A D C series motor has the following rating: 200 V , 20 A and 1200 rpm . Armature and series field resistances are 0.1 \Omega and 0.2 \Omega respectively. Magnetic circuit can be assumed to the linear. At what speed the motor will run at rated torque if a resistance of 20 \Omega is placed in parallel with the armature?
When a resistance of 20 \Omega is connected parallel to armature, as shown in Fig. 5.43, let the motor input current be I_{2} ampere.
Voltage across armature =V-I_{2} R_{s e}=200-0.2 I_{2}
Diverter current, I_{D}=\frac{200-0.2 I_{2}}{20}=10-0.01 I_{2}
\text { Armature current, } \quad I_{a 2}=I_{2}-I_{D}=I_{2}-\left(10-0.01 I_{2}\right)=1.01 I_{2}-10
\text { At rated torque i.e., } \quad T_{2}=T_{1}
\text { or } \left.\quad I_{a 2} \Phi_{2}=I_{\alpha 1} \Phi_{1} \text { (since } T \propto \Phi_{\alpha}\right)
\text { or } \quad I_{a 2}=I_{a 1} \times \frac{\Phi_{1}}{\Phi_{2}}
\text { or } \quad 1.01 I_{2}-10=I_{1} \times \frac{I_{1}}{I_{2}} \text { (series } \Phi \propto I \text { ) }
\begin{array}{l}\text { or }\\1.01 I_{2}-10=\frac{20 \times 20}{I_{2}}\end{array}
\begin{array}{l}\text { or }\\I_{2}=25.45 A\end{array}
\text { Armature current, } I_{a 2}=1.01 I_{2}-10=1.01 \times 25.45-10=15.7 A
\text { Back emf, } E_{b 2}=V-I_{2} R_{s e}-I_{a 2} R_{a}=200-25.25 \times 0.2-15.7 \times 0.1=193.34 V
\text { Then, } N_{2}=N_{1} \times \frac{E_{b 2}}{E_{b 1}} \times \frac{\Phi_{1}}{\Phi_{2}}=1200 \times \frac{193.34}{214} \times \frac{20}{25.45}= 8 5 2 \text { rpm }
