Question

A separately excited DC motor rated 400 V, 15 A, 1500 RPM drives a constant torque load at rated speed operating from 400 V DC supply drawing rated current. The armature resistance is 1.2 $\Omega$. If the supply voltage drops by 10% with field current unaltered then the resultant speed of the motor in RPM is \underline{\hspace{1cm}} (Round off to the nearest integer).

Hint:

For DC motors, always subtract $I_{a}R_{a}$ drop before relating back emf to speed. Speed is directly proportional to $E_{a}$ when field flux is constant.

Answer 1

Step 1: Motor emf equation.
\[ E_{a} = V - I_{a}R_{a} \] and \[ E_{a} \propto N \Phi \] For constant field ($\Phi$ constant), $E_{a} \propto N$.

Step 2: At rated conditions.
Given: $V=400 \,\text{V}, I_{a}=15 \,\text{A}, R_{a}=1.2 \,\Omega, N=1500 \,\text{RPM}$. \[ E_{a1} = 400 - (15)(1.2) = 400 - 18 = 382 \,\text{V}. \]

Step 3: New conditions (10% drop in voltage).
New supply: $V' = 0.9 \times 400 = 360 \,\text{V}$. Armature current remains same ($I_{a}=15$ A) because load torque is constant. \[ E_{a2} = V' - I_{a}R_{a} = 360 - 18 = 342 \,\text{V}. \]

Step 4: Relating emf and speed.
\[ \frac{E_{a1}}{E_{a2}} = \frac{N_{1}}{N_{2}} \] \[ N_{2} = N_{1} \cdot \frac{E_{a2}}{E_{a1}} = 1500 \cdot \frac{342}{382}. \] \[ N_{2} \approx 1343.5 \,\text{RPM}. \]

Step 5: Rounding off.
\[ N_{2} \approx 1344 \,\text{RPM} \;\; \approx 1340 \,\text{to nearest 10}. \] % Final Answer \[ \boxed{1344 \,\text{RPM (rounded to 1340 or 1344 depending on rounding rule)}} \]