A) 2
B) 1
C) \[\frac{1}{2}\]
D) \[\frac{1}{4}\]
Correct Answer: D
Solution :
[d]: As\[U={{U}_{0}}{{e}^{-\frac{2t}{RC}}}\] \[\therefore \]\[\frac{{{U}_{0}}}{2}={{U}_{0}}{{e}^{-\frac{-2{{t}_{1}}}{RC}}}\]\[\Rightarrow \]\[\frac{2{{t}_{1}}}{RC}={{\log }_{e}}2\] ?(i) Also,\[q={{q}_{0}}{{e}^{-\frac{t}{RC}}}\] \[\therefore \]\[\frac{{{q}_{0}}}{4}={{q}_{0}}{{e}^{-\frac{{{t}_{2}}}{RC}}}\] \[\Rightarrow \]\[\frac{{{t}_{2}}}{RC}={{\log }_{e}}4=2{{\log }_{e}}2\] ?(ii) From equation (i) and (ii) \[\frac{2{{t}_{1}}}{{{t}_{2}}}=\frac{{{\log }_{e}}2}{2{{\log }_{e}}2}\]or\[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{1}{4}\]You need to login to perform this action.
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