Q1. The gross mechanical power developed by a DC motor is maximum when:
A. IaRa = 0
B. Eb = V
C. Eb = V/2
D. Ra = 0
Correct Option: C
Explanation:
Mechanical power developed Pm = EbIa.
From voltage equation Eb = V − IaRa.
Using Pm = VIa − Ia2Ra and differentiating w.r.t. Ia, maximum power condition gives IaRa = V/2.
Substituting, Eb = V/2. At this condition, efficiency is less than 50%, hence not used in practice.
Q2. Mechanical power developed in a DC motor armature is:
A. VIa
B. V − IaRa
C. EbIa
D. Ia2Ra
Correct Option: C
Explanation:
Input power VIa is divided into mechanical power developed EbIa and armature copper loss Ia2Ra.
Hence, mechanical power developed in armature equals EbIa.
Q3. Voltage equation of a DC motor is:
A. V = Eb − IaRa
B. V = Eb + IaRa
C. Eb = IaRa
D. Eb = VIa
Correct Option: B
Explanation:
Applied voltage must overcome back emf and armature resistance drop.
Therefore, V = Eb + IaRa and equivalently Eb = V − IaRa.
Q4. Torque produced in a DC motor is proportional to:
A. Ia2
B. Φ2
C. ΦIa
D. N
Correct Option: C
Explanation:
Motor torque Ta = kΦIa.
Hence torque is proportional to the product of flux and armature current.
For shunt motor Φ ≈ constant → T ∝ Ia.
For unsaturated series motor → T ∝ Ia2.
Q5. Armature torque of a DC motor is also called:
A. Useful torque
B. Shaft torque
C. Gross torque
D. Stray torque
Correct Option: C
Explanation:
Armature torque is the total torque developed inside the armature before subtracting losses. Hence, it is called gross torque.
Q6. Shaft torque in a DC motor is:
A. Greater than armature torque
B. Equal to armature torque
C. Less than armature torque
D. Independent of armature torque
Correct Option: C
Explanation:
Due to iron loss, friction loss, and windage loss, part of the gross torque is lost.
Therefore, useful shaft torque is less than armature torque.
Q7. The difference between armature torque and shaft torque is due to:
A. Copper losses
B. Eddy current losses only
C. Iron and friction losses
D. Hysteresis only
Correct Option: C
Explanation:
The lost torque corresponds to iron losses, friction losses, and windage losses.
Copper losses represent power loss, not torque loss.
Q8. The condition for maximum power in a DC motor is obtained by differentiating:
A. Eb = V − IaRa
B. VIa
C. Pm = EbIa
D. Pm = VIa − Ia2Ra
Correct Option: D
Explanation:
Mechanical power Pm = VIa − Ia2Ra.
Differentiating with respect to Ia and equating to zero gives IaRa = V/2, which is the condition for maximum power.
Q9. For maximum power, the relationship between V and Eb is:
A. Eb = V
B. Eb = 0
C. Eb = V/2
D. Eb = 2V
Correct Option: C
Explanation:
From maximum power condition IaRa = V/2 and Eb = V − IaRa, we get Eb = V/2.
Q10. If flux is constant, motor torque is directly proportional to:
A. Speed
B. Eb
C. Armature current
D. Ra
Correct Option: C
Explanation:
Torque Ta ∝ ΦIa.
For constant flux, torque is directly proportional to armature current.
Q11. Expression for armature torque is:
A. Ta = kN
B. Ta = kIa2
C. Ta = kΦIa
D. Ta = kΦ2
Correct Option: C
Explanation:
General torque equation of DC motor is
Ta = 0.159 (PZ/A) ΦIa.
Grouping constants gives Ta = kΦIa.
Q12. A 220 V DC motor takes 40 A. Ra = 0.5 Ω. Back emf is:
A. 180 V
B. 190 V
C. 200 V
D. 210 V
Correct Option: C
Explanation:
Eb = V − IaRa = 220 − (40 × 0.5) = 200 V.
Q13. A 4-pole DC motor has 600 conductors, wave-wound. Φ = 0.025 Wb, Ia = 30 A. Armature torque is:
A. 143 N·m
B. 100 N·m
C. 180 N·m
D. 250 N·m
Correct Option: A
Explanation:
Ta = (PZΦIa) / (2πA)
= (4 × 600 × 0.025 × 30) / (2π × 2) ≈ 143 N·m.
Q14. The power corresponding to shaft torque is:
A. EbIa
B. VIa
C. Tshω
D. Ia2Ra
Correct Option: C
Explanation:
Useful mechanical output power equals shaft torque multiplied by angular speed:
Pout = Tshω.
Q15. Lost torque in DC motor is equal to:
A. Ta + Tsh
B. Ta − Tsh
C. Tsh − Ta
D. Zero
Correct Option: B
Explanation:
Lost torque = armature (gross) torque − shaft torque.
It corresponds to friction, windage, and core losses.
Q16. Mechanical output of motor is:
A. Taω
B. EbIa
C. VIa
D. Ia2Ra
Correct Option: A
Explanation:
Mechanical power developed equals torque multiplied by angular speed:
P = Taω.
Shaft output is this power minus rotational losses.
Q17. Back emf in DC motor is given by:
A. Eb = V + IaRa
B. Eb = V − IaRa
C. Eb = IaRa
D. Eb = VIa
Correct Option: B
Explanation:
From voltage equation V = Eb + IaRa, hence
Eb = V − IaRa.
Q18. Torque developed inside armature is called:
A. Shaft torque
B. Load torque
C. Armature torque
D. Stray torque
Correct Option: C
Explanation:
Torque produced internally by armature conductors is called armature or gross torque.
Q19. The torque available at shaft is called:
A. Gross torque
B. Armature torque
C. Shaft torque
D. Electrical torque
Correct Option: C
Explanation:
Shaft torque is the useful torque available at the shaft after subtracting all rotational losses:
Tsh = Ta − lost torque.
Q20. A motor takes 40 A from 220 V supply. Back emf = 200 V, Ra = 0.5 Ω. Mechanical output power is:
A. 6800 W
B. 7000 W
C. 7200 W
D. 8000 W
Correct Option: C
Explanation:
Power developed = EbIa = 200 × 40 = 8000 W.
Armature copper loss = Ia2Ra = 40² × 0.5 = 800 W.
Mechanical output = 8000 − 800 = 7200 W.