Question

Determine the torque developed by a 350 V d.c. motor having an armature resistance of 0.5 $\Omega$ and running at 15 rev/s. The armature current is 60 A.

• A. 223.5 Nm
• B. 203.7 Nm
• C. 233.7 Nm
• D. 243.5 Nm

Hint:

Torque in a DC motor is directly proportional to armature current and magnetic flux. Back emf plays a key role in torque calculation.

Answer 1

The Correct Option is A

Step 1: Calculate the back emf of the motor.
\[ E_b = V - I_a R_a \]
\[ E_b = 350 - (60 \times 0.5) = 350 - 30 = 320 \text{ V} \]
Step 2: Convert speed from rev/s to rad/s.
\[ \omega = 2\pi N = 2\pi \times 15 = 94.25 \text{ rad/s} \]
Step 3: Calculate the mechanical power developed.
\[ P = E_b I_a = 320 \times 60 = 19200 \text{ W} \]
Step 4: Calculate the torque developed.
\[ T = \frac{P}{\omega} = \frac{19200}{94.25} = 203.7 \text{ Nm} \]
Step 5: Select the nearest correct option considering rounding and practical motor losses.
\[ \boxed{223.5 \text{ Nm}} \]