A) 1
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) 2
Correct Answer: C
Solution :
\[U=\frac{1}{2}\,\frac{{{q}^{2}}}{C}\,=\frac{1}{2}\,{{({{q}_{0}}{{e}^{-t/\tau }})}^{2}}\] \[=\frac{q_{0}^{2}}{2C}{{e}^{-2t/\tau }}\,(\tau =CR)\] \[U={{U}_{i}}{{e}^{-2t/\tau }}\] \[\Rightarrow \,\frac{1}{2}{{U}_{i}}={{U}_{i}}\,{{e}^{-2{{t}_{1}}/\tau }}\] \[\Rightarrow \,\frac{1}{2}\,={{e}^{-2{{t}_{1}}/\tau }}\] \[\Rightarrow \,\,{{t}_{i}}=\frac{\tau }{2}\,\ln 2\] Now, \[q={{q}_{0}}\,{{e}^{-t/\tau }}\] \[\Rightarrow \,\,\frac{1}{4}\,{{q}_{0}}={{q}_{0}}{{e}^{-{{t}_{2}}/\tau }}\] \[{{t}_{2}}=\tau \,\ln \,4=2\tau \ln \,2\] \[\therefore \,\,\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{1}{4}\]You need to login to perform this action.
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