Question

A dc motor is connected to a 214 V rms (L-L), 3-\(\phi\), 50 Hz line using a 3-\(\phi\) bridge converter. If the firing angle is \(\frac{\pi}{6}\), full load armature current is 2500 A and armature resistance is 4 m\(\Omega\), then the back emf is

• A. \( 204\text{ V} \)
• B. \( 260.3\text{ V} \)
• C. \( 250.3\text{ V} \)
• D. \( 240.3\text{ V} \)

Hint:

To solve this quickly, break it into two simple steps: 1. Find the converter voltage: \( V_0 = 1.35 \times V_{L(\text{rms})} \times \cos\alpha \). Here, \( 1.35 \times 214 \times \cos(30^\circ) \approx 289 \times 0.866 = 250.3\text{ V} \). 2. Subtract the armature loss: \( I_a R_a = 2500 \times 0.004 = 10\text{ V} \). This gives \( 250.3 - 10 = 240.3\text{ V} \). Breaking the calculation down makes it much easier to solve!

Answer 1

The Correct Option is D