Question

A single-phase triac based AC voltage controller feeds a series RL load. The input AC supply is 230 V, 50 Hz. The values of R and L are 10 \(\Omega\) and 18.37 mH, respectively. The minimum triggering angle of the triac to obtain controllable output voltage is

• A. \( 15^\circ \)
  (B) \( 30^\circ \)
  (C) \( 45^\circ \)
  (D) \( 60^\circ \)
• B.
• C.
• D.

Hint:

The minimum firing angle \( \alpha \) for a triac in RL loads should exceed the load phase angle \( \phi \) to achieve proper voltage control.

Answer 1

The Correct Option is B

Step 1: Problem data.
We are given the following information: Resistance \( R = 10 \, \Omega \) Inductance \( L = 18.37 \, \text{mH} = 18.37 \times 10^{-3} \, \text{H} \) Frequency \( f = 50 \, \text{Hz} \) Step 2: Phase angle calculation.
The phase angle \( \phi \) between the voltage and the current in an R-L circuit is given by the formula: \[ \phi = \tan^{-1} \left( \frac{\omega L}{R} \right), \] where \( \omega = 2\pi f \) is the angular frequency of the AC supply. Substitute the given values into the equation: 1. First, calculate the angular frequency \( \omega \): \[ \omega = 2 \pi \cdot f = 2 \pi \cdot 50 = 314.16 \, \text{rad/s}. \] 2. Next, substitute the values of \( \omega \), \( L \), and \( R \) into the phase angle formula: \[ \phi = \tan^{-1} \left( \frac{314.16 \cdot 18.37 \times 10^{-3}}{10} \right). \] 3. Perform the multiplication: \[ \phi = \tan^{-1} \left( \frac{5.771}{10} \right) = \tan^{-1}(0.5771). \] 4. Using a calculator or a standard table for inverse tangent, we find: \[ \phi = 30^\circ. \] ### Conclusion: The phase angle \( \phi \) is \( \mathbf{30^\circ} \). This concludes the solution. The phase angle between the voltage and current is \( 30^\circ \), indicating that the current lags the voltage in this R-L circuit.