Answered: A voltage source of 220 AC is applied…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

A voltage source of 220 AC is applied on the primary of single phase transformer and secondary winding is kept open. Ammeter and watt meter show following readings: ammeter= 50 amp, wattmeter reading is 1000 watts. Determine parameters of transformer and show them on a circuit diagram.

B) A secondary winding is short circuited and small voltage of 50 V is applied on primary side of a single phase transformer. Ammeter and watt meter show following readings: ammeter= 5 amp, wattmeter reading is 100 watts.

I) On which side can you determine referred equivalent circuit. Show it on a circuit diagram.

ii) If turn ratio is 100:10, and primary side resistance is 5 ohm, Determine secondary resistance.

Expert Solution

Step 1

Given

Open circuit test results as

$No load voltage V_{0} = 220 v No load current I_{0} = 50 A No load power W_{0} = 1000 w$

Short circuit test results as

$Short circuit applied voltage V_{s c} = 50 v Short circuit ammeter reading I_{s c} = 5 A Short circuit wattmeter reading W_{s c} = 100 w$

Step 2

From open circuit test no load parameters can be determined

$No load power factor \cos ϕ_{0} = \frac{W_{0}}{V_{0} I_{0}} S u b s t i t u t e t h e g i v e n v a l u e s \cos ϕ_{0} = \frac{1000}{220 \times 50} = 0.09 ϕ_{0} = 84 . 78^{°} \sin ϕ_{0} = \sin 84 . 78^{°} = 0.99 I_{μ} = I_{0} \sin ϕ_{0} = 50 \times 0.99 = 49.79 A I_{w} = I_{0} \cos ϕ_{0} = 50 \times 0.09 = 4.54 A R_{0} = \frac{V_{0}}{I_{w}} = \frac{220}{4.54} = 48.44 Ω X_{0} = \frac{V_{0}}{I_{μ}} = \frac{220}{49.79} = 4.41 Ω$

Under no load the equivalent circuit is

From short circuit series elements can be determined

$Series resistance refered to primary side R_{01} = \frac{W_{s c}}{{I^{2}}_{s c}} R_{01} = \frac{100}{5^{2}} = 4 Ω series impedance refered to primary side Z_{01} = \frac{V_{s c}}{I_{s c}} Z_{01} = \frac{50}{5} = 10 Ω series reactance refered to primary side X_{01} = \sqrt{{Z_{01}}^{2} - {R_{01}}^{2}} X_{01} = \sqrt{10^{2} - 4^{2}} = 9.16 Ω$

The equivalent circuit of transformer under short circuit is

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