Question

A 400 V dc series motor drives a load at rated voltage by drawing 20 A current running at the speed of 2000 rpm. The total series resistance is 1.4 $\Omega$. Assuming magnetic circuit to be linear, if load torque is increased by 49%. The speed of the motor in rpm will be _____

• A. 1596
• B. 1632
• C. 1590
• D. 1611

Answer 1

The correct option is D 1611

$\text{Given, dc series motor,}$

$T\alpha I^2$

$E_{b1}=400-20\left(1.4\right)=372V$

$N_1=2000rpm$

$\text{when load torque increases to 49\%}$
$\frac{T_2}{T_1}=\frac{I_2^2}{I_1^2}$

$\frac{1.49T_1}{T_1}=\frac{I_2^2}{{20}^2}$

$I_2=24.41A$

$\text{At same voltage,}$
$E_{b2}=V-I_{a2}(R_a+R_{ac})$
$=400-24.41\left(1.4\right)$
$=365.82V$

$\text{For series motor,}$
$N\alpha \frac{E_b}{\varphi }$

$\frac{N_2}{N_1}=\frac{E_{b2}}{E_{b1}}\times \frac{I_{a1}}{I_{a2}}$

$\frac{N_2}{2000}=\frac{365.82}{372}\times \frac{20}{24.41}$

$N_2=1611rpm$