Time for Voltage Across Parallel Capacitors in RC Circuit

To find the time at which the voltage across the capacitors becomes $4\text{ V}$, we first simplify the given circuit by finding the equivalent resistance and equivalent capacitance.

Step 1: Equivalent Resistance ($R_{\text{eq}}$)

The two resistors, each of resistance $R = 2\text{ M}\Omega$, are connected in parallel. Their equivalent resistance is:
$R_{\text{eq}} = \frac{2\text{ M}\Omega}{2} = 1\text{ M}\Omega = 10^6\ \Omega$

Step 2: Equivalent Capacitance ($C_{\text{eq}}$)

The two capacitors, each of capacitance $C = 2\ \mu\text{F}$, are connected in parallel. Their equivalent capacitance is:
$C_{\text{eq}} = 2\ \mu\text{F} + 2\ \mu\text{F} = 4\ \mu\text{F} = 4 \times 10^{-6}\ \text{F}$

Step 3: Time Constant ($\tau$)

The time constant of the equivalent $RC$ circuit is:
$\tau = R_{\text{eq}} C_{\text{eq}} = (10^6\ \Omega) \times (4 \times 10^{-6}\ \text{F}) = 4\text{ seconds}$

Step 4: Voltage across the Capacitors as a Function of Time

The charging equation for the voltage $V(t)$ across the parallel combination of capacitors connected to a battery of $V_0 = 10\text{ V}$ is given by:
$V(t) = V_0 \left(1 - e^{-t/\tau}\right)$

Substitute $V(t) = 4\text{ V}$, $V_0 = 10\text{ V}$, and $\tau = 4\text{ s}$:
$4 = 10 \left(1 - e^{-t/4}\right)$

Divide both sides by $10$:
$0.4 = 1 - e^{-t/4}$
$e^{-t/4} = 0.6 = \frac{3}{5}$

Taking the reciprocal of both sides gives:
$e^{t/4} = \frac{5}{3}$

Step 5: Solving for $t$

Taking the natural logarithm on both sides:
$\frac{t}{4} = \ln\left(\frac{5}{3}\right) = \ln 5 - \ln 3$

Using the given values $\ln 5 = 1.6$ and $\ln 3 = 1.1$:
$\frac{t}{4} = 1.6 - 1.1 = 0.5$

$t = 4 \times 0.5 = 2\text{ seconds}$