Current Electricity Charging Discharging Of Capacitors Question 663
Question: For the arrangement shown in figure, the switch is closed at t = 0. The time after which the current becomes $ 2.5\mu A $ is given by (take ln2 = 0.69)
Options:
A) 10 s
B) 5 s
C) 7 s
D) 0.693 s
Show Answer
Answer:
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 $
BETA
