Question

A 220 V d.c. series motor develops its rated output at 1500 rpm while taking 25 A. Armature and series field resistance are 0.4 $\Omega$ and 0.6 $\Omega$ respectively. To obtain rated torque at 1200 rpm the external resistance added to armature is,

• A. 1.23 $\Omega$
• B. 1.34 $\Omega$
• C. 1.56 $\Omega$
• D. 1.72 $\Omega$

Answer 1

The correct option is C 1.56 $\Omega$

$\text{For dc series motor,}$
$\text{Given,}{V}_t=220V\text{,}$
$N_r=1500rpm$
$I_a=25A,$
$R_a=0.4\Omega$,
$R_{se}=0.6\Omega$

$\text{Torque;}T\propto I_a^2$
$\text{Back emf at 1500 rpm,}$

$E_{b1}=V_t-I_{a1}(R_a+R_{se})$
$=220-25(0.4+0.6)$
$E_{b1}=195V$

$\text{We know that,}{E}_b\propto N$
$\text{Back emf at 1200 rpm;}$

$\frac{E_{b2}}{195}=\frac{1200}{1500}$

$E_{b2}=\frac{12}{15}\times 195=156V$
$\text{Let us assume}{R}_{ext}\text{to be connected in series:}$
$E_{b2}=V_t-I_{a2}(R_a+R_{se}+R_{ext})$

$\text{To obtain rated torque at 1200 rpm,}$
$\text{armature current must be same;}$
$\text{i.e.}$ $I_{a2}=25A$

$\text{Now,}156=220-25(0.4+0.6+R_{ext})$
$220-156=25(1+R_{ext})$
$R_{ext}=1.56\Omega$