I need help on questions 4-6.ALSO dpes Vp=600V and Vs=120V? Or is it Vp=720V and Vs=600V? A 15 kVA, 600/120 V, 0.95 efficient conventional transformer is to be used for to supply powerfrom a 720 V source to a 600 V load at 0.8 PF. Neglect core losses. [SO 2a,b,d]1. Find the current ratings of the primary and secondary coils of transformer?2. Sketch the transformer connections.3. Calculate the maximum input) and output)currents of the autotransformer.4. Find the S rating of the formed auto-transformer.5. Calculate the rating advantage of the autotransformer over the conventional transformer.6. Calculate the efficiency of the auto-transformer

Answered: A 15 kVA, 600/120 V, 0.95 efficient…

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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I need help on questions 4-6.

ALSO dpes Vp=600V and Vs=120V? Or is it Vp=720V and Vs=600V?

A 15 kVA, 600/120 V, 0.95 efficient conventional transformer is to be used for to supply power
from a 720 V source to a 600 V load at 0.8 PF. Neglect core losses. [SO 2a,b,d]
1. Find the current ratings of the primary and secondary coils of transformer?
2. Sketch the transformer connections.
3. Calculate the maximum input) and output)currents of the autotransformer.
4. Find the S rating of the formed auto-transformer.
5. Calculate the rating advantage of the autotransformer over the conventional transformer.
6. Calculate the efficiency of the auto-transformer

Expert Solution

Step 1

In the problem, it is asked to design an auto-transformer whose primary voltage is 720 V and the secondary voltage is 600 V from a 600/120 V conventional transformer.

The current rating of the primary coil of the transformer is determined as:

$\begin{array}{rcl} I_{p} & = & \frac{15 k V A}{600 V} \\ = & 25 A \end{array}$

The current rating of the secondary coil of the transformer is determined as:

$\begin{array}{rcl} I_{s} & = & \frac{15 k V A}{120 V} \\ = & 125 A \end{array}$

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