Answer:
2.62 Ω
Explanation:
For a series motor, the field resistance is in series with the armature resistance. The back emf (e) is given by:
[tex]V=E_b+I_A(R_a+R_f)\\\\Where\ V\ is\ the\ terminal\ voltage, R_a=armature\ resistance,R_f=field \ resistance\\\\Given: V=415\ V,R_a=0.5 \Omega, R_f=0.25 \Omega,I_a=30A\\\\415=E_{b1}+30(0.5+0.25)\\\\415=E_{b1}+22.5\\\\E_{b1}=415-22.5=392.5V[/tex]
For a back emf of 392.5 V, the speed is 1000 rpm.
Speed is directly proportional to back emf. It is given as:
[tex]Nk\phi=E_b\\\\N_1k\phi_1= E_{b1}\\\\N_2k\phi_2= E_{b2}\\\\N_1=1000\ rpm, N_2=800\ rpm, E_{b1}=392.5\\\\\frac{E_{b1}}{E{b2}}= \frac{k\phi_1N_1}{k\phi_2N_2}\\\\\frac{E_{b1}}{E{b2}}= \frac{\phi_1N_1}{\phi_2N_2}\\But\ \phi\ is\ directly\ proportional\ to I_a\\\\\frac{E_{b1}}{E{b2}}= \frac{kI_{a1}N_1}{kI{a2}N_2}\\\\I_{a1}=I_{a2}\\\\\frac{E_{b1}}{E{b2}}= \frac{N_1}{N_2}\\\\E_{b2}=\frac{E_{b1}N_2}{N_1}=\frac{392.5*800}{1000}=314\ V[/tex]
Let the added resistance be R
[tex]V=E_{b2}+I_A(R_a+R_f+R)\\\\415=314+30(0.5+0.25+R)\\\\415=314+22.5+30R\\\\415=336.5+30R\\\\30R=78.5\\\\R=2.62\Omega[/tex]
