Question

In the following circuit, the average voltage \[ V_o = 400 \left(1 + \frac{\cos \alpha}{3} \right) {V}, \] where \( \alpha \) is the firing angle. If the power dissipated in the resistor is 64 W, then the closest value of \( \alpha \) in degrees is: 

• A. 35.9
• B. 46.4
• C. 41.4
• D. 0

Hint:

When a battery is in series with a resistive load and a controlled rectifier, calculate average current from power using: \[ I_{avg} = \sqrt{\frac{P}{R}} \] Then find the total output voltage as: \[ V_o = I_{avg} R + {Battery voltage} \] Finally, use the given formula for \( V_o \) to solve for \( \alpha \) by isolating \( \cos \alpha \).

Answer 1

The Correct Option is A

Understanding the Circuit The circuit consists of a three-phase half-wave controlled rectifier feeding an RL load with a battery in series. The average output voltage is given by: \[ V_o = 400 \left(1 + \frac{\cos \alpha}{3} \right) \, {V} \] Given: - Resistor \( R = 1\, \Omega \) - Power dissipated in resistor \( P = 64 \, {W} \) - Battery voltage = 500 V 
Step 1: Find average current \[ P = I_{avg}^2 R \Rightarrow 64 = I_{avg}^2 \Rightarrow I_{avg} = \sqrt{64} = 8 \, {A} \] Step 2: Voltage across resistor \[ V_R = I_{avg} \cdot R = 8 \cdot 1 = 8 \, {V} \] Step 3: Find total output voltage \( V_o \) \[ V_o = V_R + {Battery voltage} = 8 + 500 = 508 \, {V} \] Step 4: Plug into average voltage formula \[ 508 = 400 \left(1 + \frac{\cos \alpha}{3} \right) \] \[ \Rightarrow \frac{508}{400} = 1 + \frac{\cos \alpha}{3} \] \[ \Rightarrow 1.27 = 1 + \frac{\cos \alpha}{3} \] \[ \Rightarrow \frac{\cos \alpha}{3} = 0.27 \] \[ \Rightarrow \cos \alpha = 0.81 \] \[ \Rightarrow \alpha = \cos^{-1}(0.81) \approx 35.9^\circ \] \[ \boxed{\alpha \approx 35.9^\circ} \]