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from the given datashort shunt compound generatorLoad current, I_{L}=30 \mathrm{~A}load Volkage, V=220 \mathrm{~V}.Armatuere resistance, R_{a}=0.05 \Omega,Series feild resirtance, R_{\text {se }}=0.3 \mathrm{~N}Shunt feird resisfance, R_{\text {sh }}=200 \OmegaBrush drop =1 \mathrm{v} per brush.Voltage drop inserres winding\begin{array}{l}=\left(I_{L}\right)\left(R_{\text {Se }}\right) \\=(30)(0.3) \\=9 \mathrm{~V}\end{array}Voltuge across shunt feild wending\begin{aligned}& =220+9 \\& =229 \mathrm{~V} \quad\left(\because V+I_{L} R_{\text {se }}=V_{\text {Shunt }}\right. \\I_{S h} & =\frac{229}{200}=1.145 \mathrm{~A} \quad\left(\because V_{\text {Shunt }}=I_{\text {Sh Sh }}\right) \\I_{S h} & =1.145 \mathrm{~A} \quad(\because \quad)\end{aligned}current equation\begin{aligned}I_{a} & =I_{L}+I_{5 h} \\& =30+1.145 \Rightarrow I_{a}=31.145 \mathrm{~A}\end{aligned}Armature current\begin{aligned}\text { \& Armature drop } & =I_{a} R_{a} \\& =(31.145)(0.05)=1.56 \mathrm{~V} \\\text { Brash drop } & =2 \times 1=2 \mathrm{~V}=V_{\text {brush }} \\\therefore \quad E_{g} & =V+I_{s e} R_{s e}+I_{a} R_{a}+B_{r a s h} d r o p \\& =V+I_{L} R_{s e}+I_{a} R_{a}+V_{\text {brash }} \\& =220+q+1.56+2\end{aligned}\therefore Induced E.M.F E_{g}=232.56 \mathrm{~V} ...