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)
$[\mathrm{Take}\ln 5=1.6,\ln 3=1.1]$

Answer 1

$Q=Q_0(1-e^{-t/RC})$
$16=40(1-e^{-t/4})$
$e^{-\frac{t}{4}}=1-\frac{16}{40}=\frac{3}{5}$
$\ln \frac{3}{5}=-\frac{t}{4}$
$t=2s$