A) 10 s
B) 5 s
C) 7 s
D) 0.693 s
Correct Answer: C
Solution :
[c] In the case of discharging, \[I={{I}_{0}}{{e}^{-t/RC}}\] or \[2.5\times {{10}^{-6}}=\frac{{{q}_{0}}}{RC}{{e}^{-1/RC}}=5\times {{10}^{-6}}{{e}^{-t/10}}\] or \[{{e}^{t/10}}=2\] Taking log on both sides, we get \[\frac{t}{10}=ln\,2\]or \[t=10\,ln\,2=6.9\cong 7s\]
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