Solution for GATE 2023 question on Electrical Machines by Dr. Mudita Banerjee

GATE stands for Graduate Aptitude Test in Engineering. GATE stands as significant milestone for countless aspiring engineers who are willing to seek admission to postgraduate programs or pursuing career opportunities in the field of engineering.

For clearing gate, you have to prepare subjects like Network, Electrical measurements, Engineering Mathematics, Signals and System, Electrical Machines, Power System, and many more.

You may have seen the gate 2023 paper but you are not aware of the solution. Where you will get that? Don’t worry , Dr. Mudita Mam had explained in detail all the solution of questions which appeared in the GATE EE 2023 of the subject Electrical Machines.

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In this article , we will provide a well-acquainted solution of GATE EE 2023 Questions for the challenging subject Electrical Machines.

GATE EE 2023 Machines Question

Q. When the winding c-d of the single-phase, 50 Hz, two winding transformer is supplied from an AC current source of frequency 50 Hz, the rated voltage of 200 V (rms), 50 Hz is obtained at the open-circuited terminals a-b. The cross sectional area of the core is 5000 mm² and the average core length traversed by the mutual flux is 500 mm. The maximum allowable flux density in the core is B_max = 1 Wb/m² and the relative permeability of the core material is 5000. The leakage impedance of the winding a-b and winding c-d at 50 Hz are (5 + j100π x 0.16) Ω and (11.25 + j100 π x 0.36) Ω, respectively.

Considering the magnetizing characteristics to be linear and neglecting core loss, the self-inductance of the winding a-b in millihenry is

Solution :- Given that, A= 5000 mm², = 5 x 10^-3 m² , E= 200 V, f=50 Hz, B_m= 1 Wb/m² , μ_r = 5000, length (l) =500 mm

To Find :-

Step-1 :- Calculating Reluctance

S=\frac{I}{\mu {o}\mu {r}A}

Now putting the given values, in the formula of reluctance.

S=\frac{0.5}{4\pi \times 10^{-7}\times 5000\times 5\times 10^{-3}}=15915.49 AT/Wb

S= 15915.49 AT/Wb

Step-2 :- Calculating the no. of turns

E= 4.44fNB_m A

N=\frac{E}{4.44fB_{m}A}

Putting all the given values we get,

N=\frac{200}{4.44\times 50\times 1\times 5\times 10^{-3}}=180.18 = 180 turns

Step-III :- Calculate value of self-inductance

L=\frac{N^{2}}{S}=\frac{180 \times 180}{15915.49}=2.0357 \; H

Step-IV :- Calculating the value of self-inductance of the winding a-b

= L + 0.16 H

= (2.0357 + 0.16) mH

= 2.1957 H

= 2195.7 mH

The self-inductance of the winding a-b is 2195.7 mH.

These are some steps you can take to solve this type of problem.

Hope, so. You have well understood the solution of the problem.