At time \(t=0\), a battery of \(10 \mathrm{~V}\) 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 \(4 \mathrm{~V}\) ?
[Take \(: \ln 5=1.6, \ln 3=1.1]\)

Voltage across the capacitors will increase from 0 to \(10 \mathrm{~V}\) exponentially. The voltage at time \(t\) will be given by
\(
\begin{aligned}
& V=10\left(1-e^{-t / \tau} \mathrm{C}\right) \\
& \text { Here } \tau_c=C_{\text {net }} R_{\text {net }} \\
& =\left(1 \times 10^6\right)\left(4 \times 10^{-6}\right)=4 \mathrm{~s} \\
& \therefore V=10\left(1-e^{-t / 4}\right)\end{aligned}
\)
Substituting \(V=4\) volt we have,
\(
4=10\left(1-e^{-t / 4}\right)
\)
Substituting \(V=4\) volt we have,
\(
4=10\left(1-e^{-t / 4}\right)
\)
or \(e^{-t / 4}=0.6=\frac{3}{5}\)
Taking \(\log\) both sides we have,
\(
-\frac{t}{4}=\ln 3-\ln 5
\)
or \(\quad t=4(\ln 5-\ln 3)=2 \mathrm{~s}\).
Hence, the answer is 2 .