A) 2
B) 1
C) \[\frac{1}{2}\]
D) \[\frac{1}{4}\]
Correct Answer: D
Solution :
[d] \[U=\frac{{{Q}^{2}}}{2C}=\frac{Q_{0}^{2}{{e}^{-\frac{2t}{RC}}}}{2C}\] \[Q={{Q}_{0}}{{e}^{-t/RC}}\] |
\[U=\frac{{{U}_{0}}}{2}\] |
\[\frac{Q_{0}^{2}}{2\times 2C}=\frac{Q_{0}^{2}{{e}^{-\frac{2{{t}_{1}}}{RC}}}}{RC}\] \[\frac{{{Q}_{0}}}{4}={{Q}_{0}}{{e}^{-\frac{{{t}_{2}}}{RC}}}\] |
\[\frac{1}{2}={{e}^{-\frac{2{{t}_{1}}}{RC}}}\] \[{{\log }_{e}}4=\frac{{{t}_{2}}}{RC}\] |
\[{{t}_{1}}=\frac{RC{{\log }_{e}}2}{2}\] \[{{t}_{2}}=RC\,{{\log }_{e}}4\] |
\[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{1}{4}\] |
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