GATE 2025 EE Q.44: solution and answer

Official answer: ASource: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Since the magnetic circuit is linear and this is a series motor, flux is proportional to armature current: φ ∝ Ia. Hence electromagnetic torque, $T=k\varphi I_a\propto I_a^2$.

2. Initial condition (no external resistance, negligible series resistance): armature current $I_{a1}=40\text{A}$, terminal voltage $V=400\text{V}$, so back-EMF $E_1=V=400\text{V}$, at speed $N_1$.

3. The load torque varies as the cube of speed, $T_L\propto N^3$. At steady state the motor's electromagnetic torque equals the load torque at every operating speed. When the speed is halved, $N_2=\frac{N_1}{2}$, so $$\frac{T_2}{T_1}={\left(\frac{N_2}{N_1}\right)}^3=\frac{1}{8}$$

4. Since $T\propto I_a^2$: $$\frac{I_{a2}^2}{I_{a1}^2}=\frac{1}{8}⟹I_{a2}=\frac{I_{a1}}{\sqrt{8}}=\frac{40}{2.828}=14.142\text{A}$$

5. Back-EMF depends on both flux and speed, and here $\varphi \propto I_a$, so $E\propto \varphi N\propto I_{a}N$: $$\frac{E_2}{E_1}=\left(\frac{I_{a2}}{I_{a1}}\right)\left(\frac{N_2}{N_1}\right)=\left(\frac{14.142}{40}\right)\left(\frac{1}{2}\right)=0.1768$$

6. So $$E_2=400\times 0.1768=70.71\text{V}$$

7. With the external resistance $R$ now inserted in the armature circuit (series/armature resistance itself still negligible): $$V=E_2+I_{a2}R⟹400=70.71+14.142R$$

8. Solving: $$R=\frac{400-70.71}{14.142}=\frac{329.29}{14.142}=23.28\Omega$$ This matches option (A).