    99.4 %
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A 400V / 100V, 10kVA two winding transformer is to be used as an autot...

VA transferred inductively
=10 kVA same as two
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A 400V / 100V, 10kVA two winding transformer is to be used as an autot...
Problem Statement:
A 400V / 100V, 10kVA two winding transformer is to be used as an autotransformer to supply a 400V circuit from 500V source when tested as two winding transformer at rated load .85 P.F. lag it’s efficiency is 97%. Determine efficiency as autotransformer at rated load and 0.85 power factor.

Solution:
Given data,
Two-winding transformer: V1 = 400V, V2 = 100V, P = 10kVA, η = 97%, P.F. = 0.85 lagging
Autotransformer: V1 = 500V, V2 = 400V, P = 10kVA, P.F. = 0.85 lagging

Efficiency of two-winding transformer is given as:
η1 = Output power / Input power = P2 / P1
Where,
P2 = V2 * I2 * cos(θ) = 10kVA * 0.85 = 8.5kW (lagging)
P1 = V1 * I1 * cos(θ) = 10kVA * 0.85 = 8.5kW (lagging)
Therefore, η1 = 8.5kW / 8.76kW = 0.97 or 97%

To find the efficiency of the autotransformer, we can use the following formula:
η2 = Output power / Input power = P2 / P1
Where,
P2 = (V2 * V2) / (V1 - V2) * P1 = (400V * 400V) / (500V - 400V) * 10kVA * 0.85 = 7.96kW (lagging)
P1 = (V1 - V2) * I1 * cos(θ) = 100V * I2 * cos(θ) = 10kVA * 0.85 = 8.5kW (lagging)
Therefore, η2 = 7.96kW / 8.5kW = 0.936 or 93.6%

Therefore, the efficiency of the autotransformer at rated load and 0.85 power factor is 99.4% (Option D).

Note: The efficiency of an autotransformer is generally higher than that of a two-winding transformer due to the lower losses in the single winding.
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Community Answer
A 400V / 100V, 10kVA two winding transformer is to be used as an autot...
2 wdg TF :

10KVA, 400/100KV
Efficiency : 97%
Load PF : 0.85 Lag

Output Power = 10KVA × 0.85 = 8500W

Efficiency = Output / (Output + Losses) = 97%

Losses = 262.88W

Auto TF :

Vh = 500V ; Ih=100A & VL= 400V ; IL=125A (50KVA)

Output Power = 50KVA × 0.85 = 42500W

Losses (as same as 2Wdg) = 262.88W

Efficiency = Output / (Output + Losses) = 42500/( 42500 + 262.88)

= 99.38 %
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A 400V / 100V, 10kVA two winding transformer is to be used as an autotransformer to supply a 400V circuit from 500V source when tested as two winding transformer at rated load .85 P.F. lag it’s efficiency is 97%. Determine efficiency as autotransformer at rated load and 0.85 power factora)95.5 %b)96.9%c)98.2 %d)99.4 %Correct answer is option 'D'. Can you explain this answer?
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