A 6-pole, 250 V wave connected shunt motor running at 955 rpm ... | Technical Ability MCQ | Examoogle

Technical AbilityDC Machines

A 6-pole, 250 V wave connected shunt motor running at 955 rpm has 1200 armature conductors and useful flux/pole of 10m Wb. The armature and field resistance are 0.5 Ω and 250 Ω, respectively. If the motor draws 20 A from the supply mains, then the value of torque developed by the motor is _____.

62.3 N-m

65.8 N-m

45.6 N-m

57.9 N-m

Correct Answer

⚙️ TE • Technical Concept & PrincipleTechnical AbilityDC Machines

Option C

45.6 N-m

Quick Summary:

The torque developed by a DC motor is calculated using the relation $T_a = 9.55 \times \frac{E_b \times I_a}{N}$ . By determining the back EMF ( $E_b$ ) and the armature current ( $I_a$ ), the mechanical torque generated is obtained.

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

🔢 Key Formulas

$I_a = I_L - I_f = I_L - \frac{V}{R_f}$ — Armature current

$E_b = V - I_a R_a$ — Back EMF

$T_a = 9.55 \times \frac{E_b \times I_a}{N}$ — Torque developed

⚙️ Working Principle

In a DC motor, the armature current $I_a$ interacts with the magnetic field $\Phi$ to produce Lorentz force on conductors. The armature current is obtained by subtracting the field current $I_f = \frac{V}{R_f}$ from the total supply current $I_L$ . The back EMF $E_b$ is then calculated as $V - I_a R_a$ . Finally, the developed torque $T_a$ is derived from the power equation $P_{dev} = E_b \times I_a = \frac{2 \pi N T_a}{60}$ .

📌 Key Points

▸
Wave winding has number of parallel paths $A = P = 6$ , or more accurately for wave $A = 2$ . For wave-wound machines, $A = 2$ always.
▸
The power developed by the armature is $P_a = E_b \times I_a$ Watts.
▸
For this problem, $I_f = \frac{250}{250} = 1 \text{ A}$ , hence $I_a = 20 - 1 = 19 \text{ A}$ .
▸
$E_b = 250 - (19 \times 0.5) = 250 - 9.5 = 240.5 \text{ V}$ .
▸
$T_a = 9.55 \times \frac{240.5 \times 19}{955} = 45.69 \text{ N-m}$ .

✅ Advantages

▸
High starting torque characteristics
▸
Speed control flexibility in shunt motors

❌ Disadvantages / Limitations

▸
Armature reaction affects flux distribution
▸
Commutation issues at high speeds

🛠️ Applications / Uses

▸
Machine tools
▸
Centrifugal pumps and fans

📄 Additional Information

▸
The wave winding property $A=2$ is independent of the number of poles $P$ .
▸
Option C is the most accurate result based on the calculation $T_a = \frac{240.5 \times 19}{99.9} \approx 45.6 \text{ N-m}$ .

📊 Diagram / Illustration

✅

C is correct — The calculated torque developed by the DC shunt motor is 45.6 N-m based on the back EMF and armature current.

Core Concepts Used

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DC Machine Armature Reaction Back EMF calculation Torque-Speed relationship

💡 EXAM TIP

Always verify if the motor is wave or lap wound; for wave winding, the number of parallel paths is always 2, regardless of the number of poles.

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