Question

A 240 V, 70 A dc series motor has a combined armature circuit resistance $R_a+R_{se}=0.15\Omega$. The magnetization curve expressed in terms of $E_a$ versus $I_a$ at 900 rpm is given by following table.

$\begin{array}{ccccccc}{E}_a\left(V\right)& 95& 150& 188& 212& 229& 243\\ I_a\left(A\right)& 30& 52& 70& 78& 85& 90\end{array}$

A chopper whose duty ratio can be changed from 0.1 to 0.9 is used to brake the motor dynamically. What is the maximum speed (in rpm) the motor can achieve when the armature current is 70 A and braking resistance is 3 $\Omega$__________

• A. 955

Answer 1

The correct option is A 955
$\text{Armature current,}{I}_a=70A$

$\text{From the table,}$
$\text{At N = 900 rpm}$
$\text{and}$ $I_a=70A$,
$E_a=188V$
$E_a=(R_a+R_{se}+\delta R_B)\times I_a$

$\text{At maximum speed,}\delta =0.9$

$\text{Hence,}$ $E_a^{\prime }=(0.15+0.9\times 3)\times 70$

$=199.5V$

$\frac{E_a^{\prime }}{E_a}=\frac{N_{max}}{900}$

$N_{max}=\frac{199.5}{188}\times 900=955rpm$