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The current in a `LR` circuit builds up to `(3)/(40th` of its steady state value ion `4s`. The time constant. Of this circuit isA. `(1)/("in"2)s`B. `(2)/("in"2)s`C. `(3)/("in"2)s`D. `(4)/(I"n" 2)s` |
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Answer» Correct Answer - B We know that `i=i_(0)[1-e^((-Rt)/(L))]or (3)/(4)i_(0)=i_(0)[1-e^(-t//tau]` (where `tau=(L)/(R )=`time constant) `(3)/(4)=1-e^(-t//tau)ore^(-t//tau)=1-(3)/(4)=(1)/(4)` `e^(t//tau)=4or(t)/(tau)=In4` `impliestau=(t)/(In 4)=(4)/(2In2)impliestau=(2)/(In2)sec`. |
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