A 230/460 V transformer has a primary resistance of 0.2 Ω and a reactance of 0.5 Ω and corresponding values for secondary are 0.75 Ω and 1.8 Ω, respectively. Find the secondary terminal voltage when supplying 10 A and 0.8 pf lagging.
Answers
Answer:
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Given:
A 230/460 V transformer has a primary resistance of 0.2 Ω and a reactance of 0.5 Ω and corresponding values for secondary are 0.75 Ω and 1.8 Ω, respectively.
To find:
Find the secondary terminal voltage when supplying 10 A and 0.8 pf lagging.
Solution:
From given, we have,
A 230/460 V transformer has a primary resistance of 0.2 Ω and a reactance of 0.5 Ω and corresponding values for secondary are 0.75 Ω and 1.8 Ω, respectively.
The formulae used are:
K = 460 / 230 = 2
R_{02} = R_{2} + K²R_{1}
R_{02} = 0.75 + 2² × 0.2
R_{02} = 1.55 Ω
X_{02} = X_{2} + K²X_{1}
X_{02} = 1.8 + 2² × 0.5
X_{02} = 3.8 Ω
The voltage drop is,
= I{2} (R_{02} cos ∅ + X_{02} sin ∅ )
= 10 (1.55 × 0.8 + 3.8 × 0.6)
= 35.2 V
Therefore, the secondary terminal voltage is,
= 460 - 35.2
= 424.8 V
Therefore, the secondary terminal voltage when supplying 10 A and 0.8 pf lagging is 424.8 V.