GATE Electrical Engineering (EE) 2020 Solved Paper

Question : 65 of 65

Marks: +1, -0

250 {V}

dc shunt motor has an armature resistance of

0.2 Ω

and

{R}_{ {f}}=100 Ω

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

5 {A}

and runs at

1200 {rpm}

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

50 {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 {V}

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

Solution:

Given—

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

at No load:-

E_{1}=250-0.2 × 5-2=247 V

at loaded condition:

{I} {L}=50 {A}

∴ {I}_{ {a}}=50-2.5=47.5 {A}

∴ {E}_{2}=250-47.5 × 8.2-2=238.5

∵ E=\;{P φ N}/{60 A} ∝ φ N

∴ \;{E_{1}}/{E_{2}}=\;{φ_{1} ⋅ N_{1}}/{φ_{2} ⋅ N_{2}}

⇒ \;{E_{2}}/{E_{1}}=\;{φ_{2} ⋅ N_{2}}/{φ_{1} ⋅ N_{1}}

⇒ \;{238.5}/{247}=\;{0.95 φ_{1} N_{2}}/{φ_{1} .1200}

{N}_{2}=1219.688 ≈ 1220 {rpm}

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