Discussion
Turn off time of an SCR in series with RL circuit can be reduced by :
(a) increasing circuit resistance R
(b) decreasing circuit resistance R
(c) increasing circuit inductance L
(d) decreasing circuit inductance L
Answer: D
The official key is d only (decreasing the circuit inductance L).
The turn-off time of an SCR is the time the device needs, after its current has fallen to zero, to regain forward blocking. In a series RL circuit the current decays with time constant L/R, so a smaller inductance lets the current fall away faster and shortens the time before the device can start to recover. That supports (d).
What could not be confirmed: the same time-constant argument would also favour a larger R (option (a), giving 'a and d'), and no textbook or reference was found that states which of the two the question intends. The key accepts only (d), so (c), increasing L, and (b), decreasing R, are both the wrong direction and are rejected, and the official key is kept.
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