Question

A dc chopper is used for regenerative braking of a separately excited dc motor. The dc supply voltage is 400 V. The motor has $r_a=0.2\Omega ,K_m=1.2V-s/rad$. The average armature current during regenerative braking is kept constant at 300 A with negligible ripple. If the duty cycle of the chopper is 60%, then the minimum and maximum permissible braking speeds are respectively

• A. 314 rpm and 4126 rpm
• B. 477 rpm and 4126 rpm
• C. 477 rpm and 3660 rpm
• D. 512 rpm and 3660 rpm

Answer 1

The correct option is C 477 rpm and 3660 rpm

$E=V_t+I_a{R}_a$
Minimum terminal voltage = 0
Hence, $E=K_m\omega =I_a{r}_a$

${\omega }_{min}=\frac{I_a{r}_a}{K_m}=\frac{300\times 0.2}{1.2}=50rad/secor477.46rpm$
${\omega }_{max}$corresponds to maximum emf which occurs when supply voltage is maximum or duty cycle is 100%.
${\omega }_{max}=\frac{V_s+I_a{r}_a}{K_m}=\frac{400+300\times 0.2}{1.2}=383.33rad/secor3660rpm$