Products
Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

Q. 3.1

A transformer on no-load has a core-loss of 50 W,draws a current of 2 A(rms)and has an induced emf of 230 V(rms).Determine the no-load power factor,core-loss current and magnetizing current.Also calculate the no-load circuit parameters of the transformer.Neglect winding resistance and leakage flux.

Verified Solution

Power factor,
$\cos \theta _{0} =\frac{50}{2\times 230}=0.108$ lagging;
$\theta _{0} =83.76°$
Magnetizing current,
$I_{m} =I_{0} \sin\theta _{0}=2\sin \left(\cos ^{-1} 0.108 \right) =1.988$ A
Since $\theta _{0}\approx 90°$, there is hardly any difference between the magnitudes of the exciting current and its magnetizing component.
Core-loss current, $I_{i} = I_{0}\cos \theta _{0}=2\times 0.108=0.216$ A
In the no-load circuit model of Fig.3.7 core loss is given by
$G_{i} V_{1} ^{2} =P_{i}$
or $G_{i}=\frac{P_{i}}{V_{1} ^{2}} =\frac{50}{\left(230\right) ^{2} } =0.945\times 10^{-3} \mho$
Also    $I_{m} =B_{m}V_{1}$
or     $B_{m}=\frac{I_{m} }{V_{1}} =\frac{1.988}{230} =8.64\times 10^{-3} \mho$