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].You need to login to perform this action.
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