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

Concept: A three-phase fully controlled bridge rectifier (6-pulse converter) converts an AC source into a adjustable DC voltage to drive a DC motor. To solve this problem, we connect two main electrical relationships:
• Converter Output Voltage Equation: The average DC output voltage ($V_0$) produced by a 3-phase fully controlled bridge converter operating with a firing angle $\alpha$ is given by: \[ V_0 = \frac{3V_{mL}}{\pi} \cos\alpha = \frac{3\sqrt{2}V_{L(\text{rms})}}{\pi} \cos\alpha \] where $V_{L(\text{rms})}$ is the root-mean-square line-to-line input voltage.
• DC Motor Armature Loop Equation: The voltage applied across the motor armature equals the internal back EMF ($E_b$) plus the ohmic voltage drop across the internal resistance ($R_a$): \[ V_0 = E_b + I_a R_a \quad \Rightarrow \quad E_b = V_0 - I_a R_a \]

Step 1: Extracting variables from the problem description.
We are given:
• RMS Line-to-Line Voltage, \( V_{L(\text{rms})} = 214\text{ V} \)
• Firing angle, \( \alpha = \frac{\pi}{6} = 30^\circ \)
• Armature current, \( I_a = 2500\text{ A} \)
• Armature resistance, \( R_a = 4\text{ m}\Omega = 4 \times 10^{-3}\ \Omega \)

Step 2: Calculating the average DC output voltage \(V_0\).
Substitute our values into the 3-phase bridge rectifier equation: \[ V_0 = \frac{3\sqrt{2} \times 214}{\pi} \times \cos(30^\circ) \] We know that $\cos(30^\circ) = \frac{\sqrt{3}}{2}$, $\sqrt{2} \approx 1.4142$, and $\pi \approx 3.1416$: \[ V_0 = \frac{3 \times 1.4142 \times 214}{3.1416} \times \frac{\sqrt{3}}{2} \] \[ V_0 = \frac{907.916}{3.1416} \times 0.866025 \] \[ V_0 \approx 289.00 \times 0.866025 \approx 250.28\text{ V} \]

Step 3: Calculating the internal armature ohmic voltage drop.
Find the voltage drop across the armature resistance ($I_a R_a$): \[ \text{Voltage Drop} = I_a \times R_a = 2500\text{ A} \times (4 \times 10^{-3}\ \Omega) \] \[ \text{Voltage Drop} = 2500 \times 0.004 = 10\text{ V} \]

Step 4: Computing the motor back EMF \(E_b\).
Using the motor loop equation, subtract this 10 V internal drop from the converter output voltage: \[ E_b = V_0 - I_a R_a \] \[ E_b = 250.28\text{ V} - 10\text{ V} = 240.28\text{ V} \] Rounding to one decimal place gives \(240.3\text{ V}\), which matches option (4). Hence, the correct choice is option (4).