A shunt generator has a full-load current of 195 A at 250 V . The stray losses are 720 W and the shunt field coil resistance is 50 \Omega . It has a full-load efficiency of 88 \% . Find the armature resistance. Also find the current corresponding to maximum efficiency.
\begin{array}{l}\text { Total losses }=\text { Power input }-\text { power output }=55398-48750=6648 W \\\text { Shunt field current, } I_{s h}=\frac{V}{R_{s h}}=\frac{250}{50}=5 A \\\begin{array}{c}\text { Shunt field copper loss }=I_{s h}^{2} R_{s h}=(5)^{2} \times 50=1250 W \\\text { Stray loss }=720 W\end{array}\end{array}
\begin{array}{l}\text { Armature current, } I_{a}=I_{L}+I_{s h}=195+5=200 A \\\text { Armature copper loss, } I_{a}^{2} R_{a}=\text { Total losses }-\text { stray loss }-\text { shunt field copper loss }\end{array}
=6648-720-1250=4678 W
\text { or } \quad \text { Armature resistance, } P_{C}=\frac{4678}{I_{\sigma}^{2}}=\frac{4678}{(200)^{2}}=0.11695 \Omega
Constant losses, P_{C}= Shunt field loss + Stray loss =1250+720=1970 W
Armature current at which the efficiency will be maximum (say I_{a}^{\prime}),
I_{a}^{\prime}=\sqrt{\frac{P_{c}}{R_{a}}}=\sqrt{\frac{1970}{0.11695}}=130 A
Load current at which the efficiency will be maximum (say I_{L}^{\prime} ),
I_{L}^{\prime}=I_{a}^{\prime}-I_{s h}=130-4=126 A