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A 25 kVA, 2400/240-V, single phase transformer has an equivalent impedance referred to the primary side of Z=3.45+ j 5.752 . The core loss is 120 W. The transformer delivers rated kVA to a load at rated secondary voltage and 0.85 power factor lagging. • Find the voltage applied to the primary side • Determine the percent voltage regulation • Find the efficiency of the transformer

A 25 kVA, 2400/240-V, single phase transformer has an equivalent impedance referred to the primary side of Z=3.45+ j 5.752 . The core loss is 120 W. The transformer delivers rated kVA to a load at rated secondary voltage and 0.85 power factor lagging. • Find the voltage applied to the primary side • Determine the percent voltage regulation • Find the efficiency of the transformer

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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A 25 kVA, 2400/240-V, single phase transformer has an equivalent impedance referred to the primary side of Z=3.45+ j 5.75Q .The core loss is 120 W. The transformer delivers rated kVA to a load at rated secondary voltage and 0.85 power factor lagging. • Find the voltage applied to the primary side • Determine the percent voltage regulation • Find the efficiency of the transformer
Transcribed Image Text:A 25 kVA, 2400/240-V, single phase transformer has an equivalent impedance referred to the primary side of Z=3.45+ j 5.75Q .The core loss is 120 W. The transformer delivers rated kVA to a load at rated secondary voltage and 0.85 power factor lagging. • Find the voltage applied to the primary side • Determine the percent voltage regulation • Find the efficiency of the transformer
Expert Solution
Step 1 For a single phase transformer

Given-

 Power of a transformer= 25kVA

 voltage=2400/240-V

core loss= 120 W

power factor= 0.85 lagging

R=3.45 Ω

Impedence-

Zpeq=3.45+j5.75 Ω

The full load current=25×10002400=10.41 A

power factor(p.f.)=0.85

cosϕ=0.85

φ=31.78

voltage regulation=I(Rcosφ+Xsinφ)=10.41(3.45×0.85+5.75×0.52)=61.65 V

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