Question

A separately excited dc motor is driven from a three-phase fully controlled converter with a conductor for speed reversal. The line voltage is 440V at 50 Hz. The motor armature resistance is $0.12\Omega$ and the armature voltage constant is $2Volt-sec/rad$ and the armature voltage is constant at rated field current. Find the required value of firing angle delay necessary at rated field current for an armature current of $40A$ at a speed of $1500\text{rev/min}$.

• A. ${57.5}^{\circ }$
• B. ${43.4}^{\circ }$
• C. ${37.5}^{\circ }$
• D. ${67.2}^{\circ }$

Answer 1

The correct option is A ${57.5}^{\circ }$

$E=K_v\omega$
$=2\times 2\pi \times \frac{1500}{60}=314.2V$
$V_0=E+I_o{R}_a$
$=314.2+40\times 0.12=319V$
$V_0=\frac{3V_{L}M}{\pi }\text{cos}\alpha =3\times \frac{440\sqrt{2}}{\pi }\text{cos}\alpha$
$594\text{cos}\alpha =319$
$\text{cos}\alpha =\frac{319}{594}=0.537$
$\alpha ={\cos }^{-1}\left(0.537\right)={57.5}^{\circ }$