Question

A 240 V, series motor takes 40 A when giving its rated output at 1500 rpm. Its resistance is $0.3\Omega$. The value of resistance which must be added to obtain rated torque at starting will be

• A. $3\Omega$
• B. $5.7\Omega$
• C. $2.7\Omega$
• D. $6\Omega$

Answer 1

The correct option is B $5.7\Omega$

$V=240V,I_a=40A$
$N_1=1500rpm{R}_a=0.3\Omega$

$T\propto I_a^2,$ $\text{since the torque is constant,}$

$\therefore$ $I_{a1}^2=I_{a2}^2$

$\Rightarrow$ $I_{a1}=I_{a2}=40A$

During starting the induced emf is zero, hence the current is limited only by the resistance in the armature circuit.

$\therefore$ $\text{Total resistance}=\frac{240}{40}=6\Omega$

$\text{Extra resistance to be added in series with armature}=6-0.3=5.7\Omega$