Question

A dc series motor is controlled by the circuit shown below.



The armature and field resistance are 0.06 $\Omega$ and 0.04 $\Omega$ respectively. The average armature current is 200 A and chopper frequency is 500 Hz. If the back emf is 200 V, then the pulse width would be equal to

• A. 1.5 ms
• B. 2 ms
• C. 1 ms
• D. 0.5 ms

Answer 1

The correct option is C 1 ms

The given chopper is a step down chopper,

$V_0=\alpha V_s$

$V_0=I_0(R_a+R_{sc})+E_b=\alpha V_s$

$=200(0.06+0.04)+200=\alpha \cdot 440$

$\alpha =\frac{220}{440}$

$\alpha =0.5$

$T_{cn}=\alpha \cdot T$

$=0.5\times \frac{1}{500}=1ms$