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At time t=0, ...

Question

At time $t = 0$ , a battery of 10 V is connected across point A and B in the given circuit. If the capacitors have no charge initially, the time (in seconds) at which the voltage across them becomes 4 V is (answer upto two decimal places)
$[Take ln 5 = 1.6, ln 3 = 1.1]$

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Solution

$Q = Q_{0} (1 - e^{- t / R C})$
$16 = 40 (1 - e^{- t / 4})$
$e^{- \frac{t}{4}} = 1 - \frac{16}{40} = \frac{3}{5}$
$ln \frac{3}{5} = - \frac{t}{4}$
$t = 2 s$

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t = 0

, a battery of 10 V is connected across point A and B in the given circuit. If the capacitors have no charge initially, the time (in seconds) at which the voltage across them becomes 4 V is (answer upto two decimal places)

[Take ln 5 = 1.6, ln 3 = 1.1]

t = 0

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