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Question

A 200 V shunt motor takes 10 A when running on no load. At higher loads, the brush drop is 2 V and at light loads, it is negligible. The stray load loss at a line current of 100 A is 50% of the no load loss. Calculate the efficiency at a line current of 100 A if armature and field resistances are 0.2 and 100 ohm respectively.

A
74.48%
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B
81.2%
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C
78.14%
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D
66.12%
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Solution

The correct option is A 74.48%
At no load,
No load loss =200×10=2000W

If=200100=2A

I0=102=8A

Pio+Pwf=200×8(8)2×0.2=1587.2W

At load, Ia=1002=98A

I2aRa=(98)2×0.2=1920.8W

Stray load loss=0.5×2000=1000W

PL=(I2aRa+VbIa+Pst)+(Pio+Pwf+Psh)

=(1920+2×98+1000)+(1587.2+200×2)=5104W

Efficiency , η=PinPLPin

=200×1005104200×100=0.7448

%η=74.48%

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