GATE 2018 EE Q.52: solution and answer

Official answer: 823 to 827Source: Official answer key, GATE 2018 EE (IIT Guwahati)

1. Given: rated voltage $V=200\text{V}$, armature current $I_{a1}=10\text{A}$, speed $N_1=1000\text{r.p.m.}$, total series resistance $R=1\Omega$. The magnetic circuit is linear, so flux $\varphi \propto I_a$.

2. Since $T\propto \varphi I_a\propto I_a^2$ for a series motor with linear magnetics, a 44% increase in load torque requires $I_{a2}^2=1.44I_{a1}^2$, giving $I_{a2}=1.2I_{a1}=1.2\times 10=12\text{A}$.

3. Back-EMF at state 1: $E_{b1}=V-I_{a1}R=200-10\times 1=190\text{V}$.

4. Back-EMF at state 2: $E_{b2}=V-I_{a2}R=200-12\times 1=188\text{V}$.

5. Since $\varphi \propto I_a$, the back-EMF $E_b\propto \varphi N\propto I_{a}N$, so $\frac{E_{b1}}{E_{b2}}=\frac{N_1{I}_{a1}}{N_2{I}_{a2}}$.

6. Substituting values: $\frac{190}{188}=\frac{1000\times 10}{N_2\times 12}\Rightarrow N_2=\frac{188\times 10000}{190\times 12}=\frac{1880000}{2280}\approx 824.56\text{r.p.m.}$

7. Rounded to the nearest integer, $N_2\approx 825\text{r.p.m.}$