Question

At time t = 0, a battery of 10 V is connected across points A
and B in the given circuit. If the capacitors have no charge
initially, at what time (in second) does the voltage across
them become 4 V? [Take : In 5 = 1.6, In 3 = 1.1]

Answer 1

Answer: 2

Voltage across the capacitors will increase from 0 to 10 V
exponentially. The voltage at time t will be given by
$V=10(1-e^{-t/\tau }C)$
Here ${\tau }_c=C_{net}{R}_{net}$
$=(1\times 10^6)(4\times 10^{-6})=4s$
$\therefore V=10(1-e^{-t/4})$
Substituting V = 4 volt, we have
$4=10(1-e^{-t/4})or{e}^{-t/4}=0.6=\frac{3}{5}$
Taking log both sides we have,
$-\frac{t}{4}=In3-In5$
or t = 4 (In 5 - In 3)= 2 s
Hence, the answer is 2 .