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:
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:
Show Hint
- 35.9
- 46.4
- 41.4
- 0
The Correct Option is A
Solution and Explanation
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 \[ 508 = 400 \left(1 + \frac{\cos \alpha}{3} \right) \] \[ \Rightarrow \frac{508}{400} = 1 + \frac{\cos \alpha}{3} \] \[ \Rightarrow 1.27 = 1 + \frac{\cos \alpha}{3} \] \[ \Rightarrow \frac{\cos \alpha}{3} = 0.27 \] \[ \Rightarrow \cos \alpha = 0.81 \] \[ \Rightarrow \alpha = \cos^{-1}(0.81) \approx 35.9^\circ \] \[ \boxed{\alpha \approx 35.9^\circ} \]
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