GATE Electrical Engineering (EE) 2019 Solved Paper

Question : 49 of 65

Marks: +1, -0

220 {V}

DC shunt motor takes

3 {A}

at no-load. It draws

25 {A}

when running at full-load at

1500 {rpm}

. The armature and shunt resistances are

0.5 Ω

and

220 Ω

, respectively. The no-load speed in rpm (round off to two decimal places) is

Solution:

DC shunt motor:
No-load (case-1):

{V}_{ {t}}=220 {V}

{I}_{ {LI}}=3 {A}

{N}_{ {t}}=?

{I}_{\sh1} \;\; =\;{V_{ {t}}}/{ {r}_{ {\sh}}}=\;{220}/{220}=1 {A}

⇒ {I}_{\a1} = {I}_{ {LI}}- {I}_{ {\sh} 1}

=2 {A}

{E}_{ {b} 1} = {V}_{ {t}}- {I}_{ {a} 1} {r}_{ {a}}=220-(2)(0.5)=219 {V}

Loaded (case 2):

{V}_{ {t}}=220 {V}

{I}_{ {L} .2}=25 {A}

{N}_{2}=1500 {rpm}

{I}_{ {\sh} 2}=\;{ {V}_{ {t}}}/{ {r}_{ {\sh}}}=1 {A}

∴ ϕ=

constant
Since field current is constant

{I}_{ {a} 2}= {I}_{ {L} 2}- {I}_{ {\sh} 2}

=24 {A}

{E}_{ {b} 2}= {V}_{ {t}}- {I}_{a2} {r}_{ {a}}

=220-(24)(0.5)

=208 {V}

E_{b}=k_{a} ϕ ω ⇒ E_{b} ∝ N

=\;{ {E}_{ {b} 1}}/{ {E}_{ {b} 2}}=\;{ {N}_{1}}/{ {N}_{2}}=\;{219}/{208}=\;{ {N}_{1}}/{1500} ⇒ {N}_{1}=1579.32 {rpm}

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