A 200 V shunt motor takes 10 A when running on no-load. At higher loads the brush drop is 2 V and at light load it is negligible. The stray load loss at a line current of 100 Vis 50 \% of the no-load loss. Calculate the efficiency at a line current of 100 A if armature and field resistance are 0.2 W and 100 W respectively.
\begin{aligned}I_{s h} &=\frac{V}{R_{s h}}=\frac{200}{100}=2 A \\\text { Shunt field Cu loss } &=I_{s h}^{2} R_{s h}=(2)^{2} \times 100=400 W \\\text { Stray loss } &=\frac{50}{100} \times 2000=1000 W\end{aligned}
\begin{array}{l}\text { At load, armature current, } \quad I_{a}=I_{L}-I_{s h}=100-2=98 A \\\text { Armature copper loss }=I_{a}^{2} R_{a}=(98)^{2} \times 0.2=1920.8 W \\\text { Loss at brushes }=V_{b f} \times I_{a}=2 \times 98=196 W\end{array}
Total losses =Stray loss + Armature Cu loss + Shunt field Cu loss + Brush contact loss
=1000+1920.8+400+196=3516.8 W
Input to motor =V \times I_{L}=200 \times 100=20000 W
\begin{aligned}\text { Motor efficiency, } \eta &=\frac{\text { output }}{\text { input }}=\frac{\text { Input – Losses }}{\text { Input }} \times 100 \\&=\frac{20000-3516.8}{20000} \times 100= 8 2 . 4 1 6 \%\end{aligned}