Concept:A practical diode begins conducting only when the input voltage exceeds the forward cut-in threshold \( V_D \) (at angle \( \theta_1 = \sin^{-1}(V_D / V_m) \)) and stops conducting when the voltage falls back below \( V_D \) (at angle \( \theta_2 = \pi - \theta_1 \)).
Formula:$$\theta_1 = \sin^{-1}\left(\frac{V_D}{V_m}\right), \quad \theta_2 = \pi - \theta_1$$
$$\theta_{\text{cond}} = \theta_2 - \theta_1 = \pi - 2\theta_1 = \pi - 2\sin^{-1}\left(\frac{V_D}{V_m}\right)$$
Solution:- The diode remains OFF at the start of the cycle until \( V_m \sin\theta \ge V_D \).
- It turns off before the end of the half-cycle once \( V_m \sin\theta < V_D \).
- Subtracting the two non-conducting intervals (\( \theta_1 \) at each end) gives a total conduction angle of \( \pi - 2\sin^{-1}(V_D / V_m) \).
Why other options are incorrect:- Option B: \( \pi \) (\( 180^\circ \)) applies only to an ideal diode with zero threshold voltage (\( V_D = 0 \)).
- Option C: \( 2\pi \) corresponds to full-period continuous conduction.
- Option D: \( \sin^{-1}(V_D / V_m) \) is the turn-on delay angle, not the total conduction interval.
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