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GATE Electrical Engineering (EE) 2020 Solved Paper

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Question : 65 of 65

Marks: +1, -0

A

250\ \mathrm{V}

dc shunt motor has an armature resistance of

0.2\ \Omega

R_{f}=100\ \Omega

. When the motor is operated at no-load at rated voltage, it draws an armature current of

5\ \mathrm{A}

1200\ \mathrm{rpm}

. When the load is coupled to the motor, it draws total line current of

50\ \mathrm{A}

at rated voltage, with a

5\ \%

reduction in the air gap flux due to armature rection voltage drop across the brushes can be taken as

1\ \mathrm{V}

per brush under all operating conditions. The speed of the motor, (in rpm), under this loaded conditions, is closest to:

1200
1220
900
1000

Solution:

Given—

I_{f}=\;\frac{250}{100}=2.54

at No load:-

E_{1}=250-0.2\times5-2=247\ \mathrm{V}

at loaded condition:

I_{L}=50\ \mathrm{A}

\therefore I_{a}=50-2.5=47.5\ \mathrm{A}

\therefore E_{2}=250-47.5\times8.2-2=238.5

\because E=\frac{P\phi N}{60A}\propto\phi N

\therefore \frac{E_{1}}{E_{2}}=\frac{\phi_{1}\cdot N_{1}}{\phi_{2}\cdot N_{2}}

\Rightarrow \frac{E_{2}}{E_{1}}=\frac{\phi_{2}\cdot N_{2}}{\phi_{1}\cdot N_{1}}

\Rightarrow \frac{238.5}{247}=\frac{0.95\phi_{1} N_{2}}{\phi_{1}.1200}

N_{2}=1219.688\approx 1220\ \mathrm{rpm}

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