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Calculate the circuit turn-off time for Half-wave controlled rectifier for a ω=5 rad/sec for resistive load.(a) .62 sec(b) .42 sec(c) .58 sec(d) .64 secI got this question in an international level competition.This question is from Solid-State Switching Circuits topic in division Introduction to Solid-State Switching Circuits of Electric Drives

Answer»

Correct answer is (a) .62 sec

Explanation: The VALUE of the CIRCUIT turn-off for Half-wave controlled RECTIFIER for a ω=5 rad/sec for the RESISTIVE load is a π÷Ω=π÷5=.62 sec.



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