Question

A DC series motor gave the following open circuit curve at a speed of 600 rpm.

$\begin{array}{ccccccc}\text{Field current (A)}& 10& 20& 30& 40& 60& 80\\ \text{Open circuit voltage (V)}& 103.5& 158& 206& 230& 259& 282\end{array}$

Total resistance of armature and field circuit is 1 $\Omega$. The motor is connected to 250 V supply. The speed of motor when the armature current is 40 A is________rpm.
(Neglect losses in core and friction, windage loss)

• A. 547.82

Answer 1

The correct option is A 547.82
$T\propto I_a^2$

$\frac{T_1}{T_2}=\frac{I_{a1}^2}{I_{a2}^2}$

$\text{When}$ $I_{a1}=40A$

$E_b=V_t-I_{a1}(R_{se}+R_a)$
$E_b=250-40\left(1\right)$
$=210V$

and at $I_f=40A$

$E_b=230V$
So, $E_b\propto \varphi N\propto I_{f}N$

So, $\frac{210}{230}=\frac{40\times N}{40\times 600}$

$N=547.82rpm$