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The ratio of time constant in build-up and decay in the circuit shown in figure is(A) 1:1 (B) 3:2 (C) 2:3 (D) 1:3 |
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Answer» Correct option is (B) 3:2 T1 (time constant) during build up = \(\left(\frac L{2R}\right)\) T2 during decay = \(\left(\frac L{3R}\right)\) \(\therefore \frac{T_1}{T_2} = \frac32\) |
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