A 440V d.c. shunt-wound motor has a field resistance of 200 and an armature resistance of 0.6 When its speed is 20revs/second the current drawn from the supply is 12.2A. Calculate its back e.m.f. at this speed. If the speed were decreased to 19revs/second, the field flux remaining unchanged, calculate the new back e.m.f. and the new armature current.
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The equations which govern dc motor operation are:
Va = IaRa + Eg,
Eg = Kω
Is = If + Ia
If = Va/Rf
where,
armature voltage: Va = 440 v
armature current: Ia
armature resistance: Ra = 0.6 Ω
back emf: Eg
rotational velocity: ω = 20*2π = 125.66 rad/s
source current: Is
field current: If = Va/Rf = 440/600 = 2.2 A
field resistance: Rf = 200 Ω
Ia = Is - Ia = 12.2 - 2.2 = 10 A
Eg = Va - IaRa = 440 - 10*0.6 = 434 v
K = Eg/ω = 434/125.66 = 3.45
Eg(new) = Kω' = 3.45*19*2π = 412.36 v
Ia(new) = (Va - Eg) / Ra = 440 - 412.36 / 0.6 = 46 A
Va = IaRa + Eg,
Eg = Kω
Is = If + Ia
If = Va/Rf
where,
armature voltage: Va = 440 v
armature current: Ia
armature resistance: Ra = 0.6 Ω
back emf: Eg
rotational velocity: ω = 20*2π = 125.66 rad/s
source current: Is
field current: If = Va/Rf = 440/600 = 2.2 A
field resistance: Rf = 200 Ω
Ia = Is - Ia = 12.2 - 2.2 = 10 A
Eg = Va - IaRa = 440 - 10*0.6 = 434 v
K = Eg/ω = 434/125.66 = 3.45
Eg(new) = Kω' = 3.45*19*2π = 412.36 v
Ia(new) = (Va - Eg) / Ra = 440 - 412.36 / 0.6 = 46 A