question_answer

The ratio of time constants during current growth and current decay of the circuit shown in the figure is

A)  1 : 1                           

B)  3 : 2               

C)    2 : 3               

D)    1 : 3

Correct Answer: B

Solution :

During charging \[{{\tau }_{{{L}_{1}}}}=\frac{L}{2R}\] and during discharging \[{{\tau }_{{{L}_{2}}}}=\frac{L}{3R}\]             \[\therefore \]    \[\frac{{{\tau }_{{{L}_{1}}}}}{{{\tau }_{{{L}_{2}}}}}=\frac{L/2R}{L/3R}=\frac{3}{2}\] Hence, the correction option is [b].