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Question

How many time constants will elapse before the charge on a capacitor falls to $$0.1\%$$ of its maximum value in a discharging RC circuit?


Solution

$$q = Qe^{-t/RC}$$
$$q = 0. 1 \% Q$$      $$RC \Rightarrow $$ Time constant
$$= 1 \times 10^{-3}Q$$

So, $$1 \times 10^{-3} Q = Q \times e^{-t/RC}$$

$$\Rightarrow e^{-t/RC} = in\ 10^{-3}$$

$$\Rightarrow t/RC = - (-6.9) = 6.9$$

Physics
NCERT
Standard XII

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