Question

A dc series motor of total resistance 0.5 $\Omega$ runs at a speed of 800 rpm when taking 60 A from 500 V dc mains. Now another identical series motor is mechanically coupled, but connected in series with the first motor. If the set runs at 400 rpm with load torque equal to 2.5 times the original torque. Find the current drawn from 500 V dc source

• A. 50 A
• B. 80 A
• C. 90 A
• D. 85 A

Answer 1

The correct option is D 85 A

$E_b=V-I_a{R}_a=500-60\left(0.5\right)=470V$

$\omega T_e=E_b{I}_a$

$\frac{2\pi \times 800}{60}\times T_e=470\times 60$

$T_e=336.61Nm$

$\text{New load torque}=2.5\times 336.61=841.52Nm$

$\text{Torque developed by the set.}$

$841.52=\frac{2(250-I_a\times 0.5)\times I_a}{\frac{2\pi \times 400}{60}}$

$I_a^2-500I_a+35250=0$

$I_a=\frac{500\pm \sqrt{{500}^2-4\times 35250}}{2}=\frac{500\pm 330.15}{2}$

$I_a=84.92\left(or\right)415$

$I_a=85A\text{(Choosing smaller value)}$