The condition for maximum power in case of D.C. motor is
A. Back e.m.f. = 2 × supply voltage
B. Back e.m.f. = $$\frac{1}{2}$$ × supply voltage
C. Supply voltage = $$\frac{1}{2}$$ × back e.m.f.
D. Supply voltage = back e.m.f.
Answer: Option B
This Question Belongs to Electrical Engineering >> D.C. Motors
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Comments ( 4 )
The condition for maximum efficiency for a D.C. generator is
A. Eddy current losses = stray losses
B. Hysteresis losses = eddy current losses
C. Copper losses = 0
D. Variable losses = constant losses
If a D.C. motor is connected across the A.C. supply it will
A. Run at normal speed
B. Not run
C. Run at lower speed
D. Burn due to heat produced in the field winding by eddy currents
The speed of a D.C. shunt motor can be increased by
A. Increasing the resistance in armature circuit
B. Increasing the resistance in field circuit
C. Reducing the resistance in the field circuit
D. Reducing the resistance in the armature circuit
The armature voltage control of D.C. motor provides
A. Constant torque drive
B. Constant voltage drive
C. Constant current drive
D. None of the above
Eb=1/2 × armature voltage (v)
= V/2
wrong ans
i also says that this answer is may be wrong because maximum power in dc motor is back emf=supply voltage /2
I think the ans may wrong... In DC motors the condition for max power should be
Back EMF=1/2 ×supply voltage